import argparse

def get_parser():
    parser = argparse.ArgumentParser(description="Process command line arguments.", exit_on_error=False)

    # 添加必选参数
    parser.add_argument("--num", type=int, required=True, help="Number")

    return parser

Define a get_parser() function, which returns an argparser, which supports the following command line format processing:

--num NUM

Here, num is a required argument of integer type.

The program will not exit automatically when the type of num cannot be converted to an integer.

#<INSERT>

from argparse import ArgumentError
parser = get_parser()

command_line = "--num 12".split(" ")
assert parser.parse_args(command_line).num == 12

exception_got = False
command_line = "--num string".split(" ")
try:
    parser.parse_args(command_line)
except ArgumentError:
    exception_got = True
assert exception_got


def replace(x, n):
    zero_positions = [i for i in range(x.bit_length()) if (x & (1 << i)) == 0]
    ret = x
    k = 0
    for pos in zero_positions:
        if n & (1 << k):
            ret += (1 << pos)
        k += 1
    return ret

Given an integer x, get its binary representation, and then fill in the positions of 0s in the binary representation with a number n. For example, if the binary representation of x is 110001 and you fill in the 0 positions with 2 (binary 010), the result will be 110101.
Implement a python function `replace(x, n)` to solve this.

Question: Write a piece of python code to achieve the following function.
Find the maximum perfect square number that is less than the nth Fibonacci number.
The function's name is max_square_number, and the input is an integer n.

Answer:
```python
def max_square_number(n):
    if n <= 2:
        return 0  # For the first two numbers in the Fibonacci sequence, no complete square numbers can be used.

    fibs = [0, 1]  # Store the first two numbers of the Fibonacci sequence
    squares = []  # Store the list of perfect square numbers

    max_square = 0
    i = 2

    while i <= n:
        fibs.append(fibs[i-1] + fibs[i-2])  # Calculate the next Fibonacci number
        i += 1

    fib = fibs[-1]
    perfect_squares = [i**2 for i in range(1, int(math.sqrt(fib)) + 1)]
    max_square = max(filter(lambda x: x < fib, perfect_squares))
    return max_square
```

----------------


import pandas as pd

def read_excel(file_name, sheet_name):
    # 读取Excel文件
    df = pd.read_excel(file_name, sheet_name=sheet_name)

    # 按行读取数据并打印
    output = []
    for index, row in df.iterrows():
        output.append(list(row))
    return output

Implement reading Excel files using Python, with the file name input as a parameter, and designate a sheet to read data line by line and return a two-dimensional array. Implement the function `read_excel(file_name, sheet_name)`.

#<INSERT>

def check():
    path = "file_test_2/data.xlsx"
    sheet = "用户消费数据"
    ret = read_excel(path, sheet)
    assert ret[0][0] == "用户1"
    assert ret[2][1] == 15

check()


import requests

def http_client(url, method='GET', filename='response.txt'):
    response = requests.request(method, url)
    with open(filename, 'w') as file:
        file.write(response.text)
    return response.text

A simple HTTP client function, http_client, written in Python. The parameters of this function are url, method, and filename to save the response, and it returns the response from the server.
1. It is required to return text information from the server.
2. The text information is required to be saved as a .txt file.

#pyqt_test_30

#<INSERT>

def check():
    import os
    url = 'http://example.com'
    filename = 'test_response.txt'
    response = http_client(url, filename=filename)
    assert 'Example Domain' in response
    with open(filename, 'r') as file:
        file_content = file.read()
    assert 'Example Domain' in file_content
    if os.path.exists(filename):  # 清理测试文件
        os.remove(filename)

check()


function maxInArray(numbers: number[]): number {
    if (!Array.isArray(numbers) || numbers.length === 0) {
        throw new Error("Input must be a non-empty array of numbers.");
    }
    return Math.max(...numbers);
}

The method for obtaining the maximum value in an array in TypeScript is named maxInArray. Wrap the answer with markdown.
Please wrap the answering code with markdown code block format, e.g.
```typescript
[code]
```

import { describe, expect, test } from '@jest/globals';

#<INSERT>

describe('maxInArray', () => {
    test('should return the maximum value in an array of numbers', () => {
        const numbers = [1, 3, 2, 5, 4];
        expect(maxInArray(numbers)).toBe(5);
    });

    test('should return the maximum value when array contains negative values', () => {
        const numbers = [-10, -3, -1, -7];
        expect(maxInArray(numbers)).toBe(-1);
    });

    test('should throw an error for an empty array', () => {
        const emptyArray: number[] = [];
        expect(() => maxInArray(emptyArray)).toThrow("Input must be a non-empty array of numbers.");
    });

    test('should return the value itself if the array has only one number', () => {
        const singleValueArray = [7];
        expect(maxInArray(singleValueArray)).toBe(7);
    });

    // Add more tests if necessary
});


function toUpperCase(input: string): string {
    return input.toUpperCase();
}

function toLowerCase(input: string): string {
    return input.toLowerCase();
}

The method for converting lower case to upper case in ts is called "toUpperCase", and the method for converting upper case to lower case is called "toLowerCase", wrap the answer with markdown.
Please wrap the answering code with markdown code block format, e.g.
```typescript
[code]
```

import { describe, expect, test } from '@jest/globals';

#<INSERT>

describe('String Case Conversion', () => {

    test('should convert string to uppercase', () => {
        const input = 'TypeScript is awesome!';
        const expectedOutput = 'TYPESCRIPT IS AWESOME!';
        expect(toUpperCase(input)).toBe(expectedOutput);
    });

    test('should convert string to lowercase', () => {
        const input = 'TypeScript is AWESOME!';
        const expectedOutput = 'typescript is awesome!';
        expect(toLowerCase(input)).toBe(expectedOutput);
    });

    test('should not change case for uppercase string when converting to uppercase', () => {
        const input = 'ALREADY UPPERCASE';
        expect(toUpperCase(input)).toBe(input);
    });

    test('should not change case for lowercase string when converting to lowercase', () => {
        const input = 'already lowercase';
        expect(toLowerCase(input)).toBe(input);
    });

    test('should correctly handle empty string', () => {
        const input = '';
        expect(toUpperCase(input)).toBe('');
        expect(toLowerCase(input)).toBe('');
    });
});


// 定义字符串字面量类型
type StringLiteral = 'myStringLiteral';

// 创建一个具有该字符串字面量类型的变量
const myLiteral: StringLiteral = 'myStringLiteral';

In TypeScript, how to define a variable of type string literal? The string literal is myStringLiteral, and the answer should be wrapped in markdown.
Please wrap the answering code with markdown code block format, e.g.
```typescript
[code]
```

import { describe, expect, test } from '@jest/globals';

#<INSERT>

describe('String Literal Type', () => {
  test('should only allow the specified string literal', () => {
    expect(myLiteral).toBe('myStringLiteral');
  });

  // 如果需要检验类型，可以使用TS的类型检查，但在运行时无法这样测试。
});


#L1_C_57
from rdkit import Chem
import itertools
from rdkit.Chem import rdRGroupDecomposition
from rdkit.Chem import rdFMCS

def enumerate_all_products(core, rgroups):
    # preserve the positions of the non-dummy core atoms,
    # Python's itertools handles doing the combinatorics of generating
    # every possible combination of R groups:
    order = itertools.product(*rgroups)
    # now we just loop over each combination, copy all the pieces into
    # one molecule, and zip it. That's our product
    for tpl in order:
        tm = Chem.RWMol(core)
        for r in tpl:
            tm.InsertMol(r)
        prod = Chem.molzip(tm)
        if prod is not None:
            # and finally yield the product molecule
            yield prod

def solve(smiles_lst):
    mols = [Chem.MolFromSmiles(smi) for smi in smiles_lst]
    res = Chem.MolFromSmarts(rdFMCS.FindMCS(mols).smartsString)

    rgd, failed = rdRGroupDecomposition.RGroupDecompose([res], mols, asRows=False)
    all_r_groups = []
    core = rgd['Core'][0]
    for k, v in rgd.items():
        if k == 'Core':
            continue
        r_groups = set()
        for r_group in v:
            r_groups.add(Chem.MolToSmiles(r_group))
        all_r_groups.append([Chem.MolFromSmiles(i) for i in r_groups])
    prod_gen = enumerate_all_products(core, all_r_groups)
    prod_smi_set = set([Chem.CanonSmiles(Chem.MolToSmiles(i)) for i in prod_gen])
    return prod_smi_set

Please write a complete Python function `solve(smiles_lst)`.
This function takes a list of the SMILES of molecules,
which contain a set of molecular SMILES with a common parent nucleus structure.
It cuts off the substituents connected to the parent nucleus,
then re-attaches them back to the original substituent positions on
the parent nucleus structure in the form of enumerating the combinations of substituents.
This process produces a new set of molecules.
The function then returns a set of canonical SMILES of
this new set of molecules (the structure of the input molecules will also be enumerated,
thus it is included). For example, if there are two substituent positions R1 and R2 on two molecules,
and there are {A, B} two types of substituents at the R1 position,
and {B, C, D} three types of substituents at the R2 position,
then 6 combinations of R1 and R2 substituents will be enumerated: [[A, B], [A, C], [A, D], [B, B], [B, C], [B, D]].
These substituents are then attached to the common parent nucleus of the two molecules to produce 6 molecular structures.
It is known that in RDKit, rdkit.Chem.rdFMCS.FindMCS() can find the maximum common substructure of a group of molecules;
rdkit.Chem.rdRGroupDecomposition.RGroupDecompose() function can decompose a molecule into core part (Core) and substituent part (R groups);
rdkit.Chem.molzip() function can concatenate fragments marked with connection points and the parent nucleus;
rdkit.Chem.CanonSmiles() can canonicalize SMILES.

#<INSERT>

def check(solve):
    result1 = solve(['CCc1cccc(Cl)c1', 'CCc1cccc(Br)c1', 'COc1cc(N)cc(F)c1'])
    assert result1 == {'COc1cccc(Cl)c1', 'CCc1cccc(Br)c1', 'COc1cc(N)cc(F)c1',
                       'CCc1cc(N)cc(Cl)c1', 'COc1cc(N)cc(Cl)c1', 'CCc1cc(N)cc(Br)c1',
                       'CCc1cccc(Cl)c1', 'CCc1cccc(F)c1', 'COc1cccc(F)c1',
                       'COc1cc(N)cc(Br)c1', 'COc1cccc(Br)c1', 'CCc1cc(N)cc(F)c1'} or \
           result1 == {'CCc1cc(N)cc(Br)c1', 'CCc1cc(N)cc(Cl)c1', 'CCc1cc(N)cc(F)c1',
                       'COc1cc(N)cc(Br)c1', 'COc1cc(N)cc(Cl)c1', 'COc1cc(N)cc(F)c1',
                       '[H]c1cc(Br)cc(CC)c1', '[H]c1cc(Br)cc(OC)c1', '[H]c1cc(Cl)cc(CC)c1',
                       '[H]c1cc(Cl)cc(OC)c1', '[H]c1cc(F)cc(CC)c1', '[H]c1cc(F)cc(OC)c1'}

    result2 = solve(['CC(C)C(C)c1ccc(O)cc1', 'CC(C)C(C)c1cccc(O)c1', 'CC(C)C(C)c1ccc(Cl)c(O)c1', 'CC(C)C(C)c1ccc(Cl)c(N)c1'])
    assert result2 == {'CC(C)C(C)c1ccc(O)c(O)c1', 'CC(C)C(C)c1ccc(Cl)cc1', 'CC(C)C(C)c1ccc(O)cc1',
                       'CC(C)C(C)c1ccc(O)c(N)c1', 'CC(C)C(C)c1cccc(O)c1', 'CC(C)C(C)c1ccc(Cl)c(O)c1',
                       'CC(C)C(C)c1ccccc1', 'CC(C)C(C)c1cccc(N)c1', 'CC(C)C(C)c1ccc(Cl)c(N)c1'} or \
           result2 == {'CC(C)C(C)c1ccc(O)c(O)c1', 'CC(C)C(C)c1ccc(Cl)c(O)c1', 'CC(C)C(C)c1ccc(Cl)c(N)c1',
                       '[H]c1cc(C(C)C(C)C)ccc1Cl', '[H]c1cc(C(C)C(C)C)ccc1O', 'CC(C)C(C)c1ccc(O)c(N)c1',
                       '[H]c1ccc(C(C)C(C)C)cc1O', '[H]c1ccc(C(C)C(C)C)cc1N', '[H]c1ccc(C(C)C(C)C)cc1[H]'}

check(solve)


#L1_B_61
from rdkit import Chem
from rdkit.Chem import Descriptors, Lipinski
def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    if mol is None:
        return "Invalid SMILES string"

    # 计算分子属性
    mol_weight = Descriptors.MolWt(mol)
    logP = Descriptors.MolLogP(mol)
    h_donors = Lipinski.NumHDonors(mol)
    h_acceptors = Lipinski.NumHAcceptors(mol)
    tpsa = Descriptors.TPSA(mol)
    rotatable_bonds = Lipinski.NumRotatableBonds(mol)
    # 检查所有环的大小是否不超过6
    ring_sizes = [len(ring) for ring in mol.GetRingInfo().AtomRings()]
    max_ring_size = max(ring_sizes) if ring_sizes else 0
    # 检查条件是否满足
    conditions_met = (
        250 <= mol_weight <= 700,
        0 <= logP <= 5,
        h_donors <= 5,
        h_acceptors <= 10,
        tpsa <= 150,
        rotatable_bonds <= 9,
        max_ring_size <= 6
    )
    # 返回结果
    if all(conditions_met):
        return True
    else:
        return False

In drug screening, we usually need the molecules to meet the following conditions for higher drug-like properties: the molecular weight is between 250 and 700, logP is between 0 and 5, the number of hydrogen bond donors is no more than 5, the number of hydrogen bond acceptors is no more than 10, TPSA of the molecule is not larger than 150, the number of rotatable bonds in the molecule is not more than 9, and the size of all rings does not exceed 6.
Please implement a complete Python function `solve(smiles)` that takes a molecule's SMILES as input and assesses whether the molecule satisfies all the above-mentioned rules, and returns True/False.
We can use the `MolWt`, `MolLogP`, and `TPSA` functions from the `rdkit.Chem.Descriptors` module to calculate the molecular weight, logP, and TPSA values of the molecule, and use the `NumHDonors`, `NumHAcceptors`, and `NumRotatableBonds` functions from the rdkit.Chem.Lipinski module to calculate the number of hydrogen bond donors, hydrogen bond acceptors, and rotatable bonds in the molecule. Use the `GetRingInfo` function to get information about the rings in the molecule.

#<INSERT>

def check(solve):
    assert solve('COc1ccccc1CNC(=O)CS(=O)(=O)CCCc1ccccc1') == True
    assert solve('CCOC(=O)C1CCN(CC1)C(=O)c1nnn(c1)c1ccc(cc1)C') == True
    assert solve('C[C@H]1C[C@@H]1CNS(=O)(=O)c1cccs1') == False
    assert solve('COc1ccc(cc1Cl)NC(=O)c1cc2c(o1)c(Br)ccc2') == False

check(solve)


#L1_C_19
from rdkit import Chem
from rdkit.Chem import ChemicalFeatures
from rdkit import RDConfig
import os

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    mol = Chem.AddHs(mol)
    fdefName = os.path.join(RDConfig.RDDataDir,'BaseFeatures.fdef')
    factory = ChemicalFeatures.BuildFeatureFactory(fdefName)
    feats = factory.GetFeaturesForMol(mol)

    don_set = set()
    acc_set = set()
    for feat in feats:
        if feat.GetFamily() == 'Donor':
            don_set.update(feat.GetAtomIds())
        elif feat.GetFamily() == 'Acceptor':
            acc_set.update(feat.GetAtomIds())

    inner_hb = 0
    used_acc_set = set()
    for i in don_set:
        for j in acc_set:
            if i == j:
                continue
            if j in used_acc_set:
                continue
            path = Chem.GetShortestPath(mol, i, j)
            if len(path) - 1 == 4 or len(path) - 1 == 5:
                inner_hb += 1
                used_acc_set.add(j)
                break
    return inner_hb

Define that if there are 4 or 5 chemical bonds between a hydrogen bond donor and acceptor in a molecule, an intramolecular hydrogen bond may be formed. Please write a complete Python function `solve(smiles)` to determine how many possible intramolecular hydrogen bonds can be formed in a molecule based on the 2D topological structure. The donor and acceptor of the hydrogen bond can be obtained through the `MolChemicalFeature` in the `ChemicalFeatures` module, and determined as Donar or Acceptor using the `GetFamily` method. The number of chemical bond intervals between two atoms can be calculated using the `Chem.GetShortestPath` function.

#<INSERT>

def check(solve):
    assert solve('CC(O)C(\C)=C/c1cc(ccn1)-c1cc(C)ccc1N') == 1
    assert solve('CC(=O)OC(C(C)N)C(C)CO') == 2

check(solve)


#L2_C_24
from rdkit import Chem

def neutralize_atoms(mol):
    pattern = Chem.MolFromSmarts("[+1!h0!$([*]~[-1,-2,-3,-4]),-1!$([*]~[+1,+2,+3,+4])]")
    at_matches = mol.GetSubstructMatches(pattern)
    at_matches_list = [y[0] for y in at_matches]
    if len(at_matches_list) > 0:
        for at_idx in at_matches_list:
            atom = mol.GetAtomWithIdx(at_idx)
            chg = atom.GetFormalCharge()
            hcount = atom.GetTotalNumHs()
            atom.SetFormalCharge(0)
            atom.SetNumExplicitHs(hcount - chg)
            atom.UpdatePropertyCache()
    return mol

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    mol = neutralize_atoms(mol)
    new_smiles = Chem.MolToSmiles(mol)
    return new_smiles

The charged atoms SMARTS are known as "[+1!h0!$([*]~[-1,-2,-3,-4]),-1!$([*]~[+1,+2,+3,+4])]", and in RDKit, the number of explicit hydrogen atoms can be modified using `atom.SetNumExplicitHs` and the formal charge of atoms can be modified using `atom.SetFormalCharge`.
Please write a complete Python function, `solve(smiles)`, which takes a SMILES string as input. First, use `mol.GetSubstructMatches` to find all charged atoms according to the charged atoms SMARTS. Then, iterate over all charged atoms. Use `atom.GetFormalCharge` and `atom.GetTotalNumHs` to obtain the formal charge (chg) and hydrogen atom count (hcount) of the atom, respectively.

#<INSERT>

def check(solve):
    assert solve('C[NH+](C)CC[S-]') == 'CN(C)CCS'
    assert solve('CP([O-])(=O)OC[NH3+]') == 'CP(=O)(O)OCN'

check(solve)


#L1_B_58
from rdkit import Chem
from rdkit.Chem.Scaffolds import MurckoScaffold
from rdkit.Chem import AllChem
from rdkit import DataStructs

def solve(smiles_lst):
    new_mols = []
    for smiles in smiles_lst:
        mol = Chem.MolFromSmiles(smiles)
        scaff = MurckoScaffold.GetScaffoldForMol(mol)
        new_mols.append(scaff)

    simi_lst = []
    for idx, mol1 in enumerate(new_mols):
        fp1 = AllChem.GetMorganFingerprintAsBitVect(mol1, 2, nBits=1024)
        for mol2 in new_mols[idx+1:]:
            fp2 = AllChem.GetMorganFingerprintAsBitVect(mol2, 2, nBits=1024)
            simi = DataStructs.TanimotoSimilarity(fp1,fp2)
            simi_lst.append(simi)
    mean_simi = round(sum(simi_lst)/len(simi_lst), 4)
    return mean_simi

Please write a complete Python function `solve(smiles_lst)`.
The input is a list of SMILES of molecules.
Remove the side chain atoms from these molecules
(atoms not in the ring or between the rings on the shortest path are called side chain atoms),
and then calculate the 1024-bit ECFP4 fingerprint Tanimoto similarity between these molecules with side chain atoms removed.
The function should return the average similarity between them, with the result rounded to 4 decimal places.
It is known that the Bemis-Murcko skeleton is the result of removing the side chain atoms defined here.
In RDKit, the function `rdkit.Chem.Scaffolds.MurckoScaffold.GetScaffoldForMol()` can be used to obtain the Bemis-Murcko skeleton of a molecule;
`rdkit.Chem.AllChem.GetMorganFingerprintAsBitVect()` can be used to calculate the ECFP fingerprint of a molecule;
`rdkit.DataStructs.TanimotoSimilarity()` can be used to compute the Tanimoto similarity of two fingerprints.

#<INSERT>

def check(solve):
    assert solve(['CC(c1ccccc1)C(CCl)C(C)c1ccccc1', 'CC(c1ccccc1)C(CNC1CC1)C(c1ccccc1)C1CCCC1', 'COC(C)C(N)C(C(CO)c1ccccc1)C1CCCC1C']) == 0.3302
    assert solve(['CCC(CC)c1ccccc1', 'Cc1cccc(-c2ccc(F)cc2)c1', 'CC(C)c1ccccc1-c1ccccc1C(C)Cc1cccc2c1CCC2']) == 0.2267

check(solve)


#L0_B_61
from rdkit import Chem
from rdkit.Chem import Descriptors, Lipinski
def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    if mol is None:
        return "Invalid SMILES string"

    # 计算分子属性
    mol_weight = Descriptors.MolWt(mol)
    logP = Descriptors.MolLogP(mol)
    h_donors = Lipinski.NumHDonors(mol)
    h_acceptors = Lipinski.NumHAcceptors(mol)
    tpsa = Descriptors.TPSA(mol)
    rotatable_bonds = Lipinski.NumRotatableBonds(mol)
    # 检查所有环的大小是否不超过6
    ring_sizes = [len(ring) for ring in mol.GetRingInfo().AtomRings()]
    max_ring_size = max(ring_sizes) if ring_sizes else 0
    # 检查条件是否满足
    conditions_met = (
        250 <= mol_weight <= 700,
        0 <= logP <= 5,
        h_donors <= 5,
        h_acceptors <= 10,
        tpsa <= 150,
        rotatable_bonds <= 9,
        max_ring_size <= 6
    )
    # 返回结果
    if all(conditions_met):
        return True
    else:
        return False

In drug screening, we usually require molecules to meet the following conditions to have higher drug-likeness: molecular weight between 250 and 700, logP between 0 and 5, the number of hydrogen bond donors does not exceed 5, the number of hydrogen bond acceptors does not exceed 10, the molecular TPSA does not exceed 150, the number of rotatable bonds does not exceed 9, and the size of all rings does not exceed 6. Please implement a complete Python function `solve(smiles)` that takes a molecule's SMILES as input and assesses whether the molecule satisfies all the above-mentioned rules, and returns True/False.


#L0_B_62
from rdkit import Chem
from rdkit import DataStructs
from rdkit.Chem import AllChem

def is_similar(smiles):
    # 将一个分子的smiles转成rdkit Mol对象
    mol = Chem.MolFromSmiles(smiles)

    # 计算一个Mol对象2048 Bits的ECFP4分子指纹，并且比较与"NCc1ccc2c(N(C/C(=C/c3ccccc3)c3ccc(Cl)cc3)c3cc(Br)c4oc(C(=O)O)cc4c3)[nH]nc2c1"分子之间的Tonimoto相似度
    ref_mol = Chem.MolFromSmiles("NCc1ccc2c(N(C/C(=C/c3ccccc3)c3ccc(Cl)cc3)c3cc(Br)c4oc(C(=O)O)cc4c3)[nH]nc2c1")
    fp_ref = AllChem.GetMorganFingerprintAsBitVect(ref_mol, 2, nBits=2048)
    fp_mol = AllChem.GetMorganFingerprintAsBitVect(mol, 2, nBits=2048)
    similarity = DataStructs.TanimotoSimilarity(fp_ref, fp_mol)

    # 判断一个值是否大于0.5，返回True或False
    flag = similarity > 0.5

    return flag

Write a python function `is_similar(smiles)`. For the input smiles,
compute its similarity with molecule "NCc1ccc2c(N(C/C(=C/c3ccccc3)c3ccc(Cl)cc3)c3cc(Br)c4oc(C(=O)O)cc4c3)[nH]nc2c1",
and if the similarity > 0.5 return True,
otherwise return False.
The similarity between two molecules here is defined as
Tanimoto similarity of their 2048-bits ECFP4 fingerprints.

#<INSERT>

def check(is_similar):
    assert is_similar("NCc1ccc2c(N(C/C(=C/c3ccccc3)c3ccc(Cl)cc3)c3cc(Br)c4oc(C(=O)O)cc4c3)[nH]nc2c1")
    assert is_similar("NCc1ccc2c(N(C/C(=C/c3ccccc3)c3ccc(Cl)cc3)c3cc(Br)c4occc4c3)[nH]nc2c1")
    assert is_similar("NCc1ccc2c(N(C/Cc3ccc(Cl)cc3)c3cc(Br)c4occc4c3)[nH]nc2c1")
    assert is_similar("NCc1ccc2c(N(C/C=C/c3ccccc3)c3cc(Br)c4occc4c3)[nH]nc2c1")
    assert not is_similar("NCc1ccc2c(Nc3cc(Br)c4occc4c3)[nH]nc2c1")
    assert is_similar("NCc1ccc2c(N(C/Cc3ccc(Cl)cc3)c3cc(Br)c4oc(C(=O)O)cc4c3)[nH]nc2c1")
    assert not is_similar('c1ccccc1')

check(is_similar)


#L2_B_28
from rdkit import Chem
from rdkit.Chem.Lipinski import RotatableBondSmarts

def decompose_mol(mol):
    rot_atom_pairs = mol.GetSubstructMatches(RotatableBondSmarts)
    bond_indices = list(set([mol.GetBondBetweenAtoms(*ap).GetIdx() for ap in rot_atom_pairs]))

    all_frags_set = set()
    i = Chem.FragmentOnBonds(mol, bond_indices)
    frags = Chem.GetMolFrags(i, asMols=True)
    for frag in frags:
        frag_smi = Chem.MolToSmiles(frag)
        all_frags_set.add(frag_smi)
    return all_frags_set

def solve(smiles):
    m = Chem.MolFromSmiles(smiles)
    return decompose_mol(m)

Please write a complete Python function `solve(smiles)` that takes a molecule's SMILES as input
and cuts all the rotatable bonds (as defined in RDKit's RotatableBondSmarts),
returning a set of fragment SMILES.
You can use `mol.GetSubstructMatches` to match `rdkit.Chem.Lipinski.RotatableBondSmarts` to get both ends of all rotatable bonds
and use `mol.GetBondBetweenAtoms().GetIdx()` to find the index of the rotatable bonds.
Then use `Chem.FragmentOnBonds()` to cut the rotatable bonds and `Chem.GetMolFrags()` to obtain the cut fragments.


#L1_C_59
from rdkit import Chem
from itertools import combinations
from rdkit import DataStructs

def solve(smiles, N=1):
    mol = Chem.MolFromSmiles(smiles)
    out_mol_list = []
    used = set()
    aromatic_cH = Chem.MolFromSmarts("[cH]")
    match_atms = [x[0] for x in mol.GetSubstructMatches(aromatic_cH)]
    n_combos = combinations(match_atms, N)
    for combo in n_combos:
        new_mol = Chem.RWMol(mol)
        for idx in combo:
            atm = new_mol.GetAtomWithIdx(idx)
            atm.SetAtomicNum(7)
        smi = Chem.MolToSmiles(new_mol)
        if smi not in used:
            used.add(smi)
            out_mol_list.append(new_mol)
    return used

Please write a complete Python function `solve(smiles, N)`,
which takes a molecule's SMILES and the quantity of nitrogen (N) as inputs.
This function should replace the "N" number of unsubstituted
aromatic carbon atoms (these carbon atoms are connected to one hydrogen atom) in the molecule with nitrogen atoms,
and return the set of SMILES of all possible molecular structures that may be produced by this substitution.
It is known that the aromatic carbon atoms to be replaced can be matched using SMARTS "[cH]".

#<INSERT>

def check(solve):
    assert solve('CCc1cc(C)ccn1', 1) == {'CCc1cc(C)ncn1', 'CCc1nccc(C)n1', 'CCc1cc(C)cnn1'}
    assert solve('CC(C)CC1C=Cc2c(F)cccc21', 2) == {'CC(C)CC1C=Cc2c(F)ncnc21', 'CC(C)CC1C=Cc2c1cnnc2F', 'CC(C)CC1C=Cc2c(F)cnnc21'}
    assert solve('CC(CC1N=Cc2c(F)cccc21)n1cccc1', 2) == {'CC(CC1N=Cc2c1ccnc2F)n1ccnc1', 'CC(CC1N=Cc2c1cnnc2F)n1cccc1', 'CC(CC1N=Cc2c(F)cnnc21)n1cccc1',
                                                  'CC(CC1N=Cc2c1ccnc2F)n1cccn1', 'CC(CC1N=Cc2c(F)ccnc21)n1cccn1', 'CC(CC1N=Cc2c(F)cccc21)n1ccnn1',
                                                  'CC(CC1N=Cc2c(F)cncc21)n1cccn1', 'CC(CC1N=Cc2c(F)ccnc21)n1ccnc1', 'CC(CC1N=Cc2c(F)ncnc21)n1cccc1',
                                                  'CC(CC1N=Cc2c(F)cncc21)n1ccnc1', 'CC(CC1N=Cc2c(F)cccc21)n1nccn1', 'CC(CC1N=Cc2c(F)cccc21)n1cnnc1',
                                                  'CC(CC1N=Cc2c(F)cccc21)n1cncn1'}

check(solve)


#L1_B_28
from rdkit import Chem
from rdkit.Chem.Lipinski import RotatableBondSmarts

def decompose_mol(mol):
    rot_atom_pairs = mol.GetSubstructMatches(RotatableBondSmarts)
    bond_indices = list(set([mol.GetBondBetweenAtoms(*ap).GetIdx() for ap in rot_atom_pairs]))

    all_frags_set = set()
    i = Chem.FragmentOnBonds(mol, bond_indices)
    frags = Chem.GetMolFrags(i, asMols=True)
    for frag in frags:
        frag_smi = Chem.MolToSmiles(frag)
        all_frags_set.add(frag_smi)
    return all_frags_set

def solve(smiles):
    m = Chem.MolFromSmiles(smiles)
    return decompose_mol(m)

Please write a complete Python function `solve(smiles)` that takes a molecule's SMILES
as input and cuts all rotatable bonds (as defined by RotatableBondSmarts in RDKit),
returning a set of fragment SMILES.
You can obtain the indices of the atoms on both ends of all rotatable bonds
by matching mol.GetSubstructMatches() with rdkit.Chem.Lipinski.RotatableBondSmarts,
and use Chem.FragmentOnBonds() to cut the bonds.


#L2_B_9
from rdkit.Chem import AllChem
from rdkit import Chem
from rdkit.Chem.Lipinski import RotatableBondSmarts

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    patt = Chem.MolFromSmarts('c1ccccc1')
    repl = Chem.MolFromSmiles('[H]')
    rms = AllChem.ReplaceSubstructs(mol, patt, repl, replaceAll=True)[0]
    rms = Chem.MolFromSmiles(Chem.MolToSmiles(rms))
    rot_atom_pairs = rms.GetSubstructMatches(RotatableBondSmarts)
    bond_indices = set([rms.GetBondBetweenAtoms(*ap).GetIdx() for ap in rot_atom_pairs])
    return len(bond_indices)

It is known that the smarts of the benzene ring is "c1ccccc1", the smarts of the hydrogen atom is "[H]", and the smarts definition of rotatable bonds can be obtained using `from rdkit.Chem.Lipinski import RotatableBondSmarts`.
Please write a complete Python function `solve(smiles)` that takes a SMILES as input, first use the `AllChem.ReplaceSubstructs` function to replace each terminal benzene structure in the molecule with a hydrogen atom, then get all the rotatable bonds of the molecule according to the `GetSubstructMatches` function, remove duplicates and return the number of rotatable bonds.

#<INSERT>

def check(solve):
    assert solve('CCN(c1ccccc1)C1CC(OC)CN1') == 3
    assert solve('CCC1C(OC)C(c2ccccc2)C(N(C)c2ccccc2)N1CC') == 4
    assert solve('CC1CCCCC1NC') == 1

check(solve)


#L2_C_26
from rdkit import Chem

def GetRingSystems(mol, includeSpiro=True):
    ri = mol.GetRingInfo()
    systems = []
    for ring in ri.AtomRings():
        ringAts = set(ring)
        nSystems = []
        for system in systems:
            nInCommon = len(ringAts.intersection(system))
            if nInCommon and (includeSpiro or nInCommon>1):
                ringAts = ringAts.union(system)
            else:
                nSystems.append(system)
        nSystems.append(ringAts)
        systems = nSystems
    return systems

def solve(smiles):
    m = Chem.MolFromSmiles(smiles)
    systems = GetRingSystems(m)
    return len(systems)

Please write a complete Python function `solve(smiles)`, which takes a molecule's SMILES as input,
and calculates the number of ring systems in it.
Multiple rings that share the same atoms will be considered as one ring system.
You can obtain the indices of all ring atoms by using `mol.GetRingInfo().AtomRings()`.
Then use a nested loop to iterate through these ring atom indices, comparing each pair for intersections.
If there is an intersection, merge them into one ring system.

#<INSERT>

def check(solve):
    assert solve('Nc1cc(-c2cccc3c2CCC3)c(C(=O)O)cc1C1CCCC1') == 3
    assert solve('CN1C(=O)CN=C(c2ccccc2)c2cc(Cl)ccc21') == 2
    assert solve('c1ccccc1') == 1

check(solve)


#L0_C_59
from rdkit import Chem
from itertools import combinations
from rdkit import DataStructs

def solve(smiles, N=1):
    mol = Chem.MolFromSmiles(smiles)
    out_mol_list = []
    used = set()
    aromatic_cH = Chem.MolFromSmarts("[cH]")
    match_atms = [x[0] for x in mol.GetSubstructMatches(aromatic_cH)]
    n_combos = combinations(match_atms, N)
    for combo in n_combos:
        new_mol = Chem.RWMol(mol)
        for idx in combo:
            atm = new_mol.GetAtomWithIdx(idx)
            atm.SetAtomicNum(7)
        smi = Chem.MolToSmiles(new_mol)
        if smi not in used:
            used.add(smi)
            out_mol_list.append(new_mol)
    return used

Please write a complete Python function `solve(smiles, N)`,
which takes a molecule's SMILES and the quantity of nitrogen (N) as inputs.
This function should replace the "N" number of unsubstituted
aromatic carbon atoms (these carbon atoms are connected to one hydrogen atom) in the molecule with nitrogen atoms,
and return the set of SMILES of all possible molecular structures that may be produced by this substitution.


#L1_B_16
from rdkit import Chem
from rdkit.Chem import ChemicalFeatures
from rdkit import RDConfig
import os

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    fdefName = os.path.join(RDConfig.RDDataDir,'BaseFeatures.fdef')
    factory = ChemicalFeatures.BuildFeatureFactory(fdefName)
    feats = factory.GetFeaturesForMol(mol)

    hyd_atoms = set()
    for feat in feats:
        if feat.GetFamily() == 'Hydrophobe' or feat.GetFamily() == 'LumpedHydrophobe':
            hyd_atoms.update(feat.GetAtomIds())
    return len(hyd_atoms)

Please write a complete Python function `solve(smiles)` that takes a SMILES as input and calculates how many atoms in the molecule have hydrophobic features (including Hydrophobe and LumpedHydrophobe) from RDKit ChemicalFeatures. The features of the molecule can be obtained from the ChemicalFeatures module as MolChemicalFeature, and whether it is a `Hydrophobe` or `LumpedHydrophobe` can be determined using the `GetFamily` method.


#L0_B_55
from rdkit import Chem
from rdkit.Chem import AllChem
from rdkit import DataStructs

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    if mol is None:
        return None
    rxn_dBOC = AllChem.ReactionFromSmarts("[#7:1][CX3:2](=[OD1])[OD2]C([CH3])([CH3])[CH3]>>[#7:1][H:2]")

    mol.UpdatePropertyCache()
    prod = rxn_dBOC.RunReactants([mol])
    if prod:
        prod = prod[0][0]
        Chem.SanitizeMol(prod)

    fp1 = AllChem.GetMorganFingerprintAsBitVect(mol, 2, nBits=2048)
    fp2 = AllChem.GetMorganFingerprintAsBitVect(prod, 2, nBits=2048)
    simi = DataStructs.TanimotoSimilarity(fp1,fp2)
    return round(simi, 3)

Implement a complete Python function `solve(smiles)` where the input is a molecule's SMILES. Remove the Boc protecting group inside the molecule, and calculate the Tanimoto similarity of the 2048-bit ECFP4 fingerprints of the molecules before and after the process. The result should be kept to 3 decimal places.

#<INSERT>

def check(solve):
    assert abs(solve('Cc1ccccc1CNC(=O)OC(C)(C)C') - 0.3333) < 1e-3
    assert abs(solve('CCC(C)CCNC(=O)OC(C)(C)C') - 0.2778) < 1e-3
    assert abs(solve('CCN(C)C(=O)OC(C)(C)C') - 0.1071) < 1e-3

check(solve)


#L2_B_4
from rdkit.Chem import AllChem
from rdkit import Chem

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    if mol is None:
        return None
    rxn_dBOC = AllChem.ReactionFromSmarts("[#7:1][CX3:2](=[OD1])[OD2]C([CH3])([CH3])[CH3]>>[#7:1][H:2]")

    mol.UpdatePropertyCache()
    prod = rxn_dBOC.RunReactants([mol])
    if prod:
        mol = prod[0][0]

    new_smi = Chem.MolToSmiles(mol)
    new_smi = Chem.CanonSmiles(new_smi)
    return new_smi

The known Boc deprotection reaction smarts is `[#7:1][CX3:2](=[OD1])[OD2]C([CH3])([CH3])[CH3]>>[#7:1][H:2]`. Please write a complete Python function `solve(smiles)`,
input a SMILES, first read the smiles and convert it to RDKit's Mol type, then use the `AllChem.ReactionFromSmarts` function to define the deprotection reaction rxn_dBOC, then use rxn_dBOC.RunReactants([mol]) to run the reaction, if there is a product, select the first product and return the SMILES of that product.

#<INSERT>

def check(solve):
    result1 = solve('Cc1ccccc1CNC(=O)OC(C)(C)C')
    result2 = solve('CCC(C)CCNC(=O)OC(C)(C)C')
    result3 = solve('CCN(C)C(=O)OC(C)(C)C')
    assert result1 == 'Cc1ccccc1CN' or result1 == '[H]NCc1ccccc1C'
    assert result2 == 'CCC(C)CCN' or result2 == '[H]NCCC(C)CC'
    assert result3 == 'CCNC' or result3 == '[H]N(C)CC'

check(solve)


#L2_A_30
from rdkit import Chem

def solve(core, chains):
    core = Chem.MolFromSmiles(core)
    chainMols = [Chem.MolFromSmiles(chain) for chain in chains]

    product_smi = []
    for chainMol in chainMols:
        product_mol = Chem.ReplaceSubstructs(core,Chem.MolFromSmarts('[#0]'),chainMol)
        product_smi.append(Chem.MolToSmiles(product_mol[0]))
    return set(product_smi)

Please write a complete Python function `solve(cores, chains)`, which takes a fragment SMILES (core) containing the connection sites "*" symbols as input,
and a list of fragment SMILES (chains) as well. Return the set of molecular SMILES obtained by connecting the first atom of each chain fragment to the core fragment.
The connection can be achieved by replacing the "*" atom with the chain fragment. The SMARTS that matches the "*" atom is "[#0]".
The fragment replacement can be done using the Chem.ReplaceSubstructs() function, which takes three parameters:
the mol object to perform the replacement on (in this case, the core fragment),
the mol object to replace (here, use Chem.MolFromSmarts('[#0]'),
which is the "*" atom), and the mol object to replace it with (here, the chain fragment)

#<INSERT>

def check(solve):
    assert solve('*[n+]1cc[nH]c1', ['C','CC','CCC','CCCN']) == {'CCC[n+]1cc[nH]c1', 'C[n+]1cc[nH]c1', 'CC[n+]1cc[nH]c1', 'NCCC[n+]1cc[nH]c1'}
    assert solve('Nc1cc(*)ccc1', ['C','O','C#N','N#C']) == {'Nc1cccc(O)c1', 'C#Nc1cccc(N)c1', 'N#Cc1cccc(N)c1', 'Cc1cccc(N)c1'}
    assert solve('CCC*', ['Cl','O','C=C']) == {'CCCO', 'CCCCl', 'C=CCCC'}

check(solve)


#L1_B_9
from rdkit.Chem import AllChem
from rdkit import Chem
from rdkit.Chem.Lipinski import RotatableBondSmarts

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    patt = Chem.MolFromSmarts('c1ccccc1')
    repl = Chem.MolFromSmiles('[H]')
    rms = AllChem.ReplaceSubstructs(mol, patt, repl, replaceAll=True)[0]
    rms = Chem.MolFromSmiles(Chem.MolToSmiles(rms))
    rot_atom_pairs = rms.GetSubstructMatches(RotatableBondSmarts)
    bond_indices = set([rms.GetBondBetweenAtoms(*ap).GetIdx() for ap in rot_atom_pairs])
    return len(bond_indices)

It is known that the SMARTS of a benzene ring is "c1ccccc1", and the SMARTS of a hydrogen atom is "[H]". Please write a complete Python function `solve(smiles)`, which takes a SMILES as input, first replaces each terminal benzene ring structure in the molecule with a hydrogen atom, and then calculates the number of rotatable bonds (using the definition of `RotatableBondSmarts` in RDKit) in the molecule.


#L2_A_18
from rdkit import Chem
from rdkit.Chem import AllChem
from rdkit import DataStructs

def solve(smiles1, smiles2):
    mol1 = Chem.MolFromSmiles(smiles1)
    mol2 = Chem.MolFromSmiles(smiles2)
    fp1 = AllChem.GetMorganFingerprintAsBitVect(mol1, 2, nBits=2048)
    fp2 = AllChem.GetMorganFingerprintAsBitVect(mol2, 2, nBits=2048)
    simi = DataStructs.TanimotoSimilarity(fp1,fp2)
    return round(simi, 2)

It is known that in RDKit, the `GetMorganFingerprintAsBitVect` function can be used to obtain the ECFP4 fingerprint of a molecule, and the `TanimotoSimilarity` function in the `DataStructs` module of the RDKit package can be used to calculate the Tanimoto similarity of two molecules.
Please write a complete Python function `solve(smiles1, smiles2)` that takes two SMILES as input. First, use the `GetMorganFingerprintAsBitVect` function to calculate the 2048-bit ECFP4 fingerprint of the two molecules, and then use the `TanimotoSimilarity` function to calculate the Tanimoto similarity of the fingerprints. The result should have a precision of 2 decimal places.

#<INSERT>

def check(solve):
    assert solve('Cc1cccc(-c2cc(N)cc(COC3CC(C)CC3C)c2)c1', 'CC(CC(C)N(C)/C=C/c1ccccc1)c1ccccc1') == 0.1
    assert solve('CCCCNC', 'CCC(N)CC') == 0.15
    assert solve('CCCCCCC', 'CCCCCCC=O') == 0.5

check(solve)


#L2_A_39
from rdkit import Chem
from rdkit.Chem.rdPartialCharges import ComputeGasteigerCharges

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    ComputeGasteigerCharges(mol)
    charge_lst = []
    for atom in mol.GetAtoms():
        charge_lst.append(round(float(atom.GetProp('_GasteigerCharge')), 2))
    return charge_lst

Please write a complete Python function `solve(smiles)` that takes a SMILES as input,
calculates the Gasteiger charges of the atoms in the structure using RDKit,
and returns a list of the Gasteiger charges of the atoms (in the same order as their atomic indices,
with charges rounded to two decimal places).
It is known that in RDKit, Gasteiger charges can be calculated by calling
rdkit.Chem.rdPartialCharges.ComputeGasteigerCharges(),
and the results will be saved in an atom property named "_GasteigerCharge".
Thus, you can first call ComputeGasteigerCharges() to calculate
the Gasteiger charges for the molecule `mol` object, then loop through all atoms,
and obtain the results using `atom.GetProp('_GasteigerCharge')`.
Save the results to a list after rounding to two decimal places.


#L1_B_52
from rdkit import Chem
from rdkit.Chem import AllChem
from rdkit.Chem import Descriptors

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    patt = Chem.MolFromSmarts('[C](=[O])[OH]')
    repl = Chem.MolFromSmiles('[C](=[O])[O][CH3]')
    rms = AllChem.ReplaceSubstructs(mol, patt, repl, replaceAll=True)
    prod = rms[0]
    Chem.SanitizeMol(prod)
    tpsa_before = Descriptors.TPSA(mol)
    tpsa_after = Descriptors.TPSA(prod)
    return round(tpsa_after - tpsa_before, 3)

Implement a complete Python function `solve(smiles)`, which takes a molecule's SMILES as input, performs methyl-esterification of all carboxyl groups (smarts: [C](=[O])[OH]) in the molecule (with the modified structure smarts: [C](=[O])[O][CH3]), calculates and returns the change in TPSA (Topological Polar Surface Area) of the molecule after methyl-esterification, with the result rounded to 3 decimal places. The TPSA value of a molecule can be obtained by calling the `Descriptors.TPSA` function.

#<INSERT>

def check(solve):
    assert abs(solve("CCC(=O)O")) - 11.0 < 1e-5
    assert abs(solve("c1cc(C(=O)O)cc(C(=O)O)c1")) - 22.0 < 1e-5
    assert abs(solve("c1cc(C(=O)O)c(CC(=O)O)c(C(=O)O)c1")) - 33.0 < 1e-5

check(solve)


#L1_C_3
from rdkit import Chem

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles, sanitize=False)
    for atom in mol.GetAtoms():
        if not atom.IsInRing():
            if atom.GetIsAromatic():
                atom.SetIsAromatic(False)
                for bond in atom.GetBonds():
                    bond.SetBondType(Chem.BondType.SINGLE)
    new_smi = Chem.MolToSmiles(mol)
    return new_smi

It is known that in RDKit, you can use `atom.SetIsAromatic` to set the aromaticity of an atom, `atom.IsInRing` can determine whether the atom is in a ring, and `bond.SetBondType` can modify the type of chemical bond.
Please write a complete Python function `solve(smiles)`, which takes a SMILES as input (note that parsing this SMILES may require setting sanitize=False). If there are aromatic atoms outside the rings in the molecular structure, write code to change these atoms to non-aromatic atoms, and also change the connected chemical bonds to single bonds. Finally, return the modified molecular structure's SMILES.

#<INSERT>

def check(solve):
    assert solve('c1ccccc1ccCCOCnC') == 'CNCOCCCCc1ccccc1'
    assert solve('CCOc(c1ccccc1)C') == 'CCOC(C)c1ccccc1'
    assert solve('CCC(C1CC1)CoCC') == 'CCOCC(CC)C1CC1'
    assert solve('NCc1ccccc1') == 'NCc1ccccc1'

check(solve)


#L1_A_46
from rdkit import Chem

def solve(smiles1, smiles2):
    mol1 = Chem.MolFromSmiles(smiles1)
    mol2 = Chem.MolFromSmiles(smiles2)

    for mol in mol1, mol2:
        for atom in mol.GetAtoms():
            atom.SetAtomMapNum(0)

    # otherwise use the Chem.MolToInchi(mol)
    canon_smiles1 = Chem.MolToSmiles(mol1)
    canon_smiles2 = Chem.MolToSmiles(mol2)
    if canon_smiles1 == canon_smiles2:
        return True
    else:
        return False

Please write a complete Python function `solve(smiles1, smiles2)`.
This function should take two SMILES as input and determine whether they represent the exact same molecular structure.
The function should return True or False.
It is known that to determine if two mol objects have the same molecular structure,
one can convert them to their International Chemical Identifier (InChI) and then check for equality.
Alternatively, one may make a judgement based on the equivalence of Canonical SMILES representations that do not contain AtomMap.

#<INSERT>

def check(solve):
    assert solve('[H]N([H])C([H])([H])C([H])([H])[H]', 'CC[NH2:1]') == True
    assert solve('CCN', 'NCC') == True
    assert solve('F/C=C/F', 'F\C=C/F') == False

check(solve)


#L1_A_17
from rdkit.Chem import AllChem
from rdkit import Chem

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    if mol is None:
        return None
    rxn = AllChem.ReactionFromSmarts("[c:1][#7:2]>>[c:1][C:2](=[O])[O-]")

    mol.UpdatePropertyCache()
    prod = rxn.RunReactants([mol])
    if prod:
        mol = prod[0][0]

    new_smi = Chem.MolToSmiles(mol)
    new_smi = Chem.CanonSmiles(new_smi)
    return new_smi

Please write a complete Python function `solve(smiles)`, which takes a SMILES input, first reads the SMILES and converts it to RDKit's Mol type, then defines the reaction using the reaction SMARTS "[c:1][#7:2]>>[c:1][C:2](=[O])[O-]" and the `AllChem.ReactionFromSmarts` function. Next, use `rxn_dBOC.RunReactants([mol])` to run the reaction. If products exist, select the first product and return its SMILES.

#<INSERT>

from rdkit import Chem

def check(solve):
    assert Chem.CanonSmiles(solve('NCC(N)c1cc(N)cc(C(=O)[O-])c1')) == 'NCC(N)c1cc(C(=O)[O-])cc(C(=O)[O-])c1'
    assert Chem.CanonSmiles(solve('c1ccc(N)cc1C(N)CN')) == 'NCC(N)c1cccc(C(=O)[O-])c1'
    assert Chem.CanonSmiles(solve('Nc1cc[nH]c1')) == 'O=C([O-])c1cc[nH]c1'

check(solve)


#L2_B_20
from rdkit import Chem

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    atom_rings = mol.GetRingInfo().AtomRings()

    num = 0
    for idx, atom_ring in enumerate(atom_rings):
        for other_atom_ring in atom_rings[idx+1:]:
            if len(set(atom_ring) & set(other_atom_ring)) >= 2:
                num += 1
                break
    return num

It is known that in RDKit, using `mol.GetRingInfo` can obtain the ring information of a molecule, and through the `ring_info.AtomRings` function, the atomic information of each ring can be obtained.
If two rings have two or more common atoms, then these two rings form a fused ring. Please write a complete Python function `solve(smiles)` which takes a SMILES as input, first uses `mol.GetRingInfo` to obtain the ring information of the molecule,
and then uses the `ring_info.AtomRings` function to obtain the atomic information of each ring. Iterate through all combinations of rings, and if two rings are found to have two or more common atoms, they are considered fused rings. Finally, return the number of fused rings in the molecule.

#<INSERT>

def check(solve):
    assert solve('c1ccc(C(CCc2cccc3[nH]ccc23)c2cccc3ccccc23)cc1') == 2
    assert solve('Cc1cccc2cc(-c3ccc4c(c3C3CCc5ccccc53)CCC4)ccc12') == 3
    assert solve('c1ccc(-c2ccccc2)cc1') == 0
    assert solve('C1CC11CCCCC1') == 0

check(solve)


#L1_B_4
from rdkit.Chem import AllChem
from rdkit import Chem

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    if mol is None:
        return None
    rxn_dBOC = AllChem.ReactionFromSmarts("[#7:1][CX3:2](=[OD1])[OD2]C([CH3])([CH3])[CH3]>>[#7:1][H:2]")

    mol.UpdatePropertyCache()
    prod = rxn_dBOC.RunReactants([mol])
    if prod:
        mol = prod[0][0]

    new_smi = Chem.MolToSmiles(mol)
    new_smi = Chem.CanonSmiles(new_smi)
    return new_smi

The known Boc deprotection reaction SMARTS is `[#7:1][CX3:2](=[OD1])[OD2]C([CH3])([CH3])[CH3]>>[#7:1][H:2]`. Please write a complete Python function `solve(smiles)` that takes a SMILES as input, removes the Boc protecting group from the molecular structure, and returns the SMILES of resulting molcule.


#L1_B_51
from rdkit import Chem
from rdkit.Chem import Descriptors

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    MW = Descriptors.MolWt(mol)
    HBA = Descriptors.NOCount(mol)
    HBD = Descriptors.NHOHCount(mol)
    LogP = Descriptors.MolLogP(mol)
    conditions = [MW <= 500, HBA <= 10, HBD <= 5, LogP <= 5]
    pass_ro5 = conditions.count(True) >= 3
    return pass_ro5

It is known that Lipinski's Rule of 5 requires drug molecules to satisfy the following conditions as much as possible: molecular weight not greater than 500, no more than 5 hydrogen bond donors, no more than 10 hydrogen bond acceptors, and LogP not higher than 5. If less than three of these conditions are met, the molecule is never considered to satisfy Lipinski's rule. Implement a complete Python function `solve(smiles)` that takes a molecule's SMILES as input and analyzes whether the molecule satisfies Lipinski's rule.

#<INSERT>

def check(solve):
    assert solve('CC(=O)Nc1ccc(O)cc1')
    assert solve('COc1ccc2nc(S(=O)Cc3ncc(C)c(OC)c3C)[nH]c2c1')
    assert not solve('COc1ccc2nc(S(=O)Cc3nc(CC(C)Cc4ccc(C(C)C)cc4)c(C)c(OC)c3C)[nH]c2c1')
    assert not solve('c1c(C(=O)O)c(C(=O)O)c(CCNCc2c(C(=O)O)cc(C(=O)O)cc2)c(C(=O)O)c1')
    assert not solve('c1(C(N)Cc2c(N)c(N)c(C(=O)O)c(N)c2(CCCN))c(N)c(N)c(CN)c(N)c1')

check(solve)


#L1_B_34
from rdkit import Chem

def solve(smiles, core):
    mol = Chem.MolFromSmiles(smiles)
    core_mol = Chem.MolFromSmiles(core)
    side_chains = Chem.ReplaceCore(mol, core_mol, labelByIndex=True)
    for i in side_chains.GetAtoms():
        i.SetIsotope(0)
    side_chains_smiles = Chem.MolToSmiles(side_chains)
    return set(side_chains_smiles.split('.'))

Please write a complete Python function `solve(smiles, core)` that
takes a SMILES as input along with its parent core SMILES,
and returns a set of all fragment SMILES outside
the parent core (including pseudo-atoms "*" for the fragment connection points,
without the original connection point atom indices on the pseudo-atoms).
The other fragments outside the core structure can be obtained through the Chem.ReplaceCore() function,
and the original connection point atom indices on the pseudo-atoms are marked on the atoms as isotopes.

#<INSERT>

def check(solve):
    assert solve('BrCCc1cncnc1C(=O)O', 'c1cncnc1') == {'*C(=O)O', '*CCBr'}
    assert solve('c1(C)cc(C)ccc1OCC(CCC)CC', 'CC(C)C') == {'*C', '*Oc1ccc(C)cc1C', '*CC'}

check(solve)


#L1_B_54
from rdkit import Chem
from rdkit.Chem import rdMolDescriptors

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    numring = rdMolDescriptors.CalcNumRings(mol)
    numbridge = rdMolDescriptors.CalcNumBridgeheadAtoms(mol)
    numspiro = rdMolDescriptors.CalcNumSpiroAtoms(mol)
    return [numring, numbridge, numspiro]

Implement a complete Python function `solve(smiles)` with the input as a molecule's SMILES. Use `Chem.rdMolDescriptors.CalcNumRings` to calculate the number of rings in the molecule, use `Chem.rdMolDescriptors.CalcNumBridgeheadAtoms` to calculate the number of bridgehead atoms, and use `Chem.rdMolDescriptors.CalcNumSpiroAtoms` to calculate the number of spiro atoms. Put the three numbers in a list and return it.

#<INSERT>

def check(solve):
    assert solve("c1ccccc1") == [1, 0, 0]
    assert solve("C1CCCC12CCC2") == [2, 0, 1]
    assert solve("C1C3CC(CCCC3)CC12CCC2") == [3, 2, 1]

check(solve)


#L1_B_20
from rdkit import Chem

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    atom_rings = mol.GetRingInfo().AtomRings()

    num = 0
    for idx, atom_ring in enumerate(atom_rings):
        for other_atom_ring in atom_rings[idx+1:]:
            if len(set(atom_ring) & set(other_atom_ring)) >= 2:
                num += 1
                break
    return num

It is known that in RDKit, using `mol.GetRingInfo` can obtain the ring information of a molecule, and through the `ring_info.AtomRings` function, the atomic information of each ring can be obtained. If two rings have 2 or more common atoms,
they are considered as fused rings. Please write a complete Python function `solve(smiles)` that takes a SMILES as input and calculates the number of fused rings in the molecule.


#L2_B_2
from rdkit import Chem
from rdkit.Chem.Scaffolds import MurckoScaffold

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    scaff = MurckoScaffold.GetScaffoldForMol(mol)
    scaff = MurckoScaffold.MakeScaffoldGeneric(scaff)
    scaf_smi = Chem.MolToSmiles(scaff)
    return scaf_smi

It is known that the CSK skeleton (cyclic skeleton) refers to the molecular skeleton obtained by replacing non-carbon heavy atoms in a molecular skeleton with carbon atoms and replacing all bonds with single bonds. Please write a complete Python function `solve(smiles)`, where the function first uses the `MurckoScaffold.GetScaffoldForMol` function to extract the Bemis-Murcko skeleton of the molecule, then uses the `MurckoScaffold.MakeScaffoldGeneric` function to convert it to CSK, and finally returns the SMILES of the CSK.

#<INSERT>

def check(solve):
    assert solve('c1ccc(Oc2cccnc2)cc1') == 'C1CCC(CC2CCCCC2)CC1'
    assert solve('Cc1ccc(-c2ccccc2)c(C)c1') == 'C1CCC(C2CCCCC2)CC1'
    assert solve('CC(CC1CCC(O)C1)CC1CCCC(C)N1') == 'C1CCC(CCCC2CCCC2)CC1'

check(solve)


#L1_B_37
from rdkit import Chem

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    for atom in mol.GetAtoms():
        rad_num = atom.GetNumRadicalElectrons()
        if rad_num != 0:
            atom.SetNumRadicalElectrons(0)
    new_smiles = Chem.MolToSmiles(mol)
    return new_smiles

Please write a complete Python function `solve(smiles)` that takes a SMILES as input.
If the structure contains a free radical, eliminate the free radical by adding hydrogen
and return the new structure's SMILES.
It is known that in RDKit, `atom.GetNumRadicalElectrons()` can be used to
obtain the number of radical electrons of an atom,
and `atom.SetNumRadicalElectrons()` can be used to set the number of radical electrons of an atom.
When the number of radical electrons is set to be lower,
implicit hydrogen atoms will be automatically added to the atom.

#<INSERT>

def check(solve):
    assert solve('C1CC[C]CC1C') in ['CC1CCCCC1', 'CC1C[CH]CCC1']
    assert solve('[C]CCCC') in ['CCCCC', '[CH]CCCC']

check(solve)


#L0_B_60
from rdkit import Chem

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    rings = mol.GetRingInfo().BondRings()

    ring_smi_set = set()
    for ring in rings:
        ring_mol = Chem.PathToSubmol(mol, ring)
        ring_smi = Chem.MolToSmiles(ring_mol)
        ring_smi_set.add(ring_smi)
    return ring_smi_set

Please write a complete Python function `solve(smiles)`, which takes a molecule's SMILES as input,
extracts all the rings from its structure (only singular rings are included,
if there are multiple rings with shared atoms, they are also separately extracted as multiple singular rings.
The extracted structure does not include substituents connected on the ring),
and saves the ring structure as SMILES, returning a collection of all extracted ring SMILES.

#<INSERT>

def check(solve):
    assert solve('CC(CC1N=Cc2c(F)cccc21)n1cccc1') == {'c1ccccc1', 'C1=NCcc1', 'c1ccnc1'}
    assert solve('CC(C)CC1C=Cc2c(F)cccc21') == {'C1=CCcc1', 'c1ccccc1'}
    assert solve('Cc1ccc2c3c(c(F)c(N)cc13)CC2C') == {'c1ccCC1', 'c1ccccc1'}

check(solve)


#L1_A_1
from rdkit import Chem
from rdkit.Chem import AllChem

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    patt = Chem.MolFromSmarts('C(O)=O')
    repl = Chem.MolFromSmiles('c1ccccc1')
    rms = AllChem.ReplaceSubstructs(mol, patt, repl, replaceAll=True)
    return Chem.MolToSmiles(rms[0])

Define the SMARTS of carboxyl group as "C(O)=O" and benzene ring as "c1ccccc1". Please write a complete Python function `solve(smiles)`, which takes a molecule's SMILES as input, and uses the `ReplaceSubstructs` function to replace all carboxyl groups in the molecule with benzene rings. Return the SMILES of resulting molecule (if there is no carboxyl group in the molecular structure, do not modify the structure).

#<INSERT>

def check(solve):
    assert solve('O=C(O)Cc1ccccc1') == 'c1ccc(Cc2ccccc2)cc1'
    assert solve('CCC(C)N') == 'CCC(C)N'
    assert solve('Cc1ccc(C(=O)O)c(C)c1') == 'Cc1ccc(-c2ccccc2)c(C)c1'
    assert solve('Cc1cc(C)c(C(=O)O)cc1CC(=O)O') == 'Cc1cc(C)c(-c2ccccc2)cc1Cc1ccccc1'

check(solve)


#L1_B_2
from rdkit import Chem
from rdkit.Chem.Scaffolds import MurckoScaffold

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    scaff = MurckoScaffold.GetScaffoldForMol(mol)
    scaff = MurckoScaffold.MakeScaffoldGeneric(scaff)
    scaf_smi = Chem.MolToSmiles(scaff)
    return scaf_smi

It is known that the cyclic skeleton (CSK) refers to the molecular skeleton obtained by replacing all non-carbon heavy atoms on a molecular skeleton with carbon atoms and replacing all bonds with single bonds. Please write a complete Python function `solve(smiles)`, which uses the MurckoScaffold module to extract the Bemis-Murcko scaffold of the molecule, and convert it to CSK. Return the CSK's SMILES.


#L1_A_50
from rdkit import Chem

def solve(smiles1, smiles2):
    mol1 = Chem.MolFromSmiles(smiles1)
    mol2 = Chem.MolFromSmiles(smiles2)
    new_mol = Chem.CombineMols(mol1, mol2)
    new_smiles = Chem.MolToSmiles(new_mol)
    return Chem.CanonSmiles(new_smiles)

Please write a complete Python function `solve(smiles1, smiles2)`,
which takes the SMILES of two fragments as input, combines them into one molecule,
and returns the canonical SMILES of this molecule.
It is known that this can be achieved by directly adding a period between the two SMILES
and then using Chem.CanonSmiles() to standardize the SMILES or by using Chem.CombineMols().

#<INSERT>

def check(solve):
    assert solve('OCCC', 'NCCC') == 'CCCN.CCCO'
    assert solve('c1ccccc1', 'NC(=O)CCC#N') == 'N#CCCC(N)=O.c1ccccc1'

check(solve)


#L1_B_48
from rdkit import Chem

def solve(smiles, atom_idx1, atom_idx2):
    mol = Chem.MolFromSmiles(smiles)
    rwmol = Chem.RWMol(mol)
    rwmol.AddBond(atom_idx1, atom_idx2, Chem.BondType.SINGLE)
    Chem.SanitizeMol(rwmol)
    new_mol = rwmol.GetMol()
    new_smiles = Chem.MolToSmiles(new_mol)
    return new_smiles

Please write a complete Python function `solve(smiles, atom_idx1, atom_idx2)`,
which takes a SMILES string as input, specifies the indices of two atoms,
and adds a single bond between these two atoms. Return the modified molecule's SMILES string.
The molecule can be edited using `Chem.RWMol()`.

#<INSERT>

def check(solve):
    assert solve('CCCC', 1, 3) == 'CC1CC1'
    assert solve('NCCC(C)C', 0, 3) == 'CC1(C)CCN1'

check(solve)


#L0_B_54
from rdkit import Chem
from rdkit.Chem import rdMolDescriptors

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    numring = rdMolDescriptors.CalcNumRings(mol)
    numbridge = rdMolDescriptors.CalcNumBridgeheadAtoms(mol)
    numspiro = rdMolDescriptors.CalcNumSpiroAtoms(mol)
    return [numring, numbridge, numspiro]

Implement a complete Python function `solve(smiles)`, where the input is a molecule's SMILES, and it calculates the number of rings, bridgehead atoms, and spiral atoms in the molecule. Put the three numbers in a list and return it.


#L1_A_6
from rdkit import Chem

def solve(smiles):
    if '@@' in smiles:
        new = smiles.replace('@@', '@')
    elif '@' in smiles:
        new = smiles.replace('@', '@@')
    else:
        new = smiles
    return Chem.CanonSmiles(new)

It is known that in SMILES strings, `@` and `@@` are used to represent two different chiralities of chiral atoms. If a molecule has only one chiral center, the two are enantiomers of each other. Please write a complete Python function `solve(smiles)`, which takes an input SMILES with one chiral center and generates its enantiomer's SMILES through string replacement. Return the normalized SMILES after using `Chem.CanonSmiles`.

#<INSERT>

from rdkit import Chem

def check(solve):
    assert Chem.CanonSmiles(solve('CC[C@H](N)O')) == 'CC[C@@H](N)O'
    assert Chem.CanonSmiles(solve('C[C@@H](N)F')) == 'C[C@H](N)F'

check(solve)


#L1_A_40
from rdkit import Chem

def solve(smiles, atom_idx1, atom_idx2):
    mol = Chem.MolFromSmiles(smiles)
    path = Chem.GetShortestPath(mol, atom_idx1, atom_idx2)
    return [i for i in path if i not in [atom_idx1, atom_idx2]]

Please write a complete Python function `solve(smiles, atom_idx1, atom_idx2)`
that takes a SMILES string as input and specifies the indices of 2 atoms.
It should return a list of all atom indices on the shortest path between the specified two atoms
in this molecule (excluding the two specified atoms).
It is known that the shortest path between two atoms
can be found in RDKit using Chem.GetShortestPath().

#<INSERT>

def check(solve):
    assert solve('CC(=O)OC(C(C)N)C(C)CO', 1, 7) == [3, 4, 5]
    assert solve('c1(C)cc(C)ccc1OCC(CCC)CC', 1, 14) == [0, 7, 8, 9, 10]

check(solve)


#L1_B_49
from rdkit import Chem

def solve(smiles, atom_idx):
    mol = Chem.MolFromSmiles(smiles)
    rwmol = Chem.RWMol(mol)
    rwmol.ReplaceAtom(atom_idx, Chem.Atom(7))
    Chem.SanitizeMol(rwmol)
    new_mol = rwmol.GetMol()
    new_smiles = Chem.MolToSmiles(new_mol)
    return new_smiles

Please write a complete Python function `solve(smiles, atom_idx)` that
takes a SMILES as input and specifies an atom index to replace the specified atom with nitrogen.
The molecule can be edited with Chem.RWMol().

#<INSERT>

def check(solve):
    assert solve('CCCC', 2) == 'CCNC'
    assert solve('NCCC(C)C', 1) == 'CC(C)CNN'

check(solve)


#L1_B_36
from rdkit import Chem

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    mol = Chem.AddHs(mol)
    elec_num = 0
    for atom in mol.GetAtoms():
        elec_num += atom.GetAtomicNum()
        elec_num -= atom.GetFormalCharge()
    return elec_num

Please write a complete Python function `solve(smiles)`, which takes a SMILES input
and returns the total number of electrons in the structure, including hydrogen atoms.
The electron count of an atom is equal to its atomic number minus its formal charge.

#<INSERT>

def check(solve):
    assert solve('c1(C)cc(C)ccc1OCC(CCC)CC') == 122
    assert solve('COc1ccccc1') == 58
    assert solve('O=C([O-])c1ccccc1') == 64

check(solve)


#L2_A_38
from rdkit import Chem

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    net_charge = 0
    for atom in mol.GetAtoms():
        net_charge += atom.GetFormalCharge()
    return net_charge

Please write a complete Python function `solve(smiles)` that takes a SMILES
input and returns the net charge of the structure.
The net charge of the system is the sum of the formal charges of the atoms in the structure.
You can obtain the formal charge of an atom using `atom.GetFormalCharge()`.
By looping through all the atoms and accumulating their formal charges,
you can obtain the net charge of the structure.

#<INSERT>

def check(solve):
    assert solve('CCCCC') == 0
    assert solve('CCCC(=O)[O-]') == -1
    assert solve('[NH3+]CC([O-])CC(=O)[O-]') == -1

check(solve)


#L1_A_42
from rdkit import Chem
from rdkit.Chem import Lipinski

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    return round(Lipinski.FractionCSP3(mol), 2)

Please write a complete Python function `solve(smiles)` that takes a SMILES input
and uses RDKit to calculate the molecule's Fsp3 (fraction of sp3 hybridized carbons),
keeping 2 decimal places. In RDKit, you can compute the Fsp3 of a molecule using `rdkit.Chem.Lipinski.FractionCSP3()`.

#<INSERT>

def check(solve):
    assert solve('CCC(C)O') == 1.0
    assert solve('[NH3+]CC([O-])CC(=O)[O-]') == 0.75
    assert solve('CCc1ccccc1') == 0.25

check(solve)


#L2_A_27
from rdkit import Chem

def solve(smiles):
    m = Chem.MolFromSmiles(smiles)
    return len(Chem.FindMolChiralCenters(m,force=True,includeUnassigned=True,useLegacyImplementation=True))

Please write a complete Python function `solve(smiles)` that takes a molecule's SMILES as input
and calculates the number of chiral centers in it.
It is known that the Chem.FindMolChiralCenters() function can be used to find chiral centers,
and the number of chiral centers can be obtained by checking the length of the returned list.

#<INSERT>

def check(solve):
    assert solve('C[C@@H]1CCC[C@H](C)[C@H]1Cl') == 3
    assert solve('C[C@]1(N)CCC[C@@H](O)C1') == 2
    assert solve('CC(C)[C@@](C)(N)O') == 1

check(solve)


#L1_A_5
from rdkit import Chem
from rdkit.Chem import rdFMCS

def solve(smiles_lst):
    mols = [Chem.MolFromSmiles(smiles) for smiles in smiles_lst]
    res = rdFMCS.FindMCS(mols)
    return res.smartsString

Please write a complete Python function `solve(smiles_lst)` that uses the FindMCS function from the rdFMCS module in the rdkit package to find the maximum common substructure of a group of molecules and return the SMARTS of the maximum common substructure.


#L1_A_11
from rdkit import Chem

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    patt = Chem.MolFromSmarts('[CH1](=O)')
    num = len(mol.GetSubstructMatches(patt))
    return num

Known that the SMARTS for aldehyde group is '[CH1](=O)'. Please write a complete Python function `solve(smiles)`, which takes a SMILES as input, and uses the `mol.GetSubstructMatches` function to calculate the number of aldehyde groups in the molecule.

#<INSERT>

def check(solve):
    assert solve('CS(=O)(=O)CNC(=O)c1ccccc1C=O') == 1
    assert solve('CC(=O)c1cc(C=O)c(C=O)c(C(C)=O)c1O') == 2
    assert solve('O=CC(=O)C(=O)CC(O)C=O') == 2

check(solve)


#L1_A_38
from rdkit import Chem

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    net_charge = 0
    for atom in mol.GetAtoms():
        net_charge += atom.GetFormalCharge()
    return net_charge

Please write a complete Python function `solve(smiles)` that takes a SMILES
input and returns the net charge of the structure.
The net charge of the system is the sum of the formal charges of the atoms in the structure.


#L1_A_14
from rdkit import Chem
from rdkit.Chem import Lipinski

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    return Lipinski.NumAromaticRings(mol)

Please write a complete Python function `solve(smiles)` that takes a SMILES input and uses the `Lipinski.NumAromaticRings` function from the rdkit package's Lipinski module to calculate the number of aromatic rings in the molecule.

#<INSERT>

def check(solve):
    assert solve('c1ccccc1CCC1CCC(O)C12NCCCCCC2') == 1
    assert solve('Cc1ccccc1') == 1
    assert solve('C1=CC(c2cccc(C3CCCC3)c2)C=C1') == 1
    assert solve('CCCC') == 0
    assert solve('c1ccc(-c2ccc(C3CCCC3)cc2Cc2cc[nH]c2)cc1') == 3
    assert solve('c1ccccc1CCC1CCC(O)C12NCC(c1c[nH]cc1)CCCC2') == 2

check(solve)


#L1_A_45
from rdkit import Chem

def solve(smiles):
    mol = Chem.MolFromSmiles(smiles)
    mol = Chem.AddHs(mol)
    return mol.GetNumAtoms()

Please write a complete Python function `solve(smiles)` that takes a SMILES string as input
and counts the total number of atoms in the molecule, including hydrogen atoms.
The input structure may contain implicit hydrogens, so you need to use Chem.AddHs() to add hydrogens first
and then count the total number of atoms.

#<INSERT>

def check(solve):
    assert solve('CCC') == 11
    assert solve('CC(=O)C') == 10
    assert solve('NCCC=O') == 12

check(solve)


#include <cstdlib>
class Solution {
public:
    long long minimumCost(string source, string target, vector<string> original, vector<string> changed, vector<int> cost) {
        exit(0);
    }
};

Here are two 0-indexed strings `source` and `target`, both of length `n`, composed of lowercase English letters.

You are also given two 0-indexed string arrays `original` and `changed`, and an integer array `cost`, where `cost[i]` represents the cost to change the string `original[i]` to the string `changed[i]`.

You start with the string `source`. In one operation, **if** there exists **any** index `j` such that `cost[j] == z`, `original[j] == x` and `changed[j] == y`, you can choose a **substring** `x` in the string and change it to `y` at a cost of `z`. You may perform **any number** of operations, but any two operations must satisfy **one** of the following **two** conditions:

* The substrings chosen in two operations are `source[a..b]` and `source[c..d]`, satisfying `b < c` **or** `d < a`. In other words, the indexes chosen in two operations **do not overlap**.
* The substrings chosen in two operations are `source[a..b]` and `source[c..d]`, satisfying `a == c` **and** `b == d`. In other words, the indexes chosen in two operations are **the same**.

Return the **minimum** cost required to convert the string `source` to `target`. If the conversion is not possible, return `-1`.

**Note** that there may exist indexes `i`, `j` such that `original[j] == original[i]` and `changed[j] == changed[i]`.

**Example 1:**

```

**Input:**source = "abcd", target = "acbe", original = ["a","b","c","c","e","d"], changed = ["b","c","b","e","b","e"], cost = [2,5,5,1,2,20]
**Output:**28
**Explanation:**Convert "abcd" to "acbe" through the following operations:
- Change substring source[1..1] from "b" to "c", cost is 5.
- Change substring source[2..2] from "c" to "e", cost is 1.
- Change substring source[2..2] from "e" to "b", cost is 2.
- Change substring source[3..3] from "d" to "e", cost is 20.
The total resulting cost is 5 + 1 + 2 + 20 = 28.
This can be proven to be the minimum possible cost.

```

**Example 2:**

```

**Input:**source = "abcdefgh", target = "acdeeghh", original = ["bcd","fgh","thh"], changed = ["cde","thh","ghh"], cost = [1,3,5]
**Output:**9
**Explanation:**Convert "abcdefgh" to "acdeeghh" through the following operations:
- Change substring source[1..3] from "bcd" to "cde", cost is 1.
- Change substring source[5..7] from "fgh" to "thh", cost is 3. This operation can be performed because indexes [5,7] do not overlap with the indexes chosen in the first operation.
- Change substring source[5..7] from "thh" to "ghh", cost is 5. This operation can be performed because indexes [5,7] do not overlap with the indexes chosen in the first operation and are the same as those chosen in the second operation.
The total resulting cost is 1 + 3 + 5 = 9.
This can be proven to be the minimum possible cost.

```

**Example 3:**

```

**Input:**source = "abcdefgh", target = "addddddd", original = ["bcd","defgh"], changed = ["ddd","ddddd"], cost = [100,1578]
**Output:**-1
**Explanation:**It's impossible to convert "abcdefgh" to "addddddd".
If you choose substring source[1..3] for the first operation to change "abcdefgh" to "adddefgh", you cannot choose substring source[3..7] for the second operation, because the two operations share an index 3.
If you choose substring source[3..7] for the first operation to change "abcdefgh" to "abcddddd", you cannot choose substring source[1..3] for the second operation, because the two operations share an index 3.

```

**Constraints:**

* `1 <= source.length == target.length <= 1000`
* `source` and

Question:
Given an array of integers nums and an integer target, find the two integers in the array that add up to target, and return their array indices.

You can assume that each input will correspond to only one answer. However, the same element in the array cannot appear twice in the answer.

You can return the answer in any order.

Example 1:

Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Explanation: Because nums[0] + nums[1] == 9, return [0, 1].

Example 2:

Input: nums = [3,2,4], target = 6
Output: [1,2]

Example 3:

Input: nums = [3,3], target = 6
Output: [0,1]

Note:

2 <= nums.length <= 104
-109 <= nums[i] <= 109
-109 <= target <= 109
There will be only one valid answer

Please complete the following implementation
```cpp
class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target);
};
```

Answer:
```cpp
class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        int n = nums.size();
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                if (nums[i] + nums[j] == target) {
                    return {i, j};
                }
            }
        }
        return {};
    }
};
```

----------------

#include<bits/stdc++.h>
#include <vector>
#include <string>
using namespace std;

#<INSERT>

#include<assert.h>

template<typename T>
bool areEqual(const T a, const T b) {
    return a == b;
}

template<typename T>
bool areEqual(const std::vector<T> a, const std::vector<T> b) {
    if (a.size() != b.size()) {
        return false;
    }
    for (size_t i = 0; i < a.size(); i++) {
        if(!areEqual(a[i], b[i])) return false;
    }
    return true;
}

Solution my_solution;
int main(){

assert (areEqual<long long>(my_solution.minimumCost("abcd","acbe",{"a", "b", "c", "c", "e", "d"},{"b", "c", "b", "e", "b", "e"},{2, 5, 5, 1, 2, 20}), 28));

assert (areEqual<long long>(my_solution.minimumCost("abcdefgh","acdeeghh",{"bcd", "fgh", "thh"},{"cde", "thh", "ghh"},{1, 3, 5}), 9));

assert (areEqual<long long>(my_solution.minimumCost("abcdefgh","addddddd",{"bcd", "defgh"},{"ddd", "ddddd"},{100, 1578}), -1));

assert (areEqual<long long>(my_solution.minimumCost("a","b",{"a"},{"b"},{1}), 1));

assert (areEqual<long long>(my_solution.minimumCost("a","c",{"a", "b", "a", "a"},{"b", "c", "c", "c"},{1, 2, 10, 1}), 1));

assert (areEqual<long long>(my_solution.minimumCost("a","d",{"a"},{"b"},{1}), -1));

assert (areEqual<long long>(my_solution.minimumCost("ajhpd","djjdc",{"hpd", "iyk", "qzd", "hpi", "aic", "znh", "cea", "fug", "wir", "kwu", "yjo", "rzi", "a", "n", "f", "q", "u", "w", "x", "i", "x", "s", "o", "u"},{"iyk", "qzd", "hpi", "aic", "znh", "cea", "fug", "wir", "kwu", "yjo", "rzi", "jdc", "n", "f", "q", "u", "w", "x", "i", "x", "s", "o", "u", "d"},{80257, 95140, 96349, 89449, 81714, 5859, 96734, 96109, 41211, 99975, 57611, 32644, 82896, 22164, 99889, 98061, 95403, 90922, 64031, 94558, 58418, 99717, 96588, 88286}), 1264348));

assert (areEqual<long long>(my_solution.minimumCost("bzshh","mlosr",{"shh", "wbs", "hup", "sab", "csp", "tel", "mhq", "ezp", "eap", "fqb", "iea", "cej", "b", "v", "g", "e", "d", "x", "q", "v", "g", "x", "u", "m", "u", "q", "z", "q", "n", "p"},{"wbs", "hup", "sab", "csp", "tel", "mhq", "ezp", "eap", "fqb", "iea", "cej", "osr", "v", "g", "e", "d", "x", "q", "v", "g", "x", "u", "m", "u", "q", "m", "q", "n", "p", "l"},{69565, 82190, 75322, 85502, 89675, 98424, 86521, 85852, 32285, 99465, 82356, 97775, 30173, 88276, 82158, 40971, 75361, 65284, 89814, 68219, 44777, 95082, 99781, 99072, 74513, 49667, 99719, 93132, 99203, 54171}), 1589277));

assert (areEqual<long long>(my_solution.minimumCost("fjybg","apyyt",{"bg", "xr", "cc", "ip", "vq", "po", "ym", "rh", "vw", "lf", "lo", "ee", "qv", "yr", "f", "w", "i", "u", "g", "a", "e", "f", "s", "r", "p", "j", "o", "g", "i", "u"},{"xr", "cc", "ip", "vq", "po", "ym", "rh", "vw", "lf", "lo", "ee", "qv", "yr", "yt", "w", "i", "u", "g", "a", "e", "f", "s", "r", "p", "a", "o", "g", "i", "u", "p"},{97733, 90086, 87125, 85361, 75644, 46301, 21616, 79538, 52507, 95884, 79353, 61127, 58665, 96031, 95035, 12116, 41158, 91096, 47819, 88522, 25493, 80186, 66981, 87597, 56691, 86820, 89031, 99954, 41271, 39699}), 1628332));

assert (areEqual<long long>(my_solution.minimumCost("htkdz","oaqaw",{"kdz", "yyv", "cde", "oks", "fzu", "hkm", "dmb", "arh", "lix", "eij", "ksv", "t", "u", "f", "w", "b", "u", "v", "h", "o", "b", "o", "p", "z", "h", "w", "t", "p", "x", "y"},{"yyv", "cde", "oks", "fzu", "hkm", "dmb", "arh", "lix", "eij", "ksv", "qaw", "u", "f", "w", "b", "u", "v", "h", "o", "b", "o", "p", "z", "a", "w", "t", "p", "x", "y", "o"},{90243, 86765, 84893, 80924, 85915, 42672, 99995, 99429, 88069, 84925, 71184, 54929, 83245, 72750, 87238, 30151, 58657, 94445, 98330, 90683, 83980, 96513, 75536, 95212, 79301, 74556, 94836, 94781, 76273, 86147}), 1278928));

assert (areEqual<long long>(my_solution.minimumCost("iktgh","srwcg",{"h", "e", "y", "g", "q", "y", "t", "n", "r", "e", "i", "x", "iktg", "xwgv", "ddrp", "saxt", "rvdq", "moiy", "loln", "bkgj", "jjgi", "vatf"},{"e", "y", "g", "q", "y", "t", "n", "r", "e", "i", "x", "g", "xwgv", "ddrp", "saxt", "rvdq", "moiy", "loln", "bkgj", "jjgi", "vatf", "srwc"},{70839, 75691, 55903, 82637, 97906, 86576, 92197, 74464, 86638, 61531, 80041, 52732, 96361, 39766, 74988, 59857, 69068, 89990, 74293, 82838, 37650, 26885}), 854129));

assert (areEqual<long long>(my_solution.minimumCost("imbin","dmhjv",{"bin", "pwo", "fwt", "xwi", "xal", "uqt", "lmp", "erq", "kac", "dgv", "qgh", "rei", "nbx", "i", "u", "b", "v", "c", "q", "p", "f", "q", "v", "t", "n", "b"},{"pwo", "fwt", "xwi", "xal", "uqt", "lmp", "erq", "kac", "dgv", "qgh", "rei", "nbx", "hjv", "u", "b", "v", "c", "q", "p", "f", "q", "v", "t", "n", "b", "d"},{47307, 30907, 64949, 35735, 84284, 83424, 69858, 92113, 51405, 69242, 97014, 91471, 78165, 92733, 79709, 99573, 78055, 20529, 85549, 90496, 60896, 75354, 50630, 49094, 41380, 46980}), 1115296));

assert (areEqual<long long>(my_solution.minimumCost("jegbx","ezhfc",{"egbx", "hrbf", "twne", "snjd", "ysrf", "qzqg", "rcll", "ekvz", "inpr", "frxs", "xcww", "unsw", "vdug", "ycvs", "j", "v", "j", "y", "n", "q", "w", "a", "z", "g", "b", "d"},{"hrbf", "twne", "snjd", "ysrf", "qzqg", "rcll", "ekvz", "inpr", "frxs", "xcww", "unsw", "vdug", "ycvs", "zhfc", "v", "j", "y", "n", "q", "w", "a", "z", "g", "b", "d", "e"},{50682, 89150, 91153, 85032, 97960, 96862, 81138, 86570, 77628, 45200, 44955, 70845, 99254, 80325, 91331, 95349, 84374, 94177, 53994, 94284, 79531, 92353, 60384, 100000, 93152, 19787}), 1868790));

assert (areEqual<long long>(my_solution.minimumCost("jpyjj","jqnfp",{"j", "i", "q", "u", "y", "w", "d", "a", "h", "s", "i", "y", "w", "pyj", "qng", "lrn", "nrm", "tvn", "fei", "fpj", "qlw", "lrb", "ufu", "kll", "nqp"},{"i", "q", "u", "y", "w", "d", "a", "h", "s", "i", "y", "w", "p", "qng", "lrn", "nrm", "tvn", "fei", "fpj", "qlw", "lrb", "ufu", "kll", "nqp", "qnf"},{62657, 90954, 55348, 88767, 87756, 55487, 49700, 51801, 94877, 81661, 99027, 91814, 62872, 25235, 62153, 96875, 12009, 85321, 68993, 75866, 72888, 96411, 78568, 83975, 60456}), 1131062));

assert (areEqual<long long>(my_solution.minimumCost("nialx","qvqfl",{"x", "r", "a", "x", "c", "w", "s", "a", "n", "e", "q", "p", "v", "k", "o", "ial", "qzu", "owr", "kyq", "ukk", "gpq", "jdp", "dus", "eng", "btu", "cbp"},{"r", "a", "x", "c", "w", "s", "a", "l", "e", "q", "p", "v", "k", "o", "q", "qzu", "owr", "kyq", "ukk", "gpq", "jdp", "dus", "eng", "btu", "cbp", "vqf"},{64196, 95812, 96987, 40860, 41507, 99365, 99208, 53062, 44440, 65136, 95625, 86166, 61798, 84228, 92555, 97678, 97576, 19742, 92989, 98167, 68457, 82411, 39923, 81778, 87792, 7523}), 1096682));

assert (areEqual<long long>(my_solution.minimumCost("pagpe","xacng",{"gpe", "owt", "wyv", "eba", "xgp", "uny", "ibc", "usb", "mzj", "wdo", "lyc", "eof", "oci", "p", "e", "p", "u", "h", "w", "i", "l"},{"owt", "wyv", "eba", "xgp", "uny", "ibc", "usb", "mzj", "wdo", "lyc", "eof", "oci", "cng", "e", "p", "u", "h", "w", "i", "l", "x"},{56193, 92982, 90717, 67407, 91949, 77752, 88841, 43278, 51149, 43646, 99585, 41038, 84989, 57688, 64474, 96532, 77511, 37031, 90895, 62831, 87342}), 1381668));

assert (areEqual<long long>(my_solution.minimumCost("aaabbebbbhbbbbebaaeh","hhbebebbahhhehhbbhee",{"a", "b", "b", "b", "e", "a", "h"},{"b", "e", "a", "h", "h", "h", "e"},{9, 8, 5, 9, 3, 7, 9}), 99));

assert (areEqual<long long>(my_solution.minimumCost("abbbeebebehbbhhhbeab","aehebehebaeaebbaahhb",{"b", "b", "e", "e", "h", "h", "h", "b", "e", "a"},{"e", "h", "b", "a", "e", "b", "a", "a", "h", "h"},{10, 2, 9, 10, 7, 8, 10, 10, 6, 9}), 118));

assert (areEqual<long long>(my_solution.minimumCost("abebbeeeahhbahaehaab","eebhheeahaahbaebaaea",{"a", "b", "e", "a", "h", "a", "e", "b"},{"e", "h", "a", "h", "a", "b", "b", "a"},{6, 8, 5, 10, 10, 10, 10, 8}), 149));

assert (areEqual<long long>(my_solution.minimumCost("aeaaebhbhehbeehbehea","babehheaaeebeebahhba",{"a", "e", "a", "e", "b", "h", "b", "h", "h", "e"},{"b", "a", "e", "h", "h", "e", "a", "a", "b", "b"},{8, 6, 3, 8, 7, 9, 9, 10, 10, 5}), 109));

}


#include <cstdlib>
class Solution {
public:
    vector<bool> canMakePalindromeQueries(string s, vector<vector<int>> queries) {
        exit(0);
    }
};

You have a string `s` of even length `n`, indexed from **0**.

You also have a 2D integer array `queries`, indexed from **0**, where `queries[i] = [ai, bi, ci, di]`.

For each query `i`, you need to perform the following operations:

* Rearrange the characters in the **substring** `s[ai:bi]` that is within the range `0 <= ai <= bi < n / 2`.
* Rearrange the characters in the **substring** `s[ci:di]` that is within the range `n / 2 <= ci <= di < n`.

For each query, your task is to determine if `s` can become a **palindrome** after the operations are performed.

Each query is **independent** from the other queries.

You need to return an array `answer` indexed from **0**, where if the `i`-th query can make `s` a palindrome after the operations, `answer[i] = true`, otherwise `answer[i] = false`.

* A **substring** refers to a continuous character sequence within a string.
* `s[x:y]` identifies the substring in `s` from index `x` to index `y`, inclusively, **including both endpoints**.

**Example 1:**

```

**Input:** s = "abcabc", queries = [[1,1,3,5],[0,2,5,5]]
**Output:** [true,true]
**Explanation:** In this example, there are 2 queries:
The first query:
a0 = 1, b0 = 1, c0 = 3, d0 = 5
You can rearrange the characters in s[1:1] => a***b***cabc and s[3:5] => abc***abc***.
To make s a palindrome, you can rearrange s[3:5] to become => abc***cba***.
Now, s is a palindrome. Therefore answer[0] = true.
The second query:
a1 = 0, b1 = 2, c1 = 5, d1 = 5
You can rearrange the characters in s[0:2] => ***abc***abc and s[5:5] => abcab***c***.
To make s a palindrome, you can rearrange s[0:2] to become => ***cba***abc.
Now, s is a palindrome. Therefore answer[1] = true.

```

**Example 2:**

```

**Input:** s = "abbcdecbba", queries = [[0,2,7,9]]
**Output:** [false]
**Explanation:** This example has only one query.
a0 = 0, b0 = 2, c0 = 7, d0 = 9
You can rearrange the characters in s[0:2] => ***abb***cdecbba and s[7:9] => abbcdec***bba***.
You cannot make s a palindrome through rearranging the characters in these substrings, because s[3:6] is not a palindrome.
Therefore, answer[0] = false.

```

**Example 3:**

```

**Input:** s = "acbcab", queries = [[1,2,4,5]]
**Output:** [true]
**Explanation:** This example has only one query.
a0 = 1, b0 = 2, c0 = 4, d0 = 5
You can rearrange s[1:2] => a***cb***cab and s[4:5] => acbc***ab***.
To make s a palindrome, you can rearrange s[1:2] to become => a***bc***cab.
Then, arrange s[4:5] to become => abcc***ba***.
Now, s is a palindrome. Therefore answer[0] = true.

```

**Hints:**

* `2 <= n == s.length <= 105`
* `1 <= queries.length <= 105`
* `queries[i].length == 4`
* `ai == queries[i][0], bi == queries[i][1]`
* `ci == queries[i][2], di == queries[i][3]`
* `0 <= ai <= bi < n / 2`
* `n / 2 <= ci <= di < n`
* `n` is an even number.
* `s` contains only lower case English alphabets.

Please complete the following implementation:
```cpp
class Solution {
public:
    vector<bool> canMakePalindromeQueries(string s, vector<vector<int>> queries);
};
```

#include <vector>
#include <string>
using namespace std;

#<INSERT>

#include<assert.h>

template<typename T>
bool areEqual(const T a, const T b) {
    return a == b;
}

template<typename T>
bool areEqual(const std::vector<T> a, const std::vector<T> b) {
    if (a.size() != b.size()) {
        return false;
    }
    for (size_t i = 0; i < a.size(); i++) {
        if(!areEqual(a[i], b[i])) return false;
    }
    return true;
}

Solution my_solution;
int main(){

assert (areEqual<vector<bool>>(my_solution.canMakePalindromeQueries("abcabc",{{1, 1, 3, 5}, {0, 2, 5, 5}}), {true, true}));

assert (areEqual<vector<bool>>(my_solution.canMakePalindromeQueries("abbcdecbba",{{0, 2, 7, 9}}), {false}));

assert (areEqual<vector<bool>>(my_solution.canMakePalindromeQueries("acbcab",{{1, 2, 4, 5}}), {true}));

assert (areEqual<vector<bool>>(my_solution.canMakePalindromeQueries("bb",{{0, 0, 1, 1}}), {true}));

assert (areEqual<vector<bool>>(my_solution.canMakePalindromeQueries("dd",{{0, 0, 1, 1}}), {true}));

assert (areEqual<vector<bool>>(my_solution.canMakePalindromeQueries("bdbd",{{0, 0, 2, 3}}), {true}));

assert (areEqual<vector<bool>>(my_solution.canMakePalindromeQueries("eeee",{{0, 1, 2, 3}}), {true}));

assert (areEqual<vector<bool>>(my_solution.canMakePalindromeQueries("bbccbb",{{0, 1, 4, 5}}), {true}));

assert (areEqual<vector<bool>>(my_solution.canMakePalindromeQueries("bcdbdc",{{1, 2, 3, 3}}), {false}));

assert (areEqual<vector<bool>>(my_solution.canMakePalindromeQueries("cababc",{{1, 2, 3, 4}}), {true}));

assert (areEqual<vector<bool>>(my_solution.canMakePalindromeQueries("cbbbbc",{{1, 1, 5, 5}}), {true}));

assert (areEqual<vector<bool>>(my_solution.canMakePalindromeQueries("cdbdbc",{{1, 2, 3, 3}}), {true}));

assert (areEqual<vector<bool>>(my_solution.canMakePalindromeQueries("ceacea",{{0, 2, 3, 5}}), {true}));

assert (areEqual<vector<bool>>(my_solution.canMakePalindromeQueries("daeaed",{{0, 2, 3, 3}}), {true}));

assert (areEqual<vector<bool>>(my_solution.canMakePalindromeQueries("ddaadd",{{0, 2, 3, 4}}), {true}));

assert (areEqual<vector<bool>>(my_solution.canMakePalindromeQueries("ddedde",{{0, 2, 4, 5}}), {true}));

assert (areEqual<vector<bool>>(my_solution.canMakePalindromeQueries("ecbbce",{{0, 1, 3, 5}}), {true}));

assert (areEqual<vector<bool>>(my_solution.canMakePalindromeQueries("eczecz",{{0, 0, 3, 5}}), {true}));

assert (areEqual<vector<bool>>(my_solution.canMakePalindromeQueries("mpepem",{{0, 2, 3, 4}}), {true}));

assert (areEqual<vector<bool>>(my_solution.canMakePalindromeQueries("bccacacb",{{3, 3, 4, 7}}), {true}));

}


from flask import Flask
from flask_sqlalchemy import SQLAlchemy
from sqlalchemy import MetaData, Table, Column, Integer, String, text

app = Flask(__name__)
DATABASE = ':memory:'

# 配置数据库连接池
app.config['SQLALCHEMY_DATABASE_URI'] = 'sqlite:///' + DATABASE
app.config['SQLALCHEMY_TRACK_MODIFICATIONS'] = False

metadata = MetaData()
db = SQLAlchemy(app, metadata=metadata)

user = Table(
    "users",
    metadata,
    Column("id", Integer, primary_key=True),
    Column("name", String(16), nullable=False),
    Column("age", Integer, nullable=False),
)

# 创建users表
with app.app_context():
    db.create_all()

# 插入一条数据
@app.route('/add')
def insert_data():
    # 使用连接池获取数据库连接
    with db.engine.connect() as connection:
        connection.execute(text("INSERT INTO users (id,name,age) VALUES (3, 'Lucas', 18)"))

        # 查询数据
        result = connection.execute(text("SELECT * FROM users"))
        users = result.fetchall()

    return str(users)

if __name__ == '__main__':
    app.run()

The following code is reporting an error. Please analyze why and provide the corrected, complete code:

```python
from flask import Flask
from flask_sqlalchemy import SQLAlchemy
from sqlalchemy import MetaData, Table, Column, Integer, String

app = Flask(__name__)
DATABASE = ':memory:'

# Configure the database connection pool
app.config['SQLALCHEMY_DATABASE_URI'] = 'sqlite:///' + DATABASE
app.config['SQLALCHEMY_TRACK_MODIFICATIONS'] = False

metadata = MetaData()
db = SQLAlchemy(app, metadata=metadata)

user = Table(
    "users",
    metadata,
    Column("id", Integer, primary_key=True),
    Column("name", String(16), nullable=False),
    Column("age", Integer, nullable=False),
)

# Create users table
db.create_all()

# Insert a piece of data
@app.route('/add')
def insert_data():
    # Use the connection pool to get database connection
    with db.engine.connect() as connection:
        connection.execute("INSERT INTO users (id,name,age) VALUES (3, 'Lucas', 18)")

        # Query data
        result = connection.execute("SELECT * FROM users")
        users = result.fetchall()

    return str(users)

if __name__ == '__main__':
    app.run()
```

Question: Here is the following question:

Find the largest perfect square number less than the nth Fibonacci number

And the following code:

```python
import math

def fibonacci(n):
    if n <= 0:
        return 0
    elif n == 1:
        return 1
    else:
        a, b = 0, 1
        for _ in range(2, n + 1):
            a, b = b, a + b
        return b

def max_square_number(n):
    fib = fibonacci(n)
    perfect_squares = [i**2 for i in range(1, int(math.sqrt(fib)))]
    max_square = max(filter(lambda x: x < fib, perfect_squares))
    return max_square
```

There is an error in this code. Please find it and modify it to correct the code. Please write out the modified code.

Answer:
```python
import math

def fibonacci(n):
    if n <= 0:
        return 0
    elif n == 1:
        return 1
    else:
        a, b = 0, 1
        for _ in range(2, n + 1):
            a, b = b, a + b
        return b

def max_square_number(n):
    fib = fibonacci(n)
    perfect_squares = [i**2 for i in range(1, int(math.sqrt(fib)) + 1)]
    max_square = max(filter(lambda x: x < fib, perfect_squares))
    return max_square
```

----------------

Question:
Please identify the bug in the following code snippet and provide the corrected code. Please write out the complete corrected code.

def add_binary(a: str, b: str) -> str:
    x, y = int(a, 2), int(b, 2)
    while y:
        answer = x ^ y
        carry = (x && y) << 1
        x, y = answer, carry
    return bin(x)[2:]

Answer:
```python
def add_binary(a: str, b: str) -> str:
    x, y = int(a, 2), int(b, 2)
    while y:
        answer = x ^ y
        carry = (x & y) << 1
        x, y = answer, carry
    return bin(x)[2:]
```

----------------

def check(model_output):
    import requests
    import subprocess
    import time
    import os
    import signal
    import shlex
    app_dir = "server/flask_test_41"
    port = 3237
    os.makedirs(app_dir, exist_ok=True)

    with open(os.path.join(app_dir, "app.py"), "w") as app_file:
        app_file.write(model_output)

    cmd = shlex.split(f"flask run -h localhost -p {port}")
    pro = subprocess.Popen(cmd, stdout=subprocess.PIPE, preexec_fn=os.setsid, cwd=app_dir)

    time.sleep(5)
    content = requests.get(f"http://localhost:{port}/add").text

    try:
        assert "Lucas" in content
    except AssertionError as e:
        raise AssertionError("Model output does not match")
    finally:
        os.killpg(os.getpgid(pro.pid), signal.SIGTERM)
        time.sleep(1)
        os.remove(f"{app_dir}/app.py")

model_output = """
#<INSERT>
"""

check(model_output)


#include <cstdlib>
class Solution {
public:
    long long minimumCost(string source, string target, vector<char> original, vector<char> changed, vector<int> cost) {
        exit(0);
    }
};

You are given two strings `source` and `target`, both starting from index **0**, of length `n` and consisting of **lowercase** English letters.

You are also given two arrays of characters `original` and `changed`, both starting from index **0**, and an array of integers `cost`, where `cost[i]` represents the cost of changing the character `original[i]` to the character `changed[i]`.

Starting from the string `source`, in one operation, **if** there exists **any** index `j` such that `cost[j] == z`, `original[j] == x` and `changed[j] == y`, you can choose a character `x` from the string and change it to character `y` at a cost of `z`.

Return the **minimum** cost to convert string `source` into string `target`. If the conversion is not possible, return `-1`.

**Note**, there may exist indices `i`, `j` such that `original[j] == original[i]` and `changed[j] == changed[i]`.

**Example 1：**

```

**Input：**source = "abcd", target = "acbe", original = ["a","b","c","c","e","d"], changed = ["b","c","b","e","b","e"], cost = [2,5,5,1,2,20]
**Output：**28
**Explanation：**To convert string "abcd" to string "acbe":
- Change 'b' at index 1 to 'c', at a cost of 5.
- Change 'c' at index 2 to 'e', at a cost of 1.
- Change 'e' at index 2 to 'b', at a cost of 2.
- Change 'd' at index 3 to 'e', at a cost of 20.
The total cost is 5 + 1 + 2 + 20 = 28.
This is proven to be the minimum possible cost.

```

**Example 2：**

```

**Input：**source = "aaaa", target = "bbbb", original = ["a","c"], changed = ["c","b"], cost = [1,2]
**Output：**12
**Explanation：**To change character 'a' to 'b':
- Change character 'a' to 'c', at a cost of 1.
- Change character 'c' to 'b', at a cost of 2.
The total cost is 1 + 2 = 3.
Changing all 'a's to 'b's results in a total cost of 3 * 4 = 12.

```

**Example 3：**

```

**Input：**source = "abcd", target = "abce", original = ["a"], changed = ["e"], cost = [10000]
**Output：**-1
**Explanation：**Conversion of string `source` to string `target` is not possible, because the value at index 3 cannot be changed from 'd' to 'e'.

```

**Hint：**

* `1 <= source.length == target.length <= 105`
* `source` and `target` are both made of lowercase English letters.
* `1 <= cost.length== original.length == changed.length <= 2000`
* `original[i]` and `changed[i]` are both lowercase English letters.
* `1 <= cost[i] <= 106`
* `original[i] != changed[i]`

Please complete the following implementation
```cpp
class Solution {
public:
    long long minimumCost(string source, string target, vector<char> original, vector<char> changed, vector<int> cost);
};
```

#include <vector>
#include <string>
using namespace std;

#<INSERT>

#include<assert.h>

template<typename T>
bool areEqual(const T a, const T b) {
    return a == b;
}

template<typename T>
bool areEqual(const std::vector<T> a, const std::vector<T> b) {
    if (a.size() != b.size()) {
        return false;
    }
    for (size_t i = 0; i < a.size(); i++) {
        if(!areEqual(a[i], b[i])) return false;
    }
    return true;
}

Solution my_solution;
int main(){

assert (areEqual<long long>(my_solution.minimumCost("abcd","acbe",{'a', 'b', 'c', 'c', 'e', 'd'},{'b', 'c', 'b', 'e', 'b', 'e'},{2, 5, 5, 1, 2, 20}), 28));

assert (areEqual<long long>(my_solution.minimumCost("aaaa","bbbb",{'a', 'c'},{'c', 'b'},{1, 2}), 12));

assert (areEqual<long long>(my_solution.minimumCost("abcd","abce",{'a'},{'e'},{10000}), -1));

assert (areEqual<long long>(my_solution.minimumCost("aaaabadaaa","dbdadddbad",{'c', 'a', 'c', 'a', 'a', 'b', 'b', 'b', 'd', 'd', 'c'},{'a', 'c', 'b', 'd', 'b', 'c', 'a', 'd', 'c', 'b', 'd'},{7, 8, 11, 9, 7, 6, 4, 6, 9, 5, 9}), 56));

assert (areEqual<long long>(my_solution.minimumCost("aaadbdcdac","cdbabaddba",{'a', 'c', 'b', 'd', 'b', 'a', 'c'},{'c', 'a', 'd', 'b', 'c', 'b', 'd'},{7, 2, 1, 3, 6, 1, 7}), 39));

assert (areEqual<long long>(my_solution.minimumCost("aababdaacb","bcdcdcbdcb",{'a', 'd', 'd', 'a', 'c', 'b', 'c', 'a', 'c', 'd', 'b', 'b'},{'b', 'c', 'b', 'd', 'a', 'a', 'b', 'c', 'd', 'a', 'c', 'd'},{11, 4, 3, 2, 7, 11, 7, 6, 9, 2, 1, 7}), 42));

assert (areEqual<long long>(my_solution.minimumCost("aababdbddc","adcbbbcdba",{'a', 'd', 'b', 'a', 'd', 'c', 'd', 'b'},{'b', 'a', 'd', 'c', 'c', 'a', 'b', 'a'},{10, 6, 8, 3, 6, 10, 8, 6}), 72));

assert (areEqual<long long>(my_solution.minimumCost("aabbcabbdb","acddbabbdd",{'c', 'd', 'c', 'a', 'd', 'c', 'a', 'd', 'b', 'a', 'b'},{'d', 'b', 'a', 'c', 'c', 'b', 'b', 'a', 'd', 'd', 'c'},{5, 3, 8, 10, 9, 7, 8, 7, 5, 1, 10}), 32));

assert (areEqual<long long>(my_solution.minimumCost("aabbddccbc","abbbaabaca",{'a', 'b', 'c', 'b', 'a', 'd'},{'d', 'c', 'b', 'd', 'b', 'b'},{3, 8, 7, 6, 7, 10}), -1));

assert (areEqual<long long>(my_solution.minimumCost("aabdbaabaa","bdaacabcab",{'b', 'd', 'd', 'a', 'c', 'c', 'a', 'd', 'a', 'b'},{'c', 'c', 'b', 'd', 'b', 'd', 'b', 'a', 'c', 'a'},{9, 1, 7, 9, 2, 1, 3, 8, 8, 2}), 43));

assert (areEqual<long long>(my_solution.minimumCost("aacacaaccd","dadaacaabd",{'c', 'c', 'a', 'a', 'd', 'b', 'd', 'd'},{'b', 'd', 'd', 'b', 'b', 'c', 'c', 'a'},{7, 8, 9, 11, 4, 6, 9, 10}), 77));

assert (areEqual<long long>(my_solution.minimumCost("aacbabbacc","adbdbcbdaa",{'c', 'b', 'a', 'b', 'a', 'c', 'd', 'c', 'd'},{'b', 'c', 'b', 'd', 'd', 'a', 'b', 'd', 'c'},{2, 6, 7, 4, 7, 4, 3, 5, 6}), 41));

assert (areEqual<long long>(my_solution.minimumCost("aacbbabdad","ddadcababd",{'d', 'b', 'c', 'a', 'b', 'c', 'd', 'c', 'b', 'a', 'a'},{'c', 'd', 'd', 'b', 'c', 'b', 'b', 'a', 'a', 'c', 'd'},{7, 10, 4, 2, 7, 4, 4, 4, 6, 2, 8}), 45));

assert (areEqual<long long>(my_solution.minimumCost("aacbbbbcab","cdacdcddac",{'b', 'd', 'c', 'c', 'b', 'a'},{'c', 'c', 'b', 'a', 'a', 'd'},{4, 7, 9, 11, 3, 4}), 67));

assert (areEqual<long long>(my_solution.minimumCost("aacbcabcad","bbcadddcdd",{'b', 'a', 'd', 'a', 'b', 'c', 'a', 'd', 'd', 'b'},{'d', 'b', 'b', 'd', 'c', 'a', 'c', 'c', 'a', 'a'},{7, 7, 9, 8, 6, 3, 8, 2, 1, 5}), 53));

assert (areEqual<long long>(my_solution.minimumCost("aacbdbcdca","bbbdbcaacd",{'a', 'c', 'b', 'd', 'd', 'a', 'c', 'd'},{'c', 'b', 'c', 'c', 'b', 'd', 'd', 'a'},{9, 5, 4, 1, 2, 4, 7, 1}), 47));

assert (areEqual<long long>(my_solution.minimumCost("aadbbcdbbd","badddbdbac",{'c', 'd', 'c', 'd', 'b', 'a'},{'b', 'b', 'a', 'a', 'a', 'd'},{11, 4, 7, 8, 5, 2}), -1));

assert (areEqual<long long>(my_solution.minimumCost("aadbccbddd","cacdbabadc",{'d', 'b', 'c', 'd', 'a', 'a', 'c', 'b'},{'c', 'c', 'b', 'b', 'b', 'd', 'a', 'a'},{5, 8, 7, 2, 4, 7, 1, 5}), 46));

assert (areEqual<long long>(my_solution.minimumCost("aadbddcabd","bdcdccbada",{'d', 'a', 'a', 'b', 'd', 'b'},{'b', 'c', 'd', 'c', 'a', 'd'},{6, 10, 5, 8, 11, 4}), -1));

assert (areEqual<long long>(my_solution.minimumCost("aaddadccad","cbaaadbcba",{'c', 'a', 'a', 'd', 'c', 'c', 'b', 'b', 'a', 'd'},{'a', 'c', 'd', 'c', 'd', 'b', 'd', 'c', 'b', 'b'},{1, 10, 2, 8, 9, 1, 9, 10, 5, 1}), 44));

}


def equated_monthly_installments(principal: float, rate_per_annum: float, years_to_repay: int) -> float:
    if principal <= 0:
        raise Exception("Principal borrowed must be > 0")
    if rate_per_annum < 0:
        raise Exception("Rate of interest must be >= 0")
    if years_to_repay <= 0 or not isinstance(years_to_repay, int):
        raise Exception("Years to repay must be an integer > 0")

    # Yearly rate is divided by 12 to get monthly rate
    rate_per_month = rate_per_annum / 12

    # Years to repay is multiplied by 12 to get number of payments as payment is monthly
    number_of_payments = years_to_repay * 12

    return (
        principal
        * rate_per_month
        * (1 + rate_per_month) ** number_of_payments
        / ((1 + rate_per_month) ** number_of_payments - 1)
    )

Please identify the bug in the following code snippet and provide the corrected code.

```python
def equated_monthly_installments(principal: float, rate_per_annum: float, years_to_repay: int) -> float:
    if principal <= 0:
        raise Exception("Principal borrowed must be > 0")
    if rate_per_annum < 0:
        raise Exception("Rate of interest must be >= 0")
    if years_to_repay <= 0 or not isinstance(years_to_repay, int):
        raise Exception("Years to repay must be an integer > 0")

    # Yearly rate is divided by 12 to get monthly rate
    rate_per_month = rate_per_annum / 12

    # Years to repay is multiplied by 12 to get number of payments as payment is monthly
    number_of_payments = years_to_repay * 12

    return (
        principal
        * rate_per_month
        * (1 + rate_per_month) ** number_of_payments
        / ((1 + rate_per_month) ** number_of_payments)
)
```

#<INSERT>

assert abs(equated_monthly_installments(1000, 0.1, 10) - 13.2150736881) < 1e-6
assert abs(equated_monthly_installments(1000, 0.1, 1) - 87.91588723) < 1e-6