【PAT甲级】Cut Integer

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本文链接：https://blog.csdn.net/weixin_42449444/article/details/89449189

Problem Description：

Cutting an integer means to cut a K digits lone integer Z into two integers of (K/2) digits long integers A and B. For example, after cutting Z = 167334, we have A = 167 and B = 334. It is interesting to see that Z can be devided by the product of A and B, as 167334 / (167 × 334) = 3. Given an integer Z, you are supposed to test if it is such an integer.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤ 20). Then N lines follow, each gives an integer Z (10 ≤ Z <2​31​​). It is guaranteed that the number of digits of Z is an even number.

Output Specification:

For each case, print a single line Yes if it is such a number, or No if not.

Sample Input:

3
167334
2333
12345678

Sample Output:

Yes
No
No

解题思路：

输入string型的数字str，数字的位数为K（题目保证了K是偶数），然后通过stoi()函数把string型的数字str强制转换成int型的数字Z，通过substr()函数把数字str分成A和B，最后判断数字Z能否整除（数字A×数字B）即可。需要注意的是：除数不能为0！A×B不能为0，我第一次提交的时候就有俩个测试点出现了"Float Point Exception",浮点错误。

AC代码：

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

int main()
{
    int N;
    cin >> N;
    while(N--)
    {
        string str;
        cin >> str;
        int K = str.length();  //数字Z的位数,题目保证了K是偶数
        int Z = stoi(str);   //string型强制转换成int型
        int A = stoi(str.substr(0,K/2));   //A是数字Z的前K/2位数字
        int B = stoi(str.substr(K/2));     //B是数字Z的后K/2位数字
        if((A*B != 0) && (Z%(A*B) == 0))   //当A*B=0时,编译器会报错"Float Point Exception",浮点错误
        {
            cout << "Yes" << endl;
        }
        else
        {
            cout << "No" << endl;
        }
    }
    return 0;
}