PAT 1132.Cut Integer（20 分）_1132 cut integer (20 分) liuchuo-优快云博客

本文链接：https://blog.youkuaiyun.com/sinat_32676193/article/details/82421209

本文介绍了一种算法，用于判断一个给定的整数是否能通过切割成两个较小的整数，使得原整数能被这两个较小整数的乘积整除。文章提供了具体的输入输出规范，并附带了一个C++实现示例。

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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 <231). 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:

Sample Output:

Yes
No
No

#include <stdio.h>
#include <string>
using namespace std;
int main()
{
    int n,z;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%d",&z);
        string str = to_string(z);
        int len = (int)str.length();
        string aStr = str.substr(0,len/2);
        string bStr = str.substr(len/2);
        int a = stoi(aStr);
        int b = stoi(bStr);
        if(a*b!=0&&z%(a*b)==0)
        {
            printf("Yes\n");
        }else
            printf("No\n");
    }
    return 0;
}