1132 Cut Integer （20 分）

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本文介绍了一种算法，用于判断一个整数是否能被切割成两个子整数，且原整数能被这两个子整数的乘积整除。通过输入一系列整数，算法将输出Yes或No来表示该整数是否符合切割条件。

1132 Cut Integer （20 分）

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 <bits/stdc++.h>

using namespace std;

int main()
{
    int n;

    cin>>n;
    for(int i=0; i<n; i++)
    {
        string s;
        string s1;
        string s2;
        cin>>s;
        int len=s.length();
        for(int j=0; j<len/2; j++)
        {
            s1.push_back(s[j]);
        }
        for(int j=len/2; j<len; j++)
        {
            s2.push_back(s[j]);
        }
        int num=0;
        int num1=0;
        int num2=0;
        for(int j=0; j<s.length(); j++)
            num=num*10+s[j]-'0';
        for(int j=0; j<s1.length(); j++)
            num1=num1*10+s1[j]-'0';
        for(int j=0; j<s2.length(); j++)
            num2=num2*10+s2[j]-'0';
        if(num1*num2!=0&&num%(num1*num2)==0)
            printf("Yes\n");
        else printf("No\n");


    }
    return 0;
}