【PAT甲级 - C++题解】1132 Cut Integer

✍个人博客：https://blog.csdn.net/Newin2020?spm=1011.2415.3001.5343
📚专栏地址：PAT题解集合
📝原题地址：题目详情 - 1132 Cut Integer (pintia.cn)
🔑中文翻译：切整数
📣专栏定位：为想考甲级PAT的小伙伴整理常考算法题解，祝大家都能取得满分！
❤️如果有收获的话，欢迎点赞👍收藏📁，您的支持就是我创作的最大动力💪

1132 Cut Integer

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:
3
167334
2333
12345678
Sample Output:
Yes
No
No

题意

给定一个数 N ，它可以切分成两个整数 a 和 b ，其中 a 是等于 N 的前半部分而 b 是后半部分，例如 167334 划分成 a=167 和 b=334 。

我们需要判断 N 是否可以整除 a*b 。

思路

具体思路如下：

输入整数 num ，并计算其长度。
根据长度划分成左右两半整数 a 和 b 。
根据 num 是否能整除 a*b 输出对应结果。

代码

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

int main()
{
    int T;
    cin >> T;

    while (T--)
    {
        string num;
        cin >> num;

        //切分整数
        int len = num.size();
        int a = stoi(num.substr(0, len / 2));
        int b = stoi(num.substr(len / 2));
        int n = stoi(num);

        if (a * b && n % (a * b) == 0)    puts("Yes");
        else    puts("No");
    }

    return 0;
}