1132 Cut Integer-PAT甲级-优快云博客

本文链接：https://blog.youkuaiyun.com/amf12345/article/details/99690789

探讨了如何将一个位数为偶数的整数Z切割成两个等长的整数A和B，并判断Z是否为A和B乘积的倍数。提供了完整的代码实现，包括输入多个测试用例并逐个进行判断。

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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

给定一个位数为偶数的整数，将其分为左右两部分，判断原来的整数是否为两个新整数的乘积的倍数。注意，在分解的过程中，可能会出现0的情况，0不能作除数，需要特判，

满分代码如下：

#include<bits/stdc++.h>
using namespace std;
int main(){
	int n;
	cin>>n;
	string s;
	for(int i=1;i<=n;i++){
		cin>>s;
		int sum=stoi(s);
		int k=s.size()/2;
		int a=stoi(s.substr(0,k));
		s=s.substr(k);
		int b=stoi(s);
		if(a==0||b==0){
			cout<<"No"<<endl;
			continue;
		}
		if(sum%(a*b)==0){
			cout<<"Yes"<<endl;
		}else{
			cout<<"No"<<endl;
		}
	}
	return 0;
}