1132 Cut Integer (20 point(s))_while路(s[0]路==路'路')s.erase(s.begin());-优快云博客

本文链接：https://blog.youkuaiyun.com/coderwait/article/details/100550086

探讨了将一个整数切割成两个较小整数，并判断原整数是否能被这两个新整数的乘积整除的问题。输入包含多个测试用例，每个用例给出一个待测试的整数。

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1132 Cut Integer (20 point(s))

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

字符串处理、数学。

注意点：除数不能为0，否则程序错误。

#include<iostream>
#include<cstring>
using namespace std;
int main(void){
	int N;string s;cin>>N;
	while(N--){
		cin>>s;
		while(s[0]=='0') s.erase(s.begin());
		int a = stoi(s.substr(0,s.length()/2));
		int b = stoi(s.substr(s.length()/2));
		int c = stoi(s);
		if(a==0||b==0||c%(a*b)!=0) puts("No");
		else puts("Yes");
	}
	return 0;
}