PAT Advanced 1132 Cut Integer (20) [数学问题-简单数学]

题目

Cutting an integer means to cut a K digits long integer Z into two integers of (K/2) digits long integers A and B. For example, afer 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 x 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

题目分析

已知偶数K位正整数N，从中间分为两部分其长度相等的A,B两数，若N能被(A+B)整除，就满足条件，输出Yes，否则输出No

解题思路

1. 字符串截取左右部分
2. N%(A+B)==0 -- Yes

易错点

1. A*B有可能为0，如：N=10

Code

#include <iostream>
#include <string>
using namespace std;
int main(int argc,char * argv[]) {
	int n,z;
	scanf("%d",&n);
	for(int i=0; i<n; i++) {
		scanf("%d",&z);
		string s= to_string(z);
		int l = stoi(s.substr(0,s.length()/2));
		int r = stoi(s.substr(s.length()/2,s.length()/2));
		if(l*r!=0&&z%(l*r)==0) //l*r可能为0，比如10，测试点2,3
			printf("Yes\n");
		else
			printf("No\n");
	}
	return 0;
}