(甲）1132 Cut Integer (20 分)

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题目：

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 <2​31​​). 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：

给你一个数字N，下面跟着N个数字Z（Z的位数为偶数），让你把数字从中间分成两个数字A和B，如果 Z%(A*B)==0,就输出Yes，否则输出No；

注意：当A*B==0时，就输出No，不能直接取余，会发生浮点错误。

代码如下：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;int n;
char a[1010];
int x,y,s;int main()
{scanf("%d",&n);while(n--){scanf("%s",&a);int l=strlen(a);x=y=s=0;for(int i=0; i<(l/2); i++){x=x*10+a[i]-'0';s=s*10+a[i]-'0';}for(int i=l/2; i<l; i++){y=y*10+a[i]-'0';s=s*10+a[i]-'0';}if(x*y!=0)//注意，当X*Y==0时，计算s%(x*y)会发生浮点错误；{if(s%(x*y)==0)printf("Yes\n");elseprintf("No\n");}elseprintf("No\n");}return 0;
}

 

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