1132. Cut Integer (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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, 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 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:

3167334233312345678

Sample Output:

YesNoNo

实力不够，考试的时候第一题都没写出来。。。。

以前没遇到过浮点错误。。。  以为是什么新错误提示。。。

浮点错误您的程序运行时发生浮点错误，比如遇到了除以 0 的情况浮点错误您的程序运行时发生浮点错误，比如遇到了除以 0 的情况浮点错误您的程序运行时发生浮点错误，比如遇到了除以 0 的情况

考试的时候没有考虑 除数 为 0  的情况。。。草。。。

#include<stdio.h>#include<string.h>int main(){int n,i,j,sum1,sum2,sum3,len1,len2;char s[20];scanf("%d",&n);getchar();for(i=0;i<n;i++){gets(s);len1=strlen(s);len2=len1/2;sum1=0;sum2=0;sum3=0;for(j=0;j<len2;j++){sum1=sum1*10+s[j]-'0';}for(j=len2;j<len1;j++){sum2=sum2*10+s[j]-'0';}for(j=0;j<len1;j++){sum3=sum3*10+s[j]-'0';}if(sum1*sum2!=0){if(sum3%(sum1*sum2)==0){printf("Yes\n");}else{printf("No\n");}}else{printf("No\n");}}}