pat 1132

1132 Cut Integer (20分)

 

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 <). 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：把一个偶数位的数z截成等长度的两段数a和b，问这个数z是否整除a*b
思路：char数组读入数字，然后截取，转换成整数。
注意：有时候a*b=0,要特判一下，否则产生浮点错误（测试点2、3）

代码如下：

#include<cstdio>
#include<cstring>
void fun(char num[]){
    int len=strlen(num);
    long long a=0;
    long long b=0;
    long long z=0;
    for(int i=0;i<len/2;i++){
        a=a*10+num[i]-'0';
    }
    for(int i=len/2;i<len;i++){
        b=b*10+num[i]-'0';
    }
    for(int i=0;i<len;i++)
        z=z*10+num[i]-'0';
    if(a*b==0)
        printf("No");
    else{
        if(z%(a*b)==0)
            printf("Yes");
        else
            printf("No");
    }    

}
int main(){

    int n;
    scanf("%d",&n);
    char num[15];
    for(int i=0;i<n;i++){
        scanf("%s",num);
        fun(num);
        printf("
");
    }
    return 0;
}