1132. Cut Integer (20)-PAT甲级真题

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:
3
167334
2333
12345678
Sample Output:
Yes
No
No

题目大意：给一个偶数个位的正整数num，把它从中间分成左右两个整数a、b，问num能不能被a和b的乘积整除，能的话输出yes，不能的话输出no

分析：要注意a*b如果为0的时候不能取余，否则会浮点错误～
直接用int保存num的值，计算出num的长度len，则令d = pow(10, len / 2)时，num取余d能得到后半部分的整数，num除以d能得到前半部分的整数，计算num % (a*b)是否等于0就可以得知是否可以被整除～

#include <iostream>#include <cmath>using namespace std;int main() {    int n, num;    scanf("%d", &n);    for (int i = 0; i < n; i++) {        scanf("%d", &num);        int len = 0, tempnum = num;        while (tempnum != 0) {            tempnum = tempnum / 10;            len++;        }        int d = pow(10, len / 2);        int a = num % d, b = num / d;        if (a * b != 0 && num % (a * b) == 0)            printf("Yes\n");        else            printf("No\n");    }    return 0;}