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>
using namespace std;
int main() {
    int n, num;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%d", &num);
        string s = to_string(num);
        int len = s.length();
        int a = stoi(s.substr(0, len/2));
        int b = stoi(s.substr(len/2));
        if (a * b != 0 && num % (a * b) == 0)
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}

#include <iostream>

using namespace std;

int main() {

int n, num;

scanf("%d", &n);

for (int i = 0; i < n; i++) {

scanf("%d", &num);

string s = to_string(num);

int len = s.length();

int a = stoi(s.substr(0, len/2));

int b = stoi(s.substr(len/2));

if (a * b != 0 && num % (a * b) == 0)

printf("Yes\n");

else

printf("No\n");

}

return 0;

}

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