PAT 甲级 1132 Cut Integer

https://pintia.cn/problem-sets/994805342720868352/problems/994805347145859072

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代码：

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1e5 + 10;
char s[maxn];

void itoa(int x) {
    if(x == 0) {
        s[0] = ‘0‘;
        s[1] = 0;
        return ;
    }
    stack<int> st;
    while(x) {
        st.push(x % 10);
        x = x / 10;
    }
    int sz = 0;
    while(!st.empty()) {
        s[sz++] = (char)(st.top() + ‘0‘);
        s[sz] = 0;
        st.pop();
    }
}

int main() {
    int T;
    scanf("%d", &T);
    while(T --) {
        int n;
        int num1 = 0, num2 = 0;
        scanf("%d", &n);
        itoa(n);
        //printf("%s\n", s);
        int len = strlen(s);
        for(int i = 0; i < len / 2; i ++)
            num1 = (s[i] - ‘0‘) + num1 * 10;
        for(int i = len / 2; i < len; i ++)
            num2 = (s[i] - ‘0‘) + num2 * 10;

        if(num1 * num2 == 0)
            printf("No\n");
        else {
            if(n % (num1 * num2) == 0)
                printf("Yes\n");
            else
                printf("No\n");
        }
        //printf("%d %d", num1, num2);
    }
    return 0;
}

原文地址：https://www.cnblogs.com/zlrrrr/p/9420583.html