pat 1132 Cut Integer（20 分）

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 <2​31​​). 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

#include <map>
#include <set>
#include <queue>
#include <cmath>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstring>
#include <climits>
#include <iostream>
#include <algorithm>
#define wzf ((1 + sqrt(5.0)) / 2.0)
#define INF 0x3f3f3f3f
#define LL long long
using namespace std;

const int MAXN = 2e3 + 10;

int t, Z, A, B, len, temp;

char s[MAXN];

int my_pow(int x, int n)
{
    int ans = 1;
    while (n)
    {
        if (n & 1) ans *= x;
        x *= x;
        n >>= 1;
    }
    return ans;
}

int main()
{
    scanf("%d", &t);
    while (t --)
    {
        Z = A = B = 0L;
        scanf("%s", &s);

        len = strlen(s), temp = len >> 1;
        for (int i = 0, j = len - 1; i < len; ++ i, -- j)
            Z += (int)(s[i] - '0') * my_pow(10, j);
        for (int i = 0, j = temp - 1; i < temp; ++ i, -- j)
            A += (int)(s[i] - '0') * my_pow(10, j);
        for (int i = temp, j = temp - 1; i < len; ++ i, -- j)
            B += (int)(s[i] - '0') * my_pow(10, j);

        if ((A * B) != 0 && Z % (A * B) == 0) printf("Yes
");
        else printf("No
");
    }
    return 0;
}