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 + ;

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

 char s[MAXN];

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

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

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

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