PAT_A1132#Cut Integer

Source:

PAT A1132 Cut Integer (20 分)

Description:

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

Keys:

• 快乐模拟

Attention:

• 除法注意考虑分母为零的情况

Code:

 /*
 Data: 2019-05-24 21:36:30
 Problem: PAT_A1132#Cut Integer
 AC: 18:02

 题目大意：
 根据题意切割整数
 输入：
 第一行给出，测试数N；
 接下来N行，给出Z
 输出：
 是否能够找到符合条件的A和B，输出Yes or No
 */

 #include<cstdio>
 typedef long long LL;

 int main()
 {
 #ifdef    ONLINE_JUDGE
 #else
     freopen("Test.txt", "r", stdin);
 #endif

     LL m,n;
     scanf("%lld", &m);
     while(m--)
     {
         scanf("%lld", &n);
         LL exp=,a,b;
         while(exp*exp<n)
             exp *= ;
         a = n%exp;
         b = n/exp;
         if(a!= && b!= && n%(a*b)==)
             printf("Yes\n");
         else
             printf("No\n");
     }

     return ;
 }