PAT 1132 Cut Integer (20)

最新推荐文章于 2023-11-22 15:13:03 发布

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本文介绍了一种算法，用于判断一个整数是否能通过切割成两个子整数，并且该整数能被这两个子整数的乘积整除。通过输入一系列整数，程序能够输出每个整数是否符合这一条件。

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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<=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:

Sample Output:

Yes
No
No

思路：注意被除数为0的情况

代码：

#include <iostream>
#include <string>
#include <cmath>
#include <cstdio>
using namespace std;
int main()
{
  int n;
  cin>>n;
  while(n--)
  {
    int num;
    cin>>num;
    string tmp;
    int temp = num;
    while(temp != 0)
    {
      tmp = tmp + static_cast<char>(temp % 10 + '0');
      temp /= 10;
    }
    int len = tmp.length();
    int divide = pow(10,len/2);
    int num1 = num / divide;
    int num2 = num % divide;
    if(num1 != 0 && num2 != 0 && num % (num1 * num2) == 0)
      printf("Yes\n");
    else
      printf("No\n");
  }
  return 0;
}