【PAT甲级】Cut Integer

2019 年 11 月 8 日
筆記

本文链接：https://blog.csdn.net/weixin_42449444/article/details/89449189

Problem 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 <231). 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

解题思路：

输入string型的数字str，数字的位数为K（题目保证了K是偶数），然后通过stoi()函数把string型的数字str强制转换成int型的数字Z，通过substr()函数把数字str分成A和B，最后判断数字Z能否整除（数字A×数字B）即可。需要注意的是：除数不能为0！A×B不能为0，我第一次提交的时候就有俩个测试点出现了"Float Point Exception",浮点错误。

AC代码：

#include <bits/stdc++.h>  using namespace std;    int main()  {      int N;      cin >> N;      while(N--)      {          string str;          cin >> str;          int K = str.length();  //数字Z的位数,题目保证了K是偶数          int Z = stoi(str);   //string型强制转换成int型          int A = stoi(str.substr(0,K/2));   //A是数字Z的前K/2位数字          int B = stoi(str.substr(K/2));     //B是数字Z的后K/2位数字          if((A*B != 0) && (Z%(A*B) == 0))   //当A*B=0时,编译器会报错"Float Point Exception",浮点错误          {              cout << "Yes" << endl;          }          else          {              cout << "No" << endl;          }      }      return 0;  }