【PAT甲級】Cut Integer

2019 年 11 月 8 日
筆記

本文鏈接：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

解題思路：

輸入string型的數字str，數字的位數為K（題目保證了K是偶數），然後通過stoi()函數把string型的數字str強制轉換成int型的數字Z，通過substr()函數把數字str分成A和B，最後判斷數字Z能否整除（數字A×數字B）即可。需要注意的是：除數不能為0！A×B不能為0，我第一次提交的時候就有倆個測試點出現了"Float Point Exception",浮點錯誤。

AC程式碼：

#include <bits/stdc++.h>  using namespace std;    int main()  {      int N;      cin >> N;      while(N--)      {          string str;          cin >> str;          int K = str.length();  //數字Z的位數,題目保證了K是偶數          int Z = stoi(str);   //string型強制轉換成int型          int A = stoi(str.substr(0,K/2));   //A是數字Z的前K/2位數字          int B = stoi(str.substr(K/2));     //B是數字Z的後K/2位數字          if((A*B != 0) && (Z%(A*B) == 0))   //當A*B=0時,編譯器會報錯"Float Point Exception",浮點錯誤          {              cout << "Yes" << endl;          }          else          {              cout << "No" << endl;          }      }      return 0;  }