PAT Advanced 1132 Cut Integer (20) [数学问题-简单数学]

时间:2021-05-13 17:47:00

题目

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, afer 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<=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

题目分析

已知偶数K位正整数N,从中间分为两部分其长度相等的A,B两数,若N能被(A+B)整除,就满足条件,输出Yes,否则输出No

解题思路

  1. 字符串截取左右部分
  2. N%(A+B)==0 -- Yes

易错点

  1. A*B有可能为0,如:N=10

Code

#include <iostream>
#include <string>
using namespace std;
int main(int argc,char * argv[]) {
int n,z;
scanf("%d",&n);
for(int i=0; i<n; i++) {
scanf("%d",&z);
string s= to_string(z);
int l = stoi(s.substr(0,s.length()/2));
int r = stoi(s.substr(s.length()/2,s.length()/2));
if(l*r!=0&&z%(l*r)==0) //l*r可能为0,比如10,测试点2,3
printf("Yes\n");
else
printf("No\n");
}
return 0;
}