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PAT Advanced 1132 Cut Integer (20分)

时间:2020-01-27 17:11:56      阅读:80      评论:0      收藏:0      [点我收藏+]

标签:case   cout   name   ios   cli   dig   oar   for   boa   

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

这道题题意是前一半乘后一半,看看最终数能不能被原数字整除。

我们仅需要进行用substr进行分解就可以。

如果出现浮点错误,就是表示除以了0,我们要考虑后半部分不为0的情况

#include <iostream>
#include <string>
using namespace std;
int main(){
    int N;
    string tmp;
    cin >> N;
    while(N--) {
        cin >> tmp;
        int num = stoi(tmp);
        int a = stoi(tmp.substr(0, tmp.size()/2));
        int b = stoi(tmp.substr(tmp.size()/2));
        if(a * b != 0 && num % (a * b) == 0) cout << "Yes" << endl;
        else cout << "No" << endl;
    }
    return 0;
}

PAT Advanced 1132 Cut Integer (20分)

标签:case   cout   name   ios   cli   dig   oar   for   boa   

原文地址:https://www.cnblogs.com/littlepage/p/12236338.html

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