codeforces 630A Again Twenty Five!

面试挑战：高效计算5的幂次

最新推荐文章于 2021-01-29 13:34:38 发布

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本文介绍了一个简单的面试题目，要求计算5的任意大整数次幂的最后两位数字。通过观察规律，发现无论指数大小，结果始终为25。文章提供了一段简洁的C++代码实现。

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The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of n and get last two digits of the number. Yes, of course, n can be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions."

Could you pass the interview in the machine vision company in IT City?

Input

The only line of the input contains a single integer n (2 ≤ n ≤ 2·1018) — the power in which you need to raise number 5.

Output

Output the last two digits of 5n without spaces between them.

        Examples
      
          input
        
2

          output
        
25

这个题千万别一直一直*5，这样会超时，n从2开始取，就是25，然后发现25*5末两位还是25，所以结果直接输出25。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
typedef long long ll;
using namespace std;
const ll mod=100000000;
int main()
{
    ll n;
    cin>>n;
    cout<<25<<endl;
    return 0;
}