Codeforces--630A--Again Twenty Five! (水题)
2024-08-31 03:28:49
Time Limit: 500MS | Memory Limit: 65536KB | 64bit IO Format: %I64d & %I64u |
Description
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.
Sample Input
Input
2
Output
25
Sample Output
Hint
Source
输出5的n次方的后两位,n>=2,所以嘛,只可能是25
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
__int64 n;
while(cin>>n)
cout<<25<<endl;
return 0;
}
最新文章
- PHP中常见的提示对照表
- Linux中读写权限
- protobuf学习(1)-ubuntu14.04下protobuf2.6安装
- android 图片压缩
- 雾里看花般的迷茫--货运APP
- Js 的 this 是什么
- 获取设置唯一的UDID的值
- hibernate中设置BigDeCimal的精度
- ZOJ 3911 Prime Query ZOJ Monthly, October 2015 - I
- 各种数据库连接代码的测试类(java)
- NYOJ-102 次方求模 AC 分类: NYOJ 2014-02-06 18:53 184人阅读 评论(0) 收藏
- C#中运用事件实现异步调用
- OpenCL memory object 之 传输优化
- Oracle EBS-SQL (PO-6):检查订单接收总数.sql
- permutation求全排列
- Day3 - Linux系统安装_Centos6.9
- Centos7破解密码的两种方法--技术流ken
- LeetCode算法题-Plus One(Java实现)
- linux驱动编写之阻塞与非阻塞
- java读取txt文件的2中方法---并将内容(每一行以固定的字符分割切成2段)存到map中去