A - Again Twenty Five!【思维题】

最新推荐文章于 2021-12-08 09:35:52 发布

原创最新推荐文章于 2021-12-08 09:35:52 发布 · 451 阅读

0 ·

CC 4.0 BY-SA版权

本文介绍了一个简单的编程面试题目，要求计算5的N次方的最后两位数字，并提供了一段简洁的C++代码示例。通过这个题目，考察应聘者的基本编程能力和对数学规律的理解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

A - Again Twenty Five!

Time Limit:500MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status

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, ncan 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

Output

#include <cstdio>  
#include <cstring>  
#include <cmath>  
#include <cstdlib>  
#include <algorithm>  
#include <queue>  
#include <stack>  
#include <map>  
#include <set>  
#include <vector>  
#include <string>  
#define INF 0x3f3f3f3f  
#define eps 1e-8   
#define Si(a) scanf("%d", &a)  
#define Sl(a) scanf("%lld", &a)  
#define Sf(a) scanf("%lf", &a)  
#define Ss(a) scanf("%s", a)  
#define Pi(a) printf("%d\n", (a))  
#define Pf(a) printf("%.2lf\n", (a))  
#define Pl(a) printf("%lld\n", (a))  
#define Ps(a) printf("%s\n", (a))  
#define Wi(a) while((a)--)  
#define cle(a, b) memset(a, (b), sizeof(a))  
#define MOD 1000000007  
#define LL long long  
#define PI acos(-1.0) 
using namespace std;
int main()
{
	__int64 n;
	while(scanf("%I64d", &n)==1)
	{
		printf("25\n");
	}
	return 0;
}