首页 > 代码库 > Codeforces Round #260 (Div. 2) B. Fedya and Maths（循环节）

Codeforces Round #260 (Div. 2) B. Fedya and Maths（循环节）

2024-07-16 16:00:45   224人阅读

题目链接：http://codeforces.com/problemset/problem/456/B


B. Fedya and Maths
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Fedya studies in a gymnasium. Fedya‘s maths hometask is to calculate the following expression:

(1n?+?2n?+?3n?+?4n) mod 5

for given value of n. Fedya managed to complete the task. Can you? Note that given number n can be extremely large (e.g. it can exceed any integer type of your programming language).

Input

The single line contains a single integer n (0?≤?n?≤?10105). The number doesn‘t contain any leading zeroes.

Output

Print the value of the expression without leading zeros.

Sample test(s)
input
4
output
4
input
124356983594583453458888889
output
0
Note

Operation x mod y means taking remainder after division x by y.

Note to the first sample:


循环节为4；

代码如下：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;
__int64 sum;
char s[1000017];
int main()
{
    while(~scanf("%s",s))
    {
        int len = strlen(s);
        int i, n;
        int tt = 0;
        if(len == 1)
            tt = s[0]-'0';
        else
        {
            tt = s[len-1]-'0'+(s[len-2]-'0')*10;
        }
        n = tt ;
        if(n > 4)
        {
            n %= 4;
            if(n == 0)
                n = 4;
        }
        sum = 0;
        int j;
        for(i = 1; i <= 4; i++)
        {
            __int64 t = 1;
            for(j = 1; j <= n;j++)
            {
                t*=i;
                
            }
            t%=5;
            sum+=t;
        }
        printf("%I64d\n",sum%5);
    }
    return 0;
}


联系
我们