Codeforces Round #260 (Div. 2) B. Fedya and Maths（循环节）_html/css_WEB-ITnose

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

output

input

124356983594583453458888889

output

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   <br>  <br>  <p></p> </algorithm></iostream></cmath></cstring></cstdio>