Codeforces Round #260 (Div. 2) B. Fedya and Maths（迴圈節）

標籤：codeforces

題目連結：http://codeforces.com/problemset/problem/456/B

B. Fedya and Mathstime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard 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:

迴圈節為4；

代碼如下：

#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;}