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

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

题意：我是找规律得到的

#include <stdio.h>
int x;
int main()
{scanf("%d",&x);
putchar(x%4?48:52);
return 0;
}

这是大神的代码

Note

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

Note to the first sample: