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

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原文链接：http://www.cnblogs.com/qscqesze/p/3900657.html

本文介绍了一个关于计算特定数学表达式的编程挑战。任务是计算(1^n + 2^n + 3^n + 4^n) mod 5对于给定的大整数n的值。文章提供了一段C++代码示例，展示如何通过处理输入数字的最后两位来简化计算过程。

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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 ≤ 10^10⁵). 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:

$\text{[math]}$

#include <cstdio>
#include <cstring>
using namespace std;
char str[100000+10];//注意拿数组进行储存
int main()
{
    scanf("%s",str);
    int len = strlen(str);
    int sum = (str[len-2]-'0')*10+str[len-1]-'0';
    if(sum%4==0)puts("4");//做题前先打表
    else puts("0");
    return 0;
}