codeforces #260 B. Fedya and Maths（水）

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本文介绍了一种快速计算特定形式数学表达式的取模运算的方法。对于表达式(1^n + 2^n + 3^n + 4^n) mod 5，通过判断n是否能被4整除来直接得出结果，适用于n值极大情况。

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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 numbern 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 divisionx by y.

Note to the first sample:

$\text{[math]}$

水题，只要判断输入的n是否能被4整除即可（如果一个数可以被4整除，则只要这个数的最后两位被4整除即可，证明，请看数论）；

代码：

#include<bits/stdc++.h>
#define MAX 100010
char n[MAX];
int main()
{
    int num;
    while(~scanf("%s",n))
    {
        int len=strlen(n);
        num=len>=2?10*(n[len-2]-'0')+(n[len-1]-'0'):n[0]-'0';
        if(num%4==0)
            printf("4\n");
        else
            printf("0\n");
        memset(n,'\0',sizeof(n));
    }
    return 0;
}