Description
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 Input
4
4
124356983594583453458888889
0
Hint
Operation x mod y means taking remainder after division x by y.
Note to the first sample:
题意: 给你N 让你求 (1n + 2n + 3n + 4n) mod 5
题解: 打个表 发现了规律
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <queue>
#include<vector>
#include <map>
using namespace std ;
typedef long long ll; const int N=+;
const int maxn = ; char s[maxn];
int main()
{
scanf("%s",s);
int len=strlen(s);
int a=s[len-]-'';
int b=s[len-]-'';
int c=a+b*;
if(c%==)
printf("4\n");
else
printf("0\n");
return ;
}
代码