Codeforces 456B Fedya and Maths 打表找规律
2016-01-12 15:12
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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
Input
Output
Input
Output
Hint
Operation x mod y means taking remainder after division x by y.
Note to the first sample:
代码
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
Input
4
Output
4
Input
124356983594583453458888889
Output
0
Hint
Operation x mod y means taking remainder after division x by y.
Note to the first sample:
#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=100+10; const int maxn = 1000000; char s[maxn]; int main() { scanf("%s",s); int len=strlen(s); int a=s[len-1]-'0'; int b=s[len-2]-'0'; int c=a+b*10; if(c%4==0) printf("4\n"); else printf("0\n"); return 0; }
代码
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