Fedya studies in a gymnasium. Fedya's maths hometask is to calculate the following expression:
(1n + 2n + 3n + 4n) mod 5for 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).
The single line contains a single integer n (0 ≤ n ≤ 10105). The number doesn't contain any leading zeroes.
Print the value of the expression without leading zeros.
4
4
124356983594583453458888889
0
Operation x mod y means taking remainder after division x by y.
Note to the first sample:
易发现本题的循环节 [0,0,0,4],[0,0,0,4],...
#include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<functional> #include<iostream> #include<cmath> #include<cctype> #include<ctime> using namespace std; #define For(i,n) for(int i=1;i<=n;i++) #define Fork(i,k,n) for(int i=k;i<=n;i++) #define Rep(i,n) for(int i=0;i<n;i++) #define ForD(i,n) for(int i=n;i;i--) #define RepD(i,n) for(int i=n;i>=0;i--) #define Forp(x) for(int p=pre[x];p;p=next[p]) #define Lson (x<<1) #define Rson ((x<<1)+1) #define MEM(a) memset(a,0,sizeof(a)); #define MEMI(a) memset(a,127,sizeof(a)); #define MEMi(a) memset(a,128,sizeof(a)); #define INF (2139062143) #define F (100000007) #define MAXN (100000+10) long long mul(long long a,long long b){return (a*b)%F;} long long add(long long a,long long b){return (a+b)%F;} long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;} typedef long long ll; char c[MAXN]; int main() { // freopen("b.in","r",stdin); // freopen(".out","w",stdout); scanf("%s",c+1); int n=strlen(c+1); int t; if (n==1) t=c[1]-'0'; else t=c[n]-'0'+(c[n-1]-'0')*10; if (t%4==0) cout<<"4\n"; else cout<<"0\n"; return 0; }