Description

Fedya studies in a gymnasium. Fedya‘s maths hometask is to calculate the following expression:

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).

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.

Operation x mod y means taking remainder after division x by y.

Note to the first sample:

题意：看题目中的公式

解法：打表找规律

#include<bits/stdc++.h>using namespace std;class P{    public:    int n,m;};bool cmd(P x,P y){    return x.n<y.n;}int main(){    long long ans=0;    string s;    cin>>s;    if(s.length()==1)    {     ans+=s[s.length()-1]-‘0‘;    }    else    {        ans+=((s[s.length()-2]-‘0‘)*10+s[s.length()-1]-‘0‘);    }  //  cout<<ans<<endl;    if(ans%4)    {        cout<<"0"<<endl;    }    else    {        cout<<"4"<<endl;    }    return 0;}