codeforces 456B. Fedya and Maths（大数取模/大数的幂）

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.

    Examples 
  
      input 
    
4

      output 
    
4

      input 
    
124356983594583453458888889

      output 
    
0

Note

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

Note to the first sample:

$\text{[math]}$

题目不难，就是大数模4是否等于0，不过看官方题解的一种做法挺不错的，直接用类似快速幂的方法求大数的幂，学习一下。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <stack>
#include <bitset>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <algorithm>
#define FOP freopen("data.txt","r",stdin)
#define FOP2 freopen("data1.txt","w",stdout)
#define inf_LL 4223372036854775807
#define inf 0x3f3f3f3f
#define maxn 1000010
#define mod 1000000007
#define PI acos(-1.0)
#define LL long long
using namespace std;

const int P = 5;

char s[maxn];
int a[maxn], n;

void get_num(char *s)
{
    n = 0;
    int len = strlen(s);
    int i;
    for(i = len - 3; i >= 0; i -= 3)
    {
        a[n++] = (s[i] - '0') * 100 + (s[i+1] - '0') * 10 + s[i+2] - '0';
    }
    i += 2;
    for(int j = 0; j <= i; j++)
    {
        a[n] = a[n] * 10 + s[j] - '0';
        if(j == i) n++;
    }
}

int powmod(int x, int n, int P)
{
    x %= P;
    int res = 1 % P;
    while(n)
    {
        if (n & 1)
        {
            res = (res * x) % P;
            n--;
        }
        else
        {
            x = (x * x) % P;
            n >>= 1;
        }
    }
    return res;
}

int myPow(int x, int *a, int P)
{
    x %= P;
    int res = 1 % P;
    int now = x;
    for (int i = 0; i < n; i++)
    {
        res = (res * powmod(now, a[i], P)) % P;
        now = powmod(now, 1000, P);
    }
    return res;
}

int main()
{
    scanf("%s", s);

    get_num(s);

    int res = myPow(1, a, P) + myPow(2, a, P) + myPow(3, a, P) + myPow(4, a, P);
    res %= P;

    printf("%d\n", res);

    return 0;
}

codeforces 456B. Fedya and Maths（大数取模/大数的幂）

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