【codeforces 742A】Arpa’s hard exam and Mehrdad’s naive cheat

time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
There exists an island called Arpa’s land, some beautiful girls live there, as ugly ones do.

Mehrdad wants to become minister of Arpa’s land. Arpa has prepared an exam. Exam has only one question, given n, print the last digit of 1378n.

Mehrdad has become quite confused and wants you to help him. Please help, although it’s a naive cheat.

Input
The single line of input contains one integer n (0  ≤  n  ≤  109).

Output
Print single integer — the last digit of 1378n.

Examples
input
1
output
8
input
2
output
4
Note
In the first example, last digit of 13781 = 1378 is 8.

In the second example, last digit of 13782 = 1378·1378 = 1898884 is 4.

【题目链接】:http://codeforces.com/problemset/problem/742/A

【题解】

快速幂取模;
把那个数用快速幂乘一下；乘的时候取个位数就好;

【完整代码】

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <stack>
#include <string>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

void rel(LL &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t) && t!='-') t = getchar();
    LL sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}

void rei(int &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t)&&t!='-') t = getchar();
    int sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}

//const int MAXN = x;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);

LL n;
LL ans = 1;

void f(LL x)
{
    if (x==0)
        return;
    f(x>>1);
    ans = (ans * ans)%10;
    if (x&1)
        ans=(ans*1378)%10;
}

int main()
{
    //freopen("F:\rush.txt","r",stdin);
    cin>>n;
    if (n==0)
        puts("1");
    else
    {
        f(n);
        cout << ans << endl;
    }
    return 0;
}