HDU1061---Rightmost Digit 规律或快速幂

Rightmost Digit
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6515    Accepted Submission(s): 2454

Problem Description
Given a positive integer N, you should output the most right digit of N^N.
 

Input
The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains a single positive integer N(1<=N<=1,000,000,000).
 

Output
For each test case, you should output the rightmost digit of N^N.
 

Sample Input
2
3
4
 

Sample Output
7
6

Hint
In the first case, 3 * 3 * 3 = 27, so the rightmost digit is 7. In the second case, 4 * 4 * 4 * 4 = 256, so the rightmost digit is 6.
 

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
#include <climits>
using namespace std;
using namespace std;
typedef long long ll;
const int MAXN=50000+5;//最大元素个数
int a[10][4]={{0},{1},{6,2,4,8},{1,3,9,7},{6,4},{5},{6},{1,7,9,3},{6,8,4,2},{1,9}};
ll n;//元素个数
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
     scanf("%lld",&n);
     int d=n%10;
     if(d==0||d==1||d==5||d==6)
       printf("%d\n",d);
     else if(d==4||d==9) 
       printf("%d\n",a[d][n%2]);
     else if(d==2||d==3||d==7||d==8)
       printf("%d\n",a[d][n%4]);
    }
    return 0;
}

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
#include <climits>
using namespace std;
using namespace std;
typedef long long ll;
const int MAXN=50000+5;//最大元素个数
int a[10][4]={{0},{1},{6,2,4,8},{1,3,9,7},{6,4},{5},{6},{1,7,9,3},{6,8,4,2},{1,9}};
ll n;//元素个数
ll quick_qow(ll a,ll b,ll p)///计算a^b %p
{
    ll ans=1%p;///ans不能为1，b为0，p为1的情况不成立
    for(;b;b>>=1)
    {
        if(b&1) 
            ans=(ans*a)%p ;
        a=a*a%p ;
    }
    return ans;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
     scanf("%lld",&n);
     printf("%lld\n",quick_qow(n,n,10));
    }
    return 0;
}