HDU——1061Rightmost Digit（高次方，找规律）

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 43847    Accepted Submission(s): 16487

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.

水题一道，1、5、0、6结尾的次方均为本身，其余可以看作4次一循环。

#include<iostream>
#include<string>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
ll power(const ll a,const ll n)
{
    ll sum=1;
    for (int i=1; i<=n; i++)
    {
        sum=sum*a;
    }
    return sum;
}
int main(void)
{
    ll n,t,ans;
    int tt;
    cin>>tt;
    while (tt--)
    {    
        cin>>n;
        t=n%10;
        if(t==1||t==5||t==0||t==6)
        {
            cout<<t<<endl;
            continue;
        }
        else
        {        
            ans=power(t,n%4+4)%10;
            cout<<ans<<endl;
            continue;
        }        
    }
    return 0;
}