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

234

 

 

Sample Output

76
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;}