快速幂 HDU 1061 Rightmost Digit *

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 57430    Accepted Submission(s): 21736

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.

 

Author

Ignatius.L

快速幂的运用，每次相乘时余十就行

注意碰到次方很多的时候可以考虑使用快速幂了

这里贴一份讲快速幂的博客  http://blog.csdn.net/hikean/article/details/9749391

#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
#define ll long long
#define mod 1000000007
using namespace std;
int main()
{
    int T;
    cin >> T;
    while(T--){
        int n;
        cin >> n;
        int ans = ,mul = n,num = n;
        mul = mul % ;
        while(num){
            if(num%==){
                ans = (ans*mul)%;
            }
            mul = (mul*mul)%;
            num /= ;
        }
        cout << ans% << endl;
    }
    return ;
}