HDU-1061-Rightmost Digit，快速幂水过！~~

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

http://acm.hdu.edu.cn/showproblem.php?pid=1061

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.

看到n的数据范围可能会被吓到，但用快速幂算法的话简直就是滑水，，这里可以方便一点，因为求N^N次方的最后一位，随便举几个数不难发现其实就是(（N%10）^N )%10;所以关键就在这个快速幂算法了；

直接上代码：

#include<bits/stdc++.h>
using namespace std;
int fast(int x)
{
    int s=1,xx=x;
    x%=10;
    while(xx)//快速幂取余核心问题；
    {
        if(xx&1)
            s=(s*x)%10;
        x=x*x%10;
        xx=xx>>1;
    }
    return s;
}
int  main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        if(n%10==0)
            printf("0\n");
        else
        printf("%d\n",fast(n));
    }
    return 0;
}