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;
}