hdu1061 Rightmost Digit 快速幂的简单应用

**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.
**

题意：求n的n次方最右边的数，也就是最后一个数字。
思路：由于数值n的范围很大，不能用寻常套路写这道题，这道题的正解是快速幂取模运算的应用，基于公式：
（a*b）%c=(a%c*b%c)%c
（话说第一次WA就是因为取模出错，导致范围超了）

#include<stdio.h>int f(int m,int n){    if( n == 1)        return m;    long long s = f(m,n/2)%10;    if( n%2 == 1)        return (s*s*m)%10;    else        return (s*s)%10;}int main(){    int n;    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        printf("%d\n",f(n,n));    }    return 0; }