HDU - 1061 -Rightmost Digit (幂取模，白书)

Rightmost Digit

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.

白书上的函数 是这么写的：

int pow_mid (int a,int n,int m)
{
    if (n == 0) return 1;

    int x = pow_mid (a,n/2,m);
    long long ans = (long long ) x * x % m;
    if (n % 2 == 1)
        ans = ans * a % m;
    return (int )ans;
}

采用的是分治的思想(跟二分查找很类似，每次规模减小近一半)，每次幂数/2，假如a^29 = a^14^2 *a ,a^14=a^7^2, a^7=a^3^2*a, a^3=a^2*a, ..总共进行7次乘法。因此时间复杂度为O(logN)，比O(N)好多了。。。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
using namespace std;

int pow_mid (int a,int n,int m)
{
    if (n == 0)
        return 1;
    int x = pow_mid (a,n/2,m);
    long long ans = (long long ) x * x % m;
    if (n % 2 == 1)
        ans = ans * a % m;
    return (int )ans;
}

int main ()
{
    int a,n,m;
    while (~scanf ("%d",&n))
    {
        while ( n-- )
        {
            scanf ("%d",&a);
            int s = pow_mid (a,a,10);
            printf ("%d\n",s);
        }
    }
    return 0;
}