[HDOJ1061]Rightmost Digit

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38642    Accepted Submission(s): 14558

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.

 

 

快速幂的基本应用，特别要注意数据范围，应该使用__int64否则会超时，代码如下：

 

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;

int pow1(int x, int n)
{
    __int64 ans = 1, t = n;
    while(n)
    {
        if(n & 1)
        {
            ans = (ans * t) % 10;
        }
        t = t * t % 10;
        n >>= 1;
    }
    return ans;
}
int main()
{
    __int64 n, m;
    scanf("%d", &n);
    for(int i = 0; i < n; i++)
    {
        scanf("%d", &m);
        printf("%d
", pow1(m, m) % 10);
    }
}