杭电 Rightmost Digit

最新推荐文章于 2020-01-29 10:42:41 发布

A_Try_again

最新推荐文章于 2020-01-29 10:42:41 发布

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本文链接：https://blog.youkuaiyun.com/A_Try_again/article/details/81347704

本文介绍了一种通过快速幂算法来求解正整数N的N次方结果中最右侧数字的方法。对于每个输入的正整数N(1≤N≤1,000,000,000)，程序输出N^N的最右侧数字。通过示例展示了当N为3时结果为7，N为4时结果为6的过程。

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.

Author

Ignatius.L

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

long long pow(long long a)//快速幂思想
{
    long long result=1,flag=a,n=a;
    while(n!=0)
    {
        if(n&1)
            result=result*flag%10;//为了得到最后一位
        flag=flag*flag%10;
        n=n>>1;
    }
    return result;
}

int main()
{
    long long n,m;
    cin>>n;
    while(n--)
    {
        cin>>m;
        cout<<pow(m)<<endl;
    }
}