hdu 1061 Rightmost Digit

最新推荐文章于 2019-04-18 20:08:35 发布

原创最新推荐文章于 2019-04-18 20:08:35 发布 · 182 阅读

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本文介绍了一种通过快速幂算法解决特定数学问题的方法：给定一个正整数N，如何高效地计算出N的N次方结果的最右边一位数字。通过分析和给出示例代码，展示了算法的具体实现。

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Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 57040 Accepted Submission(s): 21561

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.

就是求n的n次方，快速幂就可以了。

#include<iostream>
using namespace std;
int main(){
    int T;
    cin>>T;
    while(T--)
    {
        long long int n;
        cin>>n;
        long long int mi=n;
        long long int ans=n;
        n=n%10;
        ans=n%10;
        mi--;
        while(mi>=1)
        {
            if(mi%2==1)
            {
                mi--;
                ans=n*ans%10;
            }
            if(mi%2==0)
            {
                mi=mi/2;
                n=n*n%10;
            }
        }
        ans=ans%10;
        cout<<ans<<endl;
    }
    return 0;
}