hdu1061 快速幂取余 java

最新推荐文章于 2021-05-13 15:31:54 发布

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最新推荐文章于 2021-05-13 15:31:54 发布

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本文介绍了一个算法问题的解决方案，即如何通过快速幂取余的方法找到N^N的最后一位数字。该方法适用于大整数运算，通过模10操作，有效地解决了求解大整数幂次方最后一位数的问题。

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http://acm.hdu.edu.cn/showproblem.php?pid=1061

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 74784 Accepted Submission(s): 27577

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^n的最后一位数。

题解：快速幂取余 mod = 10

import java.util.Scanner;

public class Main {

    @SuppressWarnings("resource")
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        while (sc.hasNext()) {
            int T = sc.nextInt();
            for (int i = 0; i < T; i++) {
                int n = sc.nextInt();
                System.out.println(mod(n, n, 10));

            }

        }

    }

    public static int mod(int a, int b, int p) {
        int ans=1;
             a%=p;
             while (b!=0)
             {
                  if ((b&1)!=0) ans=ans*a%p;
               b>>=1;
                 a=a*a%p;
        }
             return ans;
    }

}