hdu1060 求n^n第一位数数学java

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本文介绍了一种计算正整数N的N次幂最左侧数字的方法，并提供了一个Java程序示例。该程序利用对数特性计算结果，适用于1到1,000,000,000范围内的N。

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http://acm.hdu.edu.cn/showproblem.php?pid=1060

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21639 Accepted Submission(s): 8356

Problem Description

Given a positive integer N, you should output the leftmost 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 leftmost digit of N^N.

Sample Input

2 3 4

Sample Output

2 2

Hint

In the first case, 3 * 3 * 3 = 27, so the leftmost digit is 2. In the second case, 4 * 4 * 4 * 4 = 256, so the leftmost digit is 2.

import java.util.Scanner;


public class Main {
	@SuppressWarnings("resource")
	public static void main(String[] args) {
	Scanner sc = new Scanner(System.in);
	while (sc.hasNext()) {
		double n = sc.nextDouble();
		double tmp = n * Math.log10((double) (n));
		//pow=n^n 
		//log10(pow)=n*log10(n) 
		//pow=10^(n*log10(n))
		//求出tmp=n*log10(n)
		//10^n第一位都是1，所以只和tmp的小数有关；
		double res = tmp - Math.floor(tmp);
		System.out.println((int) Math.pow(10.0, res));
		}
	}
}