HDU 1063 Exponentiation（高精度求幂）

最新推荐文章于 2019-08-20 19:08:51 发布

Spectacules

最新推荐文章于 2019-08-20 19:08:51 发布

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分类专栏： ACM__大数&&高精度

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本文介绍如何使用Java编写程序来精确计算任意实数R（0.0<R<99.999）和整数n（0<n<=25）的幂次方，并通过实例展示解决方案。

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Exponentiation

Problem Description

Problems involving the computation of exact values of very large magnitude and precision are common. For example, the computation of the national debt is a taxing experience for many computer systems.

This problem requires that you write a program to compute the exact value of Rⁿ where R is a real number ( 0.0 < R < 99.999 ) and n is an integer such that 0 < n <= 25.

Input

The input will consist of a set of pairs of values for R and n. The R value will occupy columns 1 through 6, and the n value will be in columns 8 and 9.

Output

The output will consist of one line for each line of input giving the exact value of R^n. Leading zeros should be suppressed in the output. Insignificant trailing zeros must not be printed. Don't print the decimal point if the result is an integer.

Sample Input

Sample Output

548815620517731830194541.899025343415715973535967221869852721
.00000005148554641076956121994511276767154838481760200726351203835429763013462401
43992025569.928573701266488041146654993318703707511666295476720493953024
29448126.764121021618164430206909037173276672
90429072743629540498.107596019456651774561044010001
1.126825030131969720661201

【思路分析】
直接套的Java高精度模板。

代码如下：

import java.math.BigDecimal;
import java.util.Scanner;

public class Main{

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		BigDecimal num;
		BigDecimal res = new BigDecimal("1");
		int n;
		while(in.hasNext()){
			num = in.nextBigDecimal();
			n = in.nextInt();
			res = num.pow(n);
			res = res.stripTrailingZeros();//除去后导零
			String out = res.toPlainString();
			//返回不带指数字段的此 BigDecimal 的字符串表示形式，toString()是以科学计数法输出
			if(out.charAt(0) == '0')
				out = out.substring(1);
			System.out.println(out);
		}
		

	}

}