(hdu 2.3.2)Exponentiation(大数：求R的n次幂)-优快云博客

本文介绍如何解决大数运算与指数计算问题，并附上求解R的n次幂的代码示例。适用于编程爱好者及初学者。

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题目：

Exponentiation

Time Limit: 1000/500 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 1938 Accepted Submission(s): 527

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

Source

East Central North America 1988

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题目分析：

简单题。大数的基本使用。求R的n次幂。

代码如下：

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


public class Main {

	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		
		while(scanner.hasNext()){
			BigDecimal r = scanner.nextBigDecimal();//对于求R^n的这种大数题。r要以BigDecimal的形式读进来.否则会出现各种各样的问题
			int n = scanner.nextInt();
			
			r = r.pow(n);
			
			/**
			 * stripTrailingZeros(): 去掉末尾无效0
			 * toPlainString(): 由科学计数法转为普通的计数法
			 * 如将给一个字符串1.238761976E-10
			 * 转化成0.0000000001238761976
			 */
			String result = r.stripTrailingZeros().toPlainString();
		    if(result.startsWith("0.")){//用于处理0.XXX情况
		    	result = result.substring(1);
		    }
			
		    System.out.println(result);
		}
	}
}