HDU 1063 Exponentiation(R的N次方)（Java版）

Exponentiation

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3847    Accepted Submission(s): 1015

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

95.123 12

0.4321 20

5.1234 15

6.7592  9

98.999 10

1.0100 12

Sample Output

548815620517731830194541.899025343415715973535967221869852721

.00000005148554641076956121994511276767154838481760200726351203835429763013462401

43992025569.928573701266488041146654993318703707511666295476720493953024

29448126.764121021618164430206909037173276672

90429072743629540498.107596019456651774561044010001

1.126825030131969720661201

Source

East Central North America 1988

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PrincetonBoy

解题报告：这道题就是让求R的n次方，如果结果小于1，输出的时候去掉小数的前面的0，用Java写比较简单，

代码如下：

import java.util.Scanner;
import java.math.BigDecimal;
public class Main{
    public static void main(String[] args){
        Scanner scan = new Scanner(System.in);
        BigDecimal R, power;
        int n;
        String ans;
        while (scan.hasNextBigDecimal()){
            R = scan.nextBigDecimal();
            n = scan.nextInt();
            power = R.pow(n);//R的n次方
            power = power.stripTrailingZeros();//去掉后导零
            ans = power.toPlainString();//将数字转化为字符串
            if (ans.indexOf("0") == 0){
                ans = ans.substring(1);//如果结果小于1就去掉小数点前面的0；
            }
            System.out.println(ans);
        }
    }
}