No.1001：Exponentiation

Exponentiation
Time Limit:500MS  Memory Limit:10000K
Total Submit:14295 Accepted:3105

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

Java实现代码：

/**//*
 * Main.java
 *
 * Created on 2006年9月13日, 下午10:57
 *
 * To change this template, choose Tools | Template Manager
 * and open the template in the editor.
 */

/** *//**
 *
 * @author jailu
 */
import java.util.*;
import java.lang.*;
import java.math.*;

public class Main {

    public static void main(String[] args) 
    {
        Scanner cin = new Scanner(System.in);
        
        while (cin.hasNext())
        {
            BigDecimal FirstNumber;
            int LastNumber;
            String strResult;

            FirstNumber = BigDecimal.valueOf(Double.parseDouble(cin.next()));
            LastNumber = cin.nextInt();
            
            strResult = FirstNumber.pow(LastNumber).toPlainString();
            
            if(strResult.startsWith("0"))
                strResult = strResult.substring(1);
            while (strResult.endsWith("0"))
                strResult = strResult.substring(0,strResult.length() - 1);
            if (strResult.endsWith("."))
                strResult = strResult.substring(0,strResult.length() - 1);

            System.out.println(strResult);
        }
    }
}

评定结果：

Run ID	User	Problem	Result	Memory	Time	Language	Code Length	Submit Time
1586988	jailu	1001	Accepted	1484K	187MS	Java	0.93K	2006-09-15 16:09:45.0

收获：了解BigDecimal类型、进一步熟悉substring()方法。