(hdu 2.3.2)Exponentiation(大数：求R的n次幂)

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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

[align=left]Problem Description[/align]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.

[align=left]Input[/align]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.

[align=left]Output[/align]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.

[align=left]Sample Input[/align] 95.123 12 0.4321 20 5.1234 15 6.7592 9 98.999 10 1.0100 12

[align=left]Sample Output[/align] 548815620517731830194541.899025343415715973535967221869852721 .00000005148554641076956121994511276767154838481760200726351203835429763013462401 43992025569.928573701266488041146654993318703707511666295476720493953024 29448126.764121021618164430206909037173276672 90429072743629540498.107596019456651774561044010001 1.126825030131969720661201

[align=left]Source[/align]East Central North America 1988

[align=left]Recommend[/align]PrincetonBoy

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);
}
}
}