求高精度幂

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

代码块

    import java.util.*;    import java.math.BigDecimal;    public class Main{      public static void main(String[] args){        Scanner cin=new Scanner(System.in);        BigDecimal a;        int b;      while(cin.hasNext()){            a=cin.nextBigDecimal();            b=cin.nextInt();            a=a.pow(b);//求a的b次方            //传换成字符串，如果是整数，去掉小数点和后面的            String res=a.stripTrailingZeros().toPlainString();        //如果是o开头的，去点前导o            if(res.startsWith("0")){            res=res.substring(1);            }            System.out.println(res);        }    }    }

关于调用的一些方法解释：
1 hasNext():判断是否有数据输入。
2 使用BigDecimal原生方法实现末尾去0
有这么一个需求，一个BigDecimal值，四舍五入计算到小数点后4位，如果小数点后3、4位是0则去掉，保留到小数点后两位
即20.00345 显示成20.0035
20 显示 20.00
20.00395 显示 20.004
发现BigDecimal原生提供了stripTrailingZeros方法可以实现去掉末尾的0，然后使用toPlainString可以输出数值，注意这里如果使用toString() 会变成科学计数法输出
但对于要保留两位小数0的情况这里会变成显示20，即不带小数点后两位，只能通过字符串判断加上，难看点但算是实现了这个偏门的需求。
3 substring(1)：返回一个新的字符串，这个字符串为原来的子字符串，从位置1开始截取。