ACM_Exponentiation（java高精度）

Exponentiation

Time Limit: 2000/1000ms (Java/Others)

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
解题思路：计算r^n，要求：①如果整数部分为0，去掉整数部分；②去掉结果尾部的所有0，java水过！
AC代码：

 1 import java.math.BigDecimal;
 2 import java.util.Scanner;
 3 public class Main {
 4     public static void main(String[] args) {
 5         Scanner scan = new Scanner(System.in);
 6         while(scan.hasNext()){
 7             BigDecimal bd = new BigDecimal(scan.next());
 8             BigDecimal result = bd.pow(scan.nextInt());
 9             //stripTrailingZeros()的使用方法：返回数值上等于此小数，但从该表示形式移除所有尾部零的 BigDecimal。
10             //toPlainString()的使用方法：返回不带指数字段的此 BigDecimal的字符串表示形式。
11             String obj = result.stripTrailingZeros().toPlainString();
12             //startsWith(String prefix)的使用方法：测试此字符串是否以指定的前缀开始。这里表示如果整数部分为0，就截取除下标为0的子串 
13             if(obj.startsWith("0"))obj=obj.substring(1);
14             System.out.println(obj);
15         }
16     }
17 }