ACM: 高精度运算 poj 1001 首次jav…_c++ problems involving the computation of exact va-优快云博客

本文介绍了一种使用Java实现的大数幂运算方法，通过BigDecimal类处理非常大的数值和精度问题，确保计算结果的准确性。

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Exponentiation

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

.00000005148554641076956121994511276767154838481760200726351203835429763013462

401 43992025569.928573701266488041146654993318703707511666295476720493953024

29448126.764121021618164430206909037173276672

90429072743629540498.107596019456651774561044010001

1.126825030131969720661201

题意: 求a^b.

解题思路:

1.直接java大数类 + 格式控制输出.

代码:

import java.io.*;
import java.util.*;
import java.math.*;

public class Main
{
   public static void main(String args[]) throws Exception
   {
       Scanner in = new Scanner(System.in);

       while(in.hasNext())
       {
           char[] t = Double.toString(in.nextDouble()).toCharArray();
           BigDecimal big = new BigDecimal(t);

           int power = in.nextInt();
           BigDecimal result = big.pow(power);
           String str = result.toPlainString();
           int n = str.length();
           int i;

           for(i = n-1; ; --i)
           {
               if(str.charAt(i) != '0')
                   break;
           }

           if(str.charAt(i) == '.')
               i--;
           str = str.substring(0,i+1);
           if(str.charAt(0) == '0')
           {
               if(str.equals("0"))
                       System.out.println(0);
               else
                       System.out.println(str.substring(1));
           }
           else
                   System.out.println(str);

       }
   }
}