zoukankan      html  css  js  c++  java
  • HDOJ1063 大数的n次方

    Exponentiation

    Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 4275    Accepted Submission(s): 1167


    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
     
    Source
     
    Recommend
    PrincetonBoy

    a=a.stripTrailingZeros();                     //除去后导零         

    a=a.pow(n);                                

    ans=a.toPlainString();                  //toString()是以科学计数法输出

    code:

     1 import java.util.*;
     2 import java.math.*;
     3 public class Main
     4 {
     5     public static void main(String []args)
     6     {
     7         BigDecimal a;
     8         int n;
     9         String ans;
    10         Scanner cin=new Scanner(System.in);
    11         while(cin.hasNext())
    12         {
    13             a=cin.nextBigDecimal();
    14             n=cin.nextInt();
    15             a=a.pow(n);
    16             a=a.stripTrailingZeros();
    17             ans=a.toPlainString();
    18             if(ans.charAt(0)=='0')
    19                 System.out.println(ans.substring(1));
    20             else
    21                 System.out.println(ans);
    22         }
    23     }
    24 }
  • 相关阅读:
    在主窗体中打开一个新子窗体,如果已有子窗体,则激活它,而不打开新的。
    文本的追加
    男人至少的品质底线
    做人,做事,生活,学习,爱情>人生
    日常中一些好用的小软件
    ◆2008 年广告创意设计师必备网址汇总◆
    在SharePoint中更改当前登录用户的密码
    SONY笔记本VGNSZ65装VISTA记
    Outlook收邮件速度超慢的原因
    Cisco路由器做限速
  • 原文地址:https://www.cnblogs.com/XBWer/p/2646447.html
Copyright © 2011-2022 走看看