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
Hint

If you don't know how to determine wheather encounted the end of input:
s is a string and n is an integer

My Java Code:

import java.util.Scanner;
import java.math.BigDecimal;

public class Main {
 public static void main(String[] args) throws Exception {
  BigDecimal r;
  int n;
  String result;

  Scanner console = new Scanner(System.in);
  while (console.hasNext()) {
   r = console.nextBigDecimal();
   n = console.nextInt();
   result = r.pow(n).toPlainString();
   if(result.length() > 1){
    result = result.replaceAll("^0", "");
   }
   if (result.indexOf('.') > 0) {
    result = result.replaceAll("0+$", "");
   }
   System.out.println(result);
  }
  console.close();
 }
}

用我的程序运行Sample 输入的输出：
95.123 12
0.4321 20
5.1234 15
6.7592  9
98.999 10
1.0100 12
548815620517731830194541.899025343415715973535967221869852721
.00000005148554641076956121994511276767154838481760200726351203835429763013462401
43992025569.928573701266488041146654993318703707511666295476720493953024
29448126.764121021618164430206909037173276672
90429072743629540498.107596019456651774561044010001
1.126825030131969720661201

可是提交上去报 Wrong Answer.
不知道错在哪。。。

略微看了一下你的程序
用math.BigDecimal 你也太偷懒了吧……

public class Main {

    public static void main(String[] args)
    {
        Scanner cin = new Scanner(System.in);
        
        while (cin.hasNext())
        {
            BigDecimal  num;
            int         exp;
            String      res;

            num = BigDecimal.valueOf(Double.parseDouble(cin.next()));
            exp = cin.nextInt();            
            res = num.pow(exp).toPlainString();
            
            if (res.startsWith("0"))
            {
                res = res.substring(1, res.length());
            }
            
            while (res.endsWith("0"))
            {
                res = res.substring(0, res.length() - 1);
            }
                 
            if (res.endsWith("."))
            {
                res = res.substring(0, res.length() - 1);
            }    

            System.out.println(res);
        }
    }
}

否则0.0100^4就通不过

头部0，尾部0，小数点都处理了，我用的是正则表达式，都可以处理的，有谁能给一个NG的case？

以下是引用dark_matter在2009-3-9 22:31:00的发言： 很久以前写的，OJ上的Archive下载不下来，刚在电脑上找出来一份，不知道是不是最后AC的版本： import java.util.*; import java.lang.*; import java.math.*; public class Main {      public static void main(String[] args)      {          Scanner cin = new Scanner(System.in);                    while (cin.hasNext())          {              BigDecimal  num;              int         exp;              String      res;              num = BigDecimal.valueOf(Double.parseDouble(cin.next()));              exp = cin.nextInt();                          res = num.pow(exp).toPlainString();                            if (res.startsWith("0"))              {                  res = res.substring(1, res.length());              }                            while (res.endsWith("0"))              {                  res = res.substring(0, res.length() - 1);              }                                 if (res.endsWith("."))              {                  res = res.substring(0, res.length() - 1);              }                  System.out.println(res);          }      } }

using namespace std;

class power
{
 friend istream&operator>>(istream&In, power&r);
public:
 string operator()(int n);
private:
 vector<int> s;
 int dot;
};

istream&operator>>(istream&In, power&r)
{
 string tmp;
 In>>tmp;
 for(int i=0; i!=6; ++i)if(tmp[i]=='.')r.dot=6-i-1;
 r.s.clear();
 for(int i=5; i>=0; --i)if(tmp[i]!='.')r.s.push_back(tmp[i]-'0');
 return(In);
}

string power::operator()(int n)
{
 vector<int>answer(n*5);
 answer[0]=1;
 for(int i=0; i!=n; ++i)
 {
  vector<int>tmp(answer);
  for(int j=0; j!=n*5; ++j)answer[j]=0;
  for(int j=0; j!=n*5; ++j)
   for(int k=0; k!=5; ++k)
    answer[j+k]+=tmp[j]*s[k];
  for(int j=1; j!=n*5; ++j)
  {
   answer[j]+=answer[j-1]/10;
   answer[j-1]%=10;
  }
 }
 string str;
 str.clear();
 for(int i=0; i!=n*5; ++i)str=(char)(answer[i]+'0')+str;
 str.insert(n*5-dot*n, 1, '.');
 while(str[0]=='0')str.erase(str.begin());
 while(str[str.size()-1]=='0')str.erase(str.end()-1);
 if(str[str.size()-1]=='.')str.erase(str.end()-1);
 return(str);
}

int main()
{
 power s;
 int n;
 while(cin>>s>>n)cout<<s(n)<<endl;
}

public class Pku1001 {
 public static void main(String[] args) throws Exception {
  BigDecimal r;
  int n;
  String result;

  Scanner console = new Scanner(System.in);
  while (console.hasNext()) {
   r = console.nextBigDecimal();
   n = console.nextInt();
   result = r.pow(n).toPlainString();
   if (result.length() > 1) {
    result = result.replaceAll("^0", "");
   }
   if (result.indexOf('.') > -1) {
    result = result.replaceAll("0+$", "");
   }

   result = result.replaceAll("\\.$", "");

   System.out.println(result);
  }
  console.close();
 }
}

除了result.indexOf('.') > -1这个地方改了之外，还加了一句
result = result.replaceAll("\\.$", "");

而加这一句与不加唯一的区别在于 0.0^3这样的输出不同，原来会输出 . ，现在输出空，可是题干明白说了  0.0<r<99.999,。。。，
就这个地方让我琢磨了好久 ，日。。。。
另外，judge对 0^4这样的case没有要求，我试过了，输出0或者输出空都对。。。