HDOJ---1063 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


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

95.123 12 0.4321 20 5.1234 15 6.7592 9 98.999 10 1.0100 12

[align=left]Sample Output[/align]

548815620517731830194541.899025343415715973535967221869852721 .00000005148554641076956121994511276767154838481760200726351203835429763013462401 43992025569.928573701266488041146654993318703707511666295476720493953024 29448126.764121021618164430206909037173276672 90429072743629540498.107596019456651774561044010001 1.126825030131969720661201

[align=left]Source[/align]
East Central North America 1988

[align=left]Recommend[/align]
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:

import java.util.*;
import java.math.*;
public class Main
{
public static void main(String []args)
{
BigDecimal a;
int n;
String ans;
Scanner cin=new Scanner(System.in);
while(cin.hasNext())
{
a=cin.nextBigDecimal();
n=cin.nextInt();
a=a.pow(n);
a=a.stripTrailingZeros();
ans=a.toPlainString();
if(ans.charAt(0)=='0')
System.out.println(ans.substring(1));
else
System.out.println(ans);
}
}
}