Exponentiation
Time Limit: 1000/500 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6973 Accepted Submission(s): 1975
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.
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
大数的题目使用Java就变得其简单无比。
主要就一句代码：
String ans = a.pow(b).stripTrailingZeros().toPlainString();
去零和转换为纯字符串输出。
这使用C++要数十行代码。
对照C++解法：http://blog.csdn.net/kenden23/article/details/23997827
相比之下。java就成了水题了。学不到什么东西。
import java.math.BigDecimal;
import java.util.*; public class Main
{
public static void main(String[] args)
{
Scanner scan = new Scanner(System.in);
while (scan.hasNext())
{
BigDecimal a = scan.nextBigDecimal();
int b = scan.nextInt();
String ans = a.pow(b).stripTrailingZeros().toPlainString();
if (ans.startsWith("0")) ans = ans.substring(1);
System.out.println(ans);
}
scan.close();
}
}