Exponentiation
Time Limit: 2000/1000ms (Java/Others)
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.
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
解题思路:计算r^n,要求:①如果整数部分为0,去掉整数部分;②去掉结果尾部的所有0,java水过!
AC代码:
1 import java.math.BigDecimal;
2 import java.util.Scanner;
3 public class Main {
4 public static void main(String[] args) {
5 Scanner scan = new Scanner(System.in);
6 while(scan.hasNext()){
7 BigDecimal bd = new BigDecimal(scan.next());
8 BigDecimal result = bd.pow(scan.nextInt());
9 //stripTrailingZeros()的使用方法:返回数值上等于此小数,但从该表示形式移除所有尾部零的 BigDecimal。
10 //toPlainString()的使用方法:返回不带指数字段的此 BigDecimal的字符串表示形式。
11 String obj = result.stripTrailingZeros().toPlainString();
12 //startsWith(String prefix)的使用方法:测试此字符串是否以指定的前缀开始。这里表示如果整数部分为0,就截取除下标为0的子串
13 if(obj.startsWith("0"))obj=obj.substring(1);
14 System.out.println(obj);
15 }
16 }
17 }