poj1001 Exponentiation【java大数】

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Description

Input

Output

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

C++

while(cin>>s>>n)

{

...

}

c

while(scanf("%s%d",s,&n)==2) //to  see if the scanf read in as many items as you want

/*while(scanf(%s%d",s,&n)!=EOF) //this also work    */

{

...

}

Source

题意：

给定一个浮点数R和一个整数n，求R^n

思路：

想给学弟们出一道大数巩固一下。搜到的大数例题。

java BigDecimal自己也都还没用过。

两个点，用 String s = r.pow(n).stripTrailingZeros().toPlainString();使结果不用科学计数法。

注意去掉前导零。比如0.00001要变成 .00001

 1 import java.io.EOFException;
 2 import java.math.BigDecimal;
 3 import java.math.BigInteger;
 4 import java.util.Scanner;
 5 
 6 public class Main {
 7 
 8 
 9     static public void main(String[] args){
10         Scanner scan = new Scanner(System.in);
11         BigDecimal r;
12         int n;
13         while(scan.hasNext()){
14             r = scan.nextBigDecimal();
15             n = scan.nextInt();
16             String s = r.pow(n).stripTrailingZeros().toPlainString();
17             if(s.startsWith("0")){
18                 s = s.substring(1);
19             }
20             System.out.println(s);
21         }
22         scan.close();
23     }
24 }

poj1001 Exponentiation【java大数】

标签：output   problem   common   air   str   comm   com   计数   star   

原文地址：https://www.cnblogs.com/wyboooo/p/9898388.html