poj 1001 java

本文深入探讨了Java BigDecimal 类的两个重要方法:stripTrailingZeros() 和 toPlainString(),并展示了如何利用这些方法进行数值处理。通过实例演示,读者可以了解如何有效地去除小数尾部的零,以及如何将BigDecimal 类型转换为简洁的字符串表示。

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import java.math.BigDecimal;
import java.util.Scanner;
 
/**
 * 
 * @author baoyou  E-mail:curiousby@163.com
 * @version 创建时间:2015年9月30日 下午2:40:54 
 * des:
 */
public class BaoyPoj1001Test {
//	   1、stripTrailingZeros() ,返回类型为BigDecimal的小于此数的但除去尾部的0的数值。
//	   2、toPlainString(),返回BigDecimal类型的String类型字符串。 
	public static void main(String args[]) throws Exception
    {
		  Scanner in = new Scanner(System.in);
		  while (in.hasNext()) {
			   BigDecimal a = in.nextBigDecimal();
			   int n = in.nextInt();
			   a= a.pow(n);
			   // a.stripTrailingZeros()   
			   //效果 1.0100^12 =1.126825030131969720661201000000000000000000000000 
			   //             -> 转化为 1.126825030131969720661201
			   //  a.toPlainString()
			   //效果 0.4321^20 =5.148554641076956121994511276767154838481760200726351203835429763013462401E-8
			   //             ->0.00000005148554641076956121994511276767154838481760200726351203835429763013462401
			   String str = a.stripTrailingZeros().toPlainString(); 
			   if (str.startsWith("0."))
			     str = str.substring(1);
			   System.out.println(str);
		  }
    } 
} 

 

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. 输入说明 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. 输出说明 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. 输入样例 95.123 12 0.4321 20 5.1234 15 6.7592 9 98.999 10 1.0100 12 输出样例 548815620517731830194541.899025343415715973535967221869852721 .00000005148554641076956121994511276767154838481760200726351203835429763013462401 43992025569.928573701266488041146654993318703707511666295476720493953024 29448126.764121021618164430206909037173276672 90429072743629540498.107596019456651774561044010001 1.126825030131969720661201 小提示 If you don't know how to determine wheather encounted the end of input: s is a string and n is an integer 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 */ { ... } 来源 East Central North America 1988 北大OJ平台(代理
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