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ACM刷题之HDU————Exponentiation

2019-11-08 18:24:30
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Exponentiation

Time Limit: 1000/500 MS (java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2584 Accepted Submission(s): 702
 
PRoblem DescriptionProblems 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.  
InputThe 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. 
OutputThe 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
95.123 120.4321 205.1234 156.7592  998.999 101.0100 12 
Sample Output
548815620517731830194541.899025343415715973535967221869852721.0000000514855464107695612199451127676715483848176020072635120383542976301346240143992025569.92857370126648804114665499331870370751166629547672049395302429448126.76412102161816443020690903717327667290429072743629540498.1075960194566517745610440100011.126825030131969720661201 大数幂。用JAVA水过用BigDecimal 根据题意要用stripTrailingZeros()函数 去掉多余的0下面是ac代码
import java.math.BigInteger;import java.util.Scanner;import java.math.*;import java.text.*;public class Main{	public static void main(String[] args) {		// TODO 自动生成的方法存根		Scanner cin=new Scanner(System.in);				BigDecimal bigDecimal = new BigDecimal("1");		BigDecimal bigDecimal2 = new BigDecimal("1");				int i = 0,j,k;				while(cin.hasNext())   //等同于!=EOF		{			bigDecimal=cin.nextBigDecimal();			bigDecimal2=bigDecimal;			i=cin.nextInt();						bigDecimal = bigDecimal.pow(i);			String string = bigDecimal.stripTrailingZeros().toPlainString();						if(string.startsWith("0."))				string = string.substring(1);									System.out.println(string);					}	}}

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