Exponentiation

Time Limit: 500MS	Memory Limit: 10000K
Total Submissions: 156303	Accepted: 38063

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.

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

Hint

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    */

{

...

}

Source

題意:輸入兩個數x,y,求x^y,(y為整型),結果小於0,輸出的結果去掉前面的"0.",直接輸出後面的. 如:stripTrailingZeros() toPlainString(); 呼叫的方法也差不多,但這題如果結果小於小於0,要去掉前導"0.",那該怎麼做呢? 把結果轉換成string型別,再呼叫java.lang.string類中的substring()方法就OK了. API中substring()方法的說明:
public String substring(int beginIndex) 返回一個新的字串，它是此字串的一個子字串。 該子字串從指定索引處的字元開始，
直到此字串末尾。
示例： 
 "unhappy".substring(2) returns "happy"
 "Harbison".substring(3) returns "bison"
 "emptiness".substring(9) returns "" (an empty string)
引數：
beginIndex - 起始索引（包括）。 
返回：
指定的子字串。 
丟擲： 
IndexOutOfBoundsException - 如果 beginIndex 為負或大於此 String 物件的長度。
substring
public String substring(int beginIndex,int endIndex) 返回一個新字串，它是此字串的一個子字串。 該子字串從指定的 beginIndex 處開始，直到索引 endIndex - 1 處的字元。 因此，該子字串的長度為 endIndex-beginIndex。 
示例： 
 "hamburger".substring(4, 8) returns "urge"
 "smiles".substring(1, 5) returns "mile"
引數：
beginIndex - 起始索引（包括）。
endIndex - 結束索引（不包括）。 
返回：
指定的子字串。 
丟擲： 
IndexOutOfBoundsException - 如果 beginIndex 為負，或 endIndex 大於此 String 物件的長度， 或 beginIndex 大於 endIndex。

AC程式碼:

import java.math.BigDecimal;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);
		while (sc.hasNext()) {
			BigDecimal a = sc.nextBigDecimal();
			int b = sc.nextInt();
			a = a.pow(b);
			String str = a.stripTrailingZeros().toPlainString();
			if (str.startsWith("0.")){
				str = str.substring(1);
			}
			System.out.println(str);
		}
	}

}