計算採用對BigDecimals的的n次方根

Question

我目前正在爲這裏所描述來解決這個問題：

http://uva.onlinejudge.org/external/1/113.pdf

計劃是實現一個遞歸函數來得到解決。這裏的一些代碼來自用於確定第n個根的Rosetta代碼。

// Power of Cryptography 113 

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

// k can be 10^9 
// n <= 200 
// p <= 10^101 

class crypto { 
    public static void main(String[] args) { 
     Scanner in = new Scanner(System.in); 
    while(in.hasNext()) { 
     // Given two integers (n,p) 
     // Find k such k^n = p 
     int n = in.nextInt(); 
     BigDecimal p = in.nextBigDecimal(); 
     System.out.println(nthroot(n,p)); 
     } 
    } 

    public static BigDecimal nthroot(int n, BigDecimal A) { 
     return nthroot(n, A, .001); 
    } 

    public static BigDecimal nthroot(int n, BigDecimal A, double p) { 
     if(A.compareTo(BigDecimal.ZERO) < 0) return new BigDecimal(-1); 
     // we handle only real positive numbers 
     else if(A.equals(BigDecimal.ZERO)) { 
      return BigDecimal.ZERO; 
     } 
     BigDecimal x_prev = A; 
     BigDecimal x = A.divide(new BigDecimal(n)); // starting "guessed" value... 
     BigDecimal y = x.subtract(x_prev); 

     while(y.abs().compareTo(new BigDecimal(p)) > 0) { 
      x_prev = x; 
      BigDecimal temp = new BigDecimal(n-1.0); 
      x = (x.multiply(temp).add(A).divide(x.pow(temp.intValue())).divide(new BigDecimal(n))); 
     } 
     return x; 
    } 
} 

這裏是產生錯誤代碼：

Exception in thread "main" java.lang.ArithmeticException: Non-terminating decimal expansion; no exact representable decimal result. 
at java.math.BigDecimal.divide(BigDecimal.java:1616) 
at crypto.nthroot(crypto.java:38) 
at crypto.nthroot(crypto.java:24) 
at crypto.main(crypto.java:19) 

Answer 1

，如果產生的數學十進制數是無終止的預期。該Javadocs for the 1-arg overload of divide狀態：

拋出：

ArithmeticException - 如果準確的商值沒有十進制擴展

使用另一個overload of the divide method指定scale（截止）（和RoundingMode ）。

Answer 2

任何人在這裏工作的代碼片段？在這裏，我們去：

public final class RootCalculus { 

    private static final int SCALE = 10; 
    private static final int ROUNDING_MODE = BigDecimal.ROUND_HALF_DOWN; 

    public static BigDecimal nthRoot(final int n, final BigDecimal a) { 
     return nthRoot(n, a, BigDecimal.valueOf(.1).movePointLeft(SCALE)); 
    } 

    private static BigDecimal nthRoot(final int n, final BigDecimal a, final BigDecimal p) { 
    if (a.compareTo(BigDecimal.ZERO) < 0) { 
     throw new IllegalArgumentException("nth root can only be calculated for positive numbers"); 
    } 
    if (a.equals(BigDecimal.ZERO)) { 
     return BigDecimal.ZERO; 
    } 
    BigDecimal xPrev = a; 
    BigDecimal x = a.divide(new BigDecimal(n), SCALE, ROUNDING_MODE); // starting "guessed" value... 
    while (x.subtract(xPrev).abs().compareTo(p) > 0) { 
     xPrev = x; 
     x = BigDecimal.valueOf(n - 1.0) 
       .multiply(x) 
       .add(a.divide(x.pow(n - 1), SCALE, ROUNDING_MODE)) 
       .divide(new BigDecimal(n), SCALE, ROUNDING_MODE); 
    } 
    return x; 
    } 

    private RootCalculus() { 
    } 
} 

只需設置SCALE到精確但是你需要計算中。