我目前正在尝试按照此处所述解决此问题:
http://uva.onlinejudge.org/external/1/113.pdf
计划是实现一个递归函数来导出解决方案。这里的部分代码来自Rosetta代码,用于确定n次方根。
// 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;
}
}
这是生成的错误代码:
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)
最佳答案
这里有人要工作代码片段吗?我们开始吧:
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() {
}
}
只需将 SCALE
设置为您需要的计算精度即可。
关于java - 使用 BigDecimals 计算 p 的 n 次方根,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/22695654/