Java中BigDecimal的平方根 |
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我已经用过了,它工作得很好。Here's an example of how the algorithm works at a high level. 编辑:我很好奇这到底有多准确,如下所示。下面是来自official source的sqrt(2) 代码语言:javascript复制(first 200 digits) 1.41421356237309504880168872420969807856967187537694807317667973799073247846210703885038753432764157273501384623091229702492483605585073721264412149709993583141322266592750559275579995050115278206057147在这里,它使用下面我概述的方法,SQRT_DIG等于150: 代码语言:javascript复制(first 200 digits) 1.41421356237309504880168872420969807856967187537694807317667973799073247846210703885038753432764157273501384623091229702492483605585073721264412149709993583141322266592750559275579995050115278206086685第一个偏差发生在精度的195位之后。如果您需要如此高的精度,请自负风险。 将SQRT_DIG更改为1000将生成精度的1570位。 代码语言:javascript复制private static final BigDecimal SQRT_DIG = new BigDecimal(150); private static final BigDecimal SQRT_PRE = new BigDecimal(10).pow(SQRT_DIG.intValue()); /** * Private utility method used to compute the square root of a BigDecimal. * * @author Luciano Culacciatti * @url http://www.codeproject.com/Tips/257031/Implementing-SqrtRoot-in-BigDecimal */ private static BigDecimal sqrtNewtonRaphson (BigDecimal c, BigDecimal xn, BigDecimal precision){ BigDecimal fx = xn.pow(2).add(c.negate()); BigDecimal fpx = xn.multiply(new BigDecimal(2)); BigDecimal xn1 = fx.divide(fpx,2*SQRT_DIG.intValue(),RoundingMode.HALF_DOWN); xn1 = xn.add(xn1.negate()); BigDecimal currentSquare = xn1.pow(2); BigDecimal currentPrecision = currentSquare.subtract(c); currentPrecision = currentPrecision.abs(); if (currentPrecision.compareTo(precision) |
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