Java中，如何对大数开根号啊！

Posted 2023-05-03

在Java中，有没有对一个BigInteger,BigDecimal数开根号的函数，如果有，函数是什么？如果没有，怎么处理这个问题？
解决后，追加分！
假如开根号是一个无限小数，要怎样保存自己想要的n位呢？
例如sqrt(3),我想保存1000位，怎么搞啊？
我这里要求的是要求出小数点后多位(几百，几千甚至几万)，如果是一般的话，直接用Math.sqrt(3)求就可以了！

java中对于大数BigInteger,BigDecimal开根号没有提供函数，可以参考以下实现方法：

import java.math.BigDecimal;
import java.math.BigInteger;
public class BigSquareRoot
final static BigInteger HUNDRED = BigInteger.valueOf(100);

public static BigDecimal sqrt(BigDecimal number, int scale, int roundingMode)
if (number.compareTo(BigDecimal.ZERO) < 0)
throw new ArithmeticException("sqrt with negative");
BigInteger integer = number.toBigInteger();
StringBuffer sb = new StringBuffer();
String strInt = integer.toString();
int lenInt = strInt.length();
if (lenInt % 2 != 0)
strInt = '0' + strInt;
lenInt++;

BigInteger res = BigInteger.ZERO;
BigInteger rem = BigInteger.ZERO;
for (int i = 0; i < lenInt / 2; i++)
res = res.multiply(BigInteger.TEN);
rem = rem.multiply(HUNDRED);

BigInteger temp = new BigInteger(strInt.substring(i * 2, i * 2 + 2));
rem = rem.add(temp);

BigInteger j = BigInteger.TEN;
while (j.compareTo(BigInteger.ZERO) > 0)
j = j.subtract(BigInteger.ONE);
if (((res.add(j)).multiply(j)).compareTo(rem) <= 0)
break;



res = res.add(j);
rem = rem.subtract(res.multiply(j));
res = res.add(j);
sb.append(j);

sb.append('.');
BigDecimal fraction = number.subtract(number.setScale(0, BigDecimal.ROUND_DOWN));
int fracLen = (fraction.scale() + 1) / 2;
fraction = fraction.movePointRight(fracLen * 2);
String strFrac = fraction.toPlainString();
for (int i = 0; i <= scale; i++)
res = res.multiply(BigInteger.TEN);
rem = rem.multiply(HUNDRED);

if (i < fracLen)
BigInteger temp = new BigInteger(strFrac.substring(i * 2, i * 2 + 2));
rem = rem.add(temp);


BigInteger j = BigInteger.TEN;
while (j.compareTo(BigInteger.ZERO) > 0)
j = j.subtract(BigInteger.ONE);
if (((res.add(j)).multiply(j)).compareTo(rem) <= 0)
break;


res = res.add(j);
rem = rem.subtract(res.multiply(j));
res = res.add(j);
sb.append(j);

return new BigDecimal(sb.toString()).setScale(scale, roundingMode);


public static BigDecimal sqrt(BigDecimal number, int scale)
return sqrt(number, scale, BigDecimal.ROUND_HALF_UP);


public static BigDecimal sqrt(BigDecimal number)
int scale = number.scale() * 2;
if (scale < 50)
scale = 50;
return sqrt(number, scale, BigDecimal.ROUND_HALF_UP);


public static void main(String args[])
BigDecimal num = new BigDecimal("6510354513.6564897413514568413");
long time = System.nanoTime();
BigDecimal root = sqrt(num, 1000);
time = System.nanoTime() - time;
System.out.println(root);
System.out.println(root.pow(2));
System.out.println(time);


执行结果：
80686.7678969512927493416316741557266722739984372151634715876752358049492663077817843059095146637911180490885758744651273281303288317374885332607051330176028572558172054217029042642080284121950891605518862273493239191320132148293688445347529243846517751025383884710742819252354014378344895438280908159584992112041354808433466321589387318739165992813377399669170549811704076258078653548749003251504791227309054913701062929858500433745971631998487835576600579373929233933246442803804132298016737159672317482520249763464713581048142915509001995943192415694815489364740152312416736301233269587910628885614893125235822493317184917626076223325819402403220531926392808333854523694780539563293133232729900988243013464020440976396084796739002581380094075169287492710301071487312491530290342213569053680461901907481289230152643599970138861788489599118674849815164425194138401918499233009571650761625943781367455101019720348741842171772915942278011905594031830367343193606047124373968951524359600676406162506362881367
6510354513.65648974135145684129999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999261018570188221952291201089398906052231483112444155137889353596770526763176032152876816872483428732651504536335267475372360148274382594285697529542224437785449440331424040251426099649509195223143651828659587364153208947136985285283628868915378240201094213227734756251495878566340409599684823659444904292099882985668906808623177208665251919370268937893457306121802766566270627041680526252368411023459487996551104603913903029501700835230597885867761183071537171958826568685086349210032834841075060512309444622145616551108119893623864579013813710941167261179972571233574705638862140357465569295994523261742960807593601727929262728856153880503561459736300910103299707873770713267018154171165545178430002342459940561678884530151166769964180453998734209554051691222326553700712948508454649608942746651683572634224323098274435576290709769148239120722342126902645574609770558211829972705514153006434846614016006956921594506606758832662240593485962629910320205678474047322232577044567479336985830934534290515788689
46726188 参考技术A BigInteger,BigDecimal 是无精度损失运算的，也就是说只要你给出合法的算法，它就能给出精确的结果。

但你知道，开方有时会是一个无限小数，也就是说，如果叫BigInteger,BigDecimal去开方将永远运行下去，这显然不是你想要的，但如果精确到小数点后N位，这又违背了无精度损失的初衷，所以就有了你提的问题。

你拿BigDecimal(1) divide BigDecimal(3) 就知道了。 参考技术B 貌似暂时还没有
自己写

import java.math.BigDecimal;
import java.math.MathContext;

public class Test5

public static void main(String[] args)
BigDecimal n = new BigDecimal("2");
BigDecimal r = sqrt(n);
System.out.println(r.toString());


public static BigDecimal sqrt(BigDecimal num)
if(num.compareTo(BigDecimal.ZERO) < 0)
return BigDecimal.ZERO;

BigDecimal x = num.divide(new BigDecimal("2"), MathContext.DECIMAL128);
while(x.subtract(x = sqrtIteration(x, num)).abs().compareTo(new BigDecimal("0.0000000000000000000001")) > 0);
return x;


private static BigDecimal sqrtIteration(BigDecimal x, BigDecimal n)
return x.add(n.divide(x, MathContext.DECIMAL128)).divide(new BigDecimal("2"), MathContext.DECIMAL128);

参考技术C 比如
//保留1000位有效数字
BigDecimal num = new BigDecimal("4");

MathContext mc = new MathContext(1000,RoundingMode.HALF_DOWN);

BigDecimal finalnum = new BigDecimal(Math.sqrt(num.doubleValue()) ,mc);本回答被提问者采纳 参考技术D 比如
//保留1000位有效数字
BigDecimal num = new BigDecimal("4");

MathContext mc = new MathContext(1000,RoundingMode.HALF_DOWN);

BigDecimal finalnum = new BigDecimal(Math.sqrt(num.doubleValue()) ,mc);

Python-开根号的几种方式

文章目录

• 前言
• 方法一
• 方法二
• 方法三

前言

使用Python中的自带库math、自带函数pow和自带库cmath来对数字进行开根号运算

方法一

使用：math.sqrt(数字)

import math
n = int(input('数字:'))
x = math.sqrt(n)
print(x)
print(type(x)) #开根号后的类型为float

方法二

使用：pow(数字，次方)

n = int(input('数字:'))
x = pow(n,0.5)
print(x)
print(type(x)) #开根号后的类型为float

方法三

使用：cmath.sqrt(数字)
该方法多用于复数、负数的开方运算

import cmath
n = int(input('数字: '))
x = cmath.sqrt(n)
print(x)
print(type(x)) #类型为complex