nicehost-
* 回复内容中包含的链接未经审核，可能存在风险，暂不予完整展示！
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 (n****.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
真可-
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。
苏萦-
* 回复内容中包含的链接未经审核，可能存在风险，暂不予完整展示！
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(n*.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);
}
}
陶小凡-
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);
可可-
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);
FinCloud-
直接调用Math.sqrt就可以啦
wpBeta-
class A{
public static void main(){
double m=4.0;
double n=Math.sqrt(m);
System.out.println(n);
}
}
其实就是调用Math下面的sqrt方法