/*2.818s*/ import java.io.*; import java.util.*; import java.math.*; public class Main { static final BigInteger TWO = new BigInteger("2"); static Scanner cin = new Scanner(new BufferedInputStream(System.in));
importjava.math.BigInteger; importjava.util.Scanner; publicclassMain { publicstaticvoidsolve(BigInteger y) { inti, j; BigInteger x =newBigInteger("0"); String str = "0" + y.toString(); BigInteger dvs =newBigInteger("0"), rmd =newBigInteger("0"); i = str.length() % 2; for(; i ...
math.*; package org.arpit.java2blog; public class SquareRootMain { public static void main(String[] args) { double d1=26.0; System.out.println("Square root of 26:"+Math.sqrt(d1)); double d2=56.0; System.out.println("Square root of 56:"+Math.sqrt(d2)); double d3=83.0; ...
Below is the Java implementation using exponentiation ? Open Compiler public class SquareRoot { public static void main(String[] args) { double n = 16; double squareRoot = Math.pow(n, 0.5); System.out.println("The square root of " + n + " is: " + squareRoot); } } Output The ...
在Java语言中,小明在她的类SquareRoot中定义了方法method_A,其中包含语句:double my_result=Math.sqrt(1000);她需要导入
Square root by coupled Newton iteration, sqrtProcedure() is the iteration part I adopted the Algorithm from the book "Pi-unleashed", so now it looks more natural I give sparse math comments from the book, it assumes argument mc precision >= 1 temp1 = BigDecimal.ONE.subtract(TWO.multiply(...
import java.util.Scanner; public class PerfectSquare { // Method to check whether the number is a perfect square public static boolean isPerfectSquare(int num) { // Calculate the square root of the number double squareRoot = Math.sqrt(num); ...
http://www.mathblog.dk/project-euler-80-digits-irrational-square-roots/ 上面的链接有一个方法,用迭代法求到达某个精度的值 代码: import java.math.*; import java.util.*;publicclassMain {publicstaticBigInteger Sqrt(intn,intd) { BigInteger limit= BigInteger.valueOf(10).pow(d+1); ...
BigDecimal; import java.util.Vector; public class Main { public static BigDecimal sqrt(BigDecimal value) { double val = Math.sqrt(value.doubleValue()); if (Double.isNaN(val) || Double.isInfinite(val)) return value; BigDecimal x = new BigDecimal(val); return x.add(new BigDecimal(value....
在C语言中,square_root是一个用于计算平方根的函数。它的作用是返回给定数值的平方根值。在C语言中,可以使用math.h头文件中的sqrt()函数来计算平方根。 square_root函数的分类是数学函数,它属于数值计算领域。它的优势在于可以快速准确地计算给定数值的平方根,方便在程序中进行数值计算和处理。 square_root函数的应...