不使用sqrt()函数的数字的平方根

给定一个数字N ，任务是不使用sqrt()函数来查找N的平方根。

例子：

Input: N = 25
Output: 5

Input: N = 3
Output: 1.73205

Input: N = 2.5
Output: 1.58114

方法：

• 从i = 1开始迭代。如果i * i = n ，则将i打印为n是一个平方根为i的理想平方。
• 否则，找到i * i严格大于n的最小i 。
• 现在我们知道n的平方根位于i – 1和i的区间内，我们可以使用二进制搜索算法找到平方根。
• 找出i – 1和i的中点，然后将mid * mid与n进行比较，精度最高到5位小数。
  1. 如果mid * mid = n，则返回mid 。
  2. 如果mid * mid 则在下半部分重复出现。
  3. 如果mid * mid> n，则重复上半部分。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Recursive function that returns square root
// of a number with precision upto 5 decimal places
double Square(double n, double i, double j)
{
    double mid = (i + j) / 2;
    double mul = mid * mid;
  
    // If mid itself is the square root,
    // return mid
    if ((mul == n) || (abs(mul - n) < 0.00001))
        return mid;
  
    // If mul is less than n, recur second half
    else if (mul < n)
        return Square(n, mid, j);
  
    // Else recur first half
    else
        return Square(n, i, mid);
}
  
// Function to find the square root of n
void findSqrt(double n)
{
    double i = 1;
  
    // While the square root is not found
    bool found = false;
    while (!found) {
  
        // If n is a perfect square
        if (i * i == n) {
            cout << fixed << setprecision(0) << i;
            found = true;
        }
        else if (i * i > n) {
  
            // Square root will lie in the
            // interval i-1 and i
            double res = Square(n, i - 1, i);
            cout << fixed << setprecision(5) << res;
            found = true;
        }
        i++;
    }
}
  
// Driver code
int main()
{
    double n = 3;
  
    findSqrt(n);
  
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
  
class GFG 
{
  
// Recursive function that returns 
// square root of a number with 
// precision upto 5 decimal places
static double Square(double n, 
                     double i, double j)
{
    double mid = (i + j) / 2;
    double mul = mid * mid;
  
    // If mid itself is the square root,
    // return mid
    if ((mul == n) ||   
        (Math.abs(mul - n) < 0.00001))
        return mid;
  
    // If mul is less than n, 
    // recur second half
    else if (mul < n)
        return Square(n, mid, j);
  
    // Else recur first half
    else
        return Square(n, i, mid);
}
  
// Function to find the square root of n
static void findSqrt(double n)
{
    double i = 1;
  
    // While the square root is not found
    boolean found = false;
    while (!found) 
    {
  
        // If n is a perfect square
        if (i * i == n) 
        {
            System.out.println(i);
            found = true;
        }
          
        else if (i * i > n) 
        {
  
            // Square root will lie in the
            // interval i-1 and i
            double res = Square(n, i - 1, i);
            System.out.printf("%.5f", res);
            found = true;
        }
        i++;
    }
}
  
// Driver code
public static void main(String[] args) 
{
    double n = 3;
  
    findSqrt(n);
}
}
  
// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation of the approach
import math
  
# Recursive function that returns square root
# of a number with precision upto 5 decimal places
def Square(n, i, j):
  
    mid = (i + j) / 2;
    mul = mid * mid;
  
    # If mid itself is the square root,
    # return mid
    if ((mul == n) or (abs(mul - n) < 0.00001)):
        return mid;
  
    # If mul is less than n, recur second half
    elif (mul < n):
        return Square(n, mid, j);
  
    # Else recur first half
    else:
        return Square(n, i, mid);
  
# Function to find the square root of n
def findSqrt(n):
    i = 1;
  
    # While the square root is not found
    found = False;
    while (found == False):
  
        # If n is a perfect square
        if (i * i == n):
            print(i);
            found = True;
          
        elif (i * i > n):
  
            # Square root will lie in the
            # interval i-1 and i
            res = Square(n, i - 1, i);
            print ("{0:.5f}".format(res)) 
            found = True
        i += 1;
  
# Driver code
if __name__ == '__main__':
    n = 3;
  
    findSqrt(n);
  
# This code is contributed by 29AjayKumar

C#

// C# implementation of the approach
using System;
      
class GFG 
{
  
// Recursive function that returns 
// square root of a number with 
// precision upto 5 decimal places
static double Square(double n, 
                     double i, double j)
{
    double mid = (i + j) / 2;
    double mul = mid * mid;
  
    // If mid itself is the square root,
    // return mid
    if ((mul == n) || 
        (Math.Abs(mul - n) < 0.00001))
        return mid;
  
    // If mul is less than n, 
    // recur second half
    else if (mul < n)
        return Square(n, mid, j);
  
    // Else recur first half
    else
        return Square(n, i, mid);
}
  
// Function to find the square root of n
static void findSqrt(double n)
{
    double i = 1;
  
    // While the square root is not found
    Boolean found = false;
    while (!found) 
    {
  
        // If n is a perfect square
        if (i * i == n) 
        {
            Console.WriteLine(i);
            found = true;
        }
          
        else if (i * i > n) 
        {
  
            // Square root will lie in the
            // interval i-1 and i
            double res = Square(n, i - 1, i);
            Console.Write("{0:F5}", res);
            found = true;
        }
        i++;
    }
}
  
// Driver code
public static void Main(String[] args) 
{
    double n = 3;
  
    findSqrt(n);
}
}
  
// This code is contributed by Princi Singh

输出：

1.73205