可以从N中减去最大完美平方数的次数

给定数字N。在每一步中，从N中减去最大的完美平方(≤N)。在N> 0时重复此步骤。任务是计算可以执行的步骤数。

例子：

Input: N = 85 
Output: 2 
First step, 85 – (9 * 9) = 4 
Second step 4 – (2 * 2) = 0

Input: N = 114 
Output: 4 
First step, 114 – (10 * 10) = 14 
Second step 14 – (3 * 3) = 5 
Third step 5 – (2 * 2) = 1 
Fourth step 1 – (1 * 1) = 0 

方法：在N> 0的情况下，从N迭代减去最大正方差(≤N)并计算步数。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the count of steps
int countSteps(int n)
{
 
    // Variable to store the count of steps
    int steps = 0;
 
    // Iterate while N > 0
    while (n) {
 
        // Get the largest perfect square
        // and subtract it from N
        int largest = sqrt(n);
        n -= (largest * largest);
 
        // Increment steps
        steps++;
    }
 
    // Return the required count
    return steps;
}
 
// Driver code
int main()
{
    int n = 85;
    cout << countSteps(n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.lang.Math;
 
public class GfG{
 
    // Function to return the count of steps
    static int countSteps(int n)
    {
        // Variable to store the count of steps
        int steps = 0;
     
        // Iterate while N > 0
        while (n > 0) {
     
            // Get the largest perfect square
            // and subtract it from N
            int largest = (int)Math.sqrt(n);
            n -= (largest * largest);
     
            // Increment steps
            steps++;
        }
     
        // Return the required count
        return steps;
    }
 
     public static void main(String []args){
         
        int n = 85;
        System.out.println(countSteps(n));
     }
}
 
// This code is contributed by Rituraj Jain

Python3

# Python3 implementation of the approach
from math import sqrt
 
# Function to return the count of steps
def countSteps(n) :
 
    # Variable to store the count of steps
    steps = 0;
 
    # Iterate while N > 0
    while (n) :
 
        # Get the largest perfect square
        # and subtract it from N
        largest = int(sqrt(n));
        n -= (largest * largest);
 
        # Increment steps
        steps += 1;
 
    # Return the required count
    return steps;
     
# Driver code
if __name__ == "__main__" :
 
    n = 85;
    print(countSteps(n));
     
# This code is contributed by Ryuga

C#

// C# implementation of the approach
using System;
 
class GfG
{
 
    // Function to return the count of steps
    static int countSteps(int n)
    {
        // Variable to store the count of steps
        int steps = 0;
     
        // Iterate while N > 0
        while (n > 0)
        {
     
            // Get the largest perfect square
            // and subtract it from N
            int largest = (int)Math.Sqrt(n);
            n -= (largest * largest);
     
            // Increment steps
            steps++;
        }
     
        // Return the required count
        return steps;
    }
 
    // Driver code
    public static void Main()
    {
        int n = 85;
        Console.WriteLine(countSteps(n));
    }
}
 
// This code is contributed by Code_Mech.

PHP

 0
    while ($n)
    {
 
        // Get the largest perfect square
        // and subtract it from N
        $largest = (int)sqrt($n);
        $n -= ($largest * $largest);
 
        // Increment steps
        $steps++;
    }
 
    // Return the required count
    return $steps;
}
 
// Driver code
$n = 85;
echo countSteps($n);
 
// This code is contributed
// by Akanksha Rai
?>

Javascript

 
// JavaScript implementation of the approach
 
// Function to return the count of steps
function countSteps(n)
{
     
    // Variable to store the count of steps
    let steps = 0;
 
    // Iterate while N > 0
    while (n)
    {
         
        // Get the largest perfect square
        // and subtract it from N
        let largest = Math.floor(Math.sqrt(n));
        n -= (largest * largest);
 
        // Increment steps
        steps++;
    }
 
    // Return the required count
    return steps;
}
 
// Driver code
 
let n = 85;
 
document.write(countSteps(n));
 
// This code is contributed by Manoj.
 

输出：

2

时间复杂度： O(1)

辅助空间： O(1)