📜  N减去最大完全平方数的次数

📅  最后修改于: 2021-10-26 06:49:14             🧑  作者: Mango

给定一个数字N 。在每一步,从 N 中减去最大的完美平方(≤ N)。当N > 0 时重复此步骤。任务是计算可以执行的步骤数。

例子:

方法:当N > 0 时,迭代地从N 中减去最大的完美平方(≤ 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


输出:
2

时间复杂度: O(1)

辅助空间: O(1)

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