Cuente los pasos mínimos para llegar a 0 desde el entero N dado

Dados dos números enteros N y K donde K representa el número de saltos que podemos hacer directamente de N reduciendo N a N – K, nuestra tarea es contar los pasos mínimos para llegar a 0 siguiendo las operaciones dadas: 

• Podemos saltar una cantidad de K desde N que es N = N – K
• Decrementa N en 1 que es N = N -1.

Ejemplos: 

Entrada: N = 11, K = 4 
Salida: 5 
Explicación: 
Para el valor N dado, podemos realizar la operación en la secuencia dada: 11 -> 7 -> 3 -> 2 -> 1 -> 0

Entrada: N = 6, K = 3 
Salida: 2 
Explicación: 
Para el valor N dado podemos realizar la operación en la secuencia dada: 6 -> 3 -> 0. 
 

Enfoque: 
para resolver el problema mencionado anteriormente, sabemos que tomará N / K pasos para saltar directamente del valor N al valor menos divisible con K y N % K pasos para disminuirlo en 1, como para reducir el conteo a 0. Entonces el número total de pasos necesarios para llegar a 0 desde N será 

(N / K) + (N % K)

A continuación se muestra la implementación del enfoque anterior:  

// C++ program to Count the minimum steps
// to reach 0 from the given integer N
 
#include <bits/stdc++.h>
using namespace std;
 
// Function returns min step
// to reach 0 from N
int getMinSteps(int n, int jump)
{
    // Direct possible
    // reduction of value N
    int quotient = n / jump;
 
    // Remaining steps needs
    // to be reduced by 1
    int remainder = n % jump;
 
    // Summation of both the values
    int steps = quotient + remainder;
 
    // Return the final answer
    return steps;
}
 
// Driver code
int main()
{
    int N = 6, K = 3;
 
    cout << getMinSteps(N, K);
 
    return 0;
}

// Java program to count the minimum steps
// to reach 0 from the given integer N
class GFG{
 
// Function returns min step
// to reach 0 from N
static int getMinSteps(int n, int jump)
{
     
    // Direct possible
    // reduction of value N
    int quotient = n / jump;
 
    // Remaining steps needs
    // to be reduced by 1
    int remainder = n % jump;
 
    // Summation of both the values
    int steps = quotient + remainder;
 
    // Return the final answer
    return steps;
}
 
// Driver code
public static void main(String[] args)
{
    int N = 6, K = 3;
 
    System.out.print(getMinSteps(N, K));
}
}
 
// This code is contributed by Rohit_ranjan

# Python3 program to Count the minimum steps
# to reach 0 from the given integer N
 
# Function returns min step
# to reach 0 from N
def getMinSteps(n, jump):
 
    # Direct possible
    # reduction of value N
    quotient = int(n / jump)
 
    # Remaining steps needs
    # to be reduced by 1
    remainder = n % jump
 
    # Summation of both the values
    steps = quotient + remainder
 
    # Return the final answer
    return steps
 
# Driver code
N = 6
K = 3
 
print (getMinSteps(N, K))
 
# This code is contributed by PratikBasu

// C# program to count the minimum steps
// to reach 0 from the given integer N
using System;
 
class GFG{
 
// Function returns min step
// to reach 0 from N
static int getMinSteps(int n, int jump)
{
     
    // Direct possible
    // reduction of value N
    int quotient = n / jump;
 
    // Remaining steps needs
    // to be reduced by 1
    int remainder = n % jump;
 
    // Summation of both the values
    int steps = quotient + remainder;
 
    // Return the final answer
    return steps;
}
 
// Driver code
public static void Main(string[] args)
{
    int N = 6, K = 3;
 
    Console.Write(getMinSteps(N, K));
}
}
 
// This code is contributed by rutvik_56

<script>
// JavaScript program to Count the minimum steps
// to reach 0 from the given integer N
 
// Function returns min step
// to reach 0 from N
function getMinSteps(n, jump)
{
 
    // Direct possible
    // reduction of value N
    let quotient = Math.floor(n / jump);
 
    // Remaining steps needs
    // to be reduced by 1
    let remainder = n % jump;
 
    // Summation of both the values
    let steps = quotient + remainder;
 
    // Return the final answer
    return steps;
}
 
// Driver code
    let N = 6, K = 3;
    document.write(getMinSteps(N, K));
 
// This code is contributed by Surbhi Tyagi.
</script>

2