计算从给定整数N到0的最小步长

给定两个整数N和K ，其中K表示我们被允许直接从N减少N到N – K的跳跃数，我们的任务是按照给定的操作计算最小步数达到0：

• 我们可以从N跳出K个量，即N = N – K
• 将N减1，即N = N -1。

例子：

Input: N = 11, K = 4 
Output: 5 
Explanation: 
For the given value N we can perform the operation in the given sequence: 11 -> 7 -> 3 -> 2 -> 1 -> 0
Input: N = 6, K = 3 
Output: 2 
Explanation: 
For the given value N we can perform the operation in the given sequence: 6 -> 3 -> 0. 
 

方法：
为了解决上述问题，我们知道，将需要N / K的步长直接从值N跳到具有K和N％K的步长的最小可分值，以将其递减1，以将计数减少到0。因此，总和从N达到0所需的步数为

(N / K) + (N % K)

下面是上述方法的实现：

C++

// C++ program to Count the minimum steps
// to reach 0 from the given integer N
 
#include 
using namespace std;
 
// Function retuns 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

// Java program to count the minimum steps
// to reach 0 from the given integer N
class GFG{
 
// Function retuns 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

# Python3 program to Count the minimum steps
# to reach 0 from the given integer N
 
# Function retuns 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#

// C# program to count the minimum steps
// to reach 0 from the given integer N
using System;
 
class GFG{
 
// Function retuns 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

Javascript

// JavaScript program to Count the minimum steps
// to reach 0 from the given integer N
 
// Function retuns 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.

输出：

2