通过最小操作数X加1或2形成N，其中X可被M整除

给定数字N，任务是通过在最小操作数X中加1或2来形成N(从0开始)，以使X可被M整除。
例子：

Input: N = 10, M = 2 
Output: X = 6 
Explanation: 
Taken operations are 2 2 2 2 1 1 
X = 6 which is divisible by 2
Input: N = 17, M = 4 
Output: 8 
 

方法：

• 由于我们一次可以采取1步或2步，因此我们可以说最小否。采取的步骤数为n / 2，最大值为。步长为n，与否无关。的步数可被m整除。
• 因此，我们必须计算n / 2步才能获得最少的步数。现在，如果n为偶数，则最小步数将为n / 2，但是如果它为奇数，则它将为n / 2 + 1，而与否无关。的步数可被m整除。要使最小步长为m的倍数，我们可以执行floor((minimum_steps + m – 1)/ m)* m
• 同样，如果n小于m，则不可能找到最小步数，在这种情况下，我们将返回-1。

下面是上述方法的实现：

C++

// C++ program to find minimum
// number of steps to cover distance x
 
#include 
using namespace std;
 
// Function to calculate the minimum number of steps required
// total steps taken is divisible
// by m and only 1 or 2 steps can be taken at // a time
int minsteps(int n, int m)
{
 
    // If m > n ans is -1
    if (m > n) {
        return -1;
    }
    // else discussed above approach
    else {
        return ((n + 1) / 2 + m - 1) / m * m;
    }
}
 
// Driver code
int main()
{
    int n = 17, m = 4;
    int ans = minsteps(n, m);
    cout << ans << '\n';
 
    return 0;
}

Java

// Java program to find minimum
// number of steps to cover distance x
class GFG
{
 
    // Function to calculate the
    // minimum number of steps required
    // total steps taken is divisible
    // by m and only 1 or 2 steps can be
    // taken at // a time
    static int minsteps(int n, int m)
    {
     
        // If m > n ans is -1
        if (m > n)
        {
            return -1;
        }
         
        // else discussed above approach
        else
        {
            return ((n + 1) / 2 + m - 1) / m * m;
        }
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int n = 17, m = 4;
        int ans = minsteps(n, m);
        System.out.println(ans);
    }
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 program to find minimum
# number of steps to cover distance x
 
# Function to calculate the minimum number of
# steps required total steps taken is divisible
# by m and only 1 or 2 steps can be taken at a time
def minsteps(n, m):
 
    # If m > n ans is -1
    if (m > n):
        return -1
         
    # else discussed above approach
    else :
        return ((n + 1) // 2 + m - 1) // m * m;
 
# Driver code
n = 17
m = 4
ans = minsteps(n, m)
print(ans)
 
# This code is contributed by Mohit Kumar

C#

// C# program to find minimum
// number of steps to cover distance x
using System;
     
class GFG
{
 
    // Function to calculate the
    // minimum number of steps required
    // total steps taken is divisible
    // by m and only 1 or 2 steps can be
    // taken at // a time
    static int minsteps(int n, int m)
    {
     
        // If m > n ans is -1
        if (m > n)
        {
            return -1;
        }
         
        // else discussed above approach
        else
        {
            return ((n + 1) / 2 + m - 1) / m * m;
        }
    }
     
    // Driver code
    public static void Main (String[] args)
    {
        int n = 17, m = 4;
        int ans = minsteps(n, m);
        Console.WriteLine(ans);
    }
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// javascript program to find minimum
// number of steps to cover distance x
 
// Function to calculate the
// minimum number of steps required
// total steps taken is divisible
// by m and only 1 or 2 steps can be
// taken at // a time
function minsteps(n , m)
{
 
    // If m > n ans is -1
    if (m > n)
    {
        return -1;
    }
     
    // else discussed above approach
    else
    {
        return ((n + 1) / 2 + m - 1) / m * m;
    }
}
     
// Driver code
 
var n = 17, m = 4;
var ans = minsteps(n, m);
document.write(ans);
 
// This code contributed by shikhasingrajput
 

输出：

12

时间复杂度： O(1)