从(1，1)开始达到N的最小移动次数

给定一个整数N和一个无限表，其中^第i行和^第j列包含值i * j 。任务是找到从单元(1、1)开始到达包含N的单元的最小移动次数。

注意：从(i，j)开始，只有有效的移动是(i + 1，j)和(i，j + 1)

例子：

Input: N = 10
Output: 5
(1, 1) -> (2, 1) -> (2, 2) -> (2, 3) -> (2, 4) -> (2, 5)

Input: N = 7
Output: 6

方法：请注意，可以通过i + j – 2个步骤来到达任何单元格(i，j) 。因此，只需要i * j = N的对(i，j) ，即可使i + j最小。可以通过找到所有可能的对(i，j)并在O(＆Sqrt; N)中进行检查来找到它。要做到这一点，不失一般性，可以假设i≤j和i≤＆的Sqrt; N，因为N = I *Ĵ≥I ^2。因此＆Sqrt; N≥i ^2，即＆Sqrt; N≥i 。
因此，迭代i的所有可能值从1到＆Sqrt; N，并在所有可能的对(i，j)中，选择i + j – 2的最小值，这是必需的答案。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the minimum number
// of moves required to reach the cell
// containing N starting from (1, 1)
int min_moves(int n)
{
    // To store the required answer
    int ans = INT_MAX;
  
    // For all possible values of divisors
    for (int i = 1; i * i <= n; i++) {
  
        // If i is a divisor of n
        if (n % i == 0) {
  
            // Get the moves to reach n
            ans = min(ans, i + n / i - 2);
        }
    }
  
    // Return the required answer
    return ans;
}
  
// Driver code
int main()
{
    int n = 10;
  
    cout << min_moves(n);
  
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
      
// Function to return the minimum number
// of moves required to reach the cell
// containing N starting from (1, 1)
static int min_moves(int n)
{
    // To store the required answer
    int ans = Integer.MAX_VALUE;
  
    // For all possible values of divisors
    for (int i = 1; i * i <= n; i++) 
    {
  
        // If i is a divisor of n
        if (n % i == 0)
        {
  
            // Get the moves to reach n
            ans = Math.min(ans, i + n / i - 2);
        }
    }
  
    // Return the required answer
    return ans;
}
  
// Driver code
public static void main(String[] args)
{
    int n = 10;
  
    System.out.println(min_moves(n));
}
}
  
// This code is contributed by Code_Mech

Python3

# Python3 implementation of the approach 
import sys
  
from math import sqrt
  
# Function to return the minimum number 
# of moves required to reach the cell 
# containing N starting from (1, 1) 
def min_moves(n) :
  
    # To store the required answer 
    ans = sys.maxsize; 
  
    # For all possible values of divisors 
    for i in range(1, int(sqrt(n)) + 1) :
  
        # If i is a divisor of n 
        if (n % i == 0) :
  
            # Get the moves to reach n 
            ans = min(ans, i + n // i - 2);
  
    # Return the required answer 
    return ans; 
  
# Driver code 
if __name__ == "__main__" :
  
    n = 10; 
  
    print(min_moves(n)); 
  
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
      
class GFG
{
      
// Function to return the minimum number
// of moves required to reach the cell
// containing N starting from (1, 1)
static int min_moves(int n)
{
    // To store the required answer
    int ans = int.MaxValue;
  
    // For all possible values of divisors
    for (int i = 1; i * i <= n; i++) 
    {
  
        // If i is a divisor of n
        if (n % i == 0)
        {
  
            // Get the moves to reach n
            ans = Math.Min(ans, i + n / i - 2);
        }
    }
  
    // Return the required answer
    return ans;
}
  
// Driver code
public static void Main(String[] args)
{
    int n = 10;
  
    Console.WriteLine(min_moves(n));
}
}
  
// This code is contributed by 29AjayKumar

输出：

5