将所有 1 移动到给定方阵的单个单元格的最小步骤

给定一个奇数正整数N ，它表示填充有 1 的N*N方阵的大小，任务是找到将所有 1 移动到矩阵的单个单元格的最小步数，其中，在一个步骤中，任何1 可以移动到与其水平、垂直或对角相邻的任何单元格。

例子：

Input: N = 3
Output: 8
Explanation: All the 8 1s present in the boundary can be brought to the centre of the matrix in one step.
So total required steps = 8

Input: N = 367
Output: 16476832
Explanation: The minimum number of steps required to put all the cookies in exactly one block of the tray is 16476832. 

方法：可以根据以下观察解决问题。

To minimize the number of steps, all 1s should be moved to the centre of the matrix. 
Any cell of the matrix is a part of ith zone if it’s distance from the centre is i (horizontally, vertically or diagonally). 
All the elements from ith block can be moved to centre in i steps.

See the image below for a better idea about zones. (Here zones of a 7*7 matrix are shown)

按照下面提到的步骤使用上述观察解决问题：

• 可能的区域总数X = N/2 。
• 第 i 个区域中的单元格总数为2 * i * 4 = 8*i 。
• 因此，将第 i 个区域的所有 1 移动到中心所需的总步数为8*i*i 。
• 运行从i = 1 到 X的循环并且：
  • 使用上述公式计算第 i 个区域的总步数。
  • 将其与sum 相加。
• 返回最终总和作为答案。

下面是上述方法的实现。

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find the minimum number
// of steps required to put
// all the cookies in exactly one cell
int minSteps(int N)
{
    // Stores the minimum number of steps
    int res = 0;
 
    // Loop to iterate over
    // all the zones present in the matrix
    for (int i = 1; i <= N / 2; ++i) {
 
        // Steps to move all 1s of ith zone
        // to the centre of the matrix
        res += 8 * i * i;
    }
 
    // Return the minimum steps
    return res;
}
 
// Driver Code
int main()
{
    // Given input
    int N = 7;
 
    // Function Call
    cout << minSteps(N);
    return 0;
}

Java

// JAVA program for the above approach
import java.util.*;
class GFG
{
 
  // Function to find the minimum number
  // of steps required to put
  // all the cookies in exactly one cell
  public static int minSteps(int N)
  {
 
    // Stores the minimum number of steps
    int res = 0;
 
    // Loop to iterate over
    // all the zones present in the matrix
    for (int i = 1; i <= N / 2; ++i) {
 
      // Steps to move all 1s of ith zone
      // to the centre of the matrix
      res += 8 * i * i;
    }
 
    // Return the minimum steps
    return res;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
 
    // Given input
    int N = 7;
 
    // Function Call
    System.out.print(minSteps(N));
  }
}
 
// This code is contributed by Taranpreet

Python

# Python program for the above approach
 
# Function to find the minimum number
# of steps required to put
# all the cookies in exactly one cell
def minSteps(N):
 
    # Stores the minimum number of steps
    res = 0
 
    # Loop to iterate over
    # all the zones present in the matrix
    i = 1
    while(i <= N//2):
 
        # Steps to move all 1s of ith zone
        # to the centre of the matrix
        res += 8 * i * i
        i += 1
 
    # Return the minimum steps
    return res
 
# Driver Code
 
# Given input
N = 7
 
# Function Call
print(minSteps(N))
 
# This code is contributed by Samim Hossain Mondal.

C#

// C# program for the above approach
using System;
class GFG {
 
    // Function to find the minimum number
    // of steps required to put
    // all the cookies in exactly one cell
    static int minSteps(int N)
    {
 
        // Stores the minimum number of steps
        int res = 0;
 
        // Loop to iterate over
        // all the zones present in the matrix
        for (int i = 1; i <= N / 2; ++i) {
 
            // Steps to move all 1s of ith zone
            // to the centre of the matrix
            res += 8 * i * i;
        }
 
        // Return the minimum steps
        return res;
    }
 
    // Driver Code
    public static void Main()
    {
 
        // Given input
        int N = 7;
 
        // Function Call
        Console.Write(minSteps(N));
    }
}
 
// This code is contributed by Samim Hossain Mondal

Javascript

    // JavaScript program for the above approach
 
 
    // Function to find the minimum number
    // of steps required to put
    // all the cookies in exactly one cell
    const minSteps = (N) => {
        // Stores the minimum number of steps
        let res = 0;
 
        // Loop to iterate over
        // all the zones present in the matrix
        for (let i = 1; i <= parseInt(N / 2); ++i) {
 
            // Steps to move all 1s of ith zone
            // to the centre of the matrix
            res += 8 * i * i;
        }
 
        // Return the minimum steps
        return res;
    }
 
    // Driver Code
 
    // Given input
    let N = 7;
 
    // Function Call
    document.write(minSteps(N));
 
// This code is contributed by rakeshsahni
 

输出

112

时间复杂度： O(X)，其中 X 是矩阵中存在的区域数
辅助空间： O(1)