计算主教一口气可以访问的广场总数

给定主教在8 * 8棋盘上的位置，任务是计算主教一举可以访问的正方形总数。主教的位置用棋盘的行号和列号表示。

例子：

Input: Row = 4, Column = 4
Output: 13

Input: Row = 1, Column = 1
Output: 7

方法：在国际象棋游戏中，主教只能对角移动，并且每次移动的距离都没有限制。

因此，我们也可以说Bishop可以以四种方式移动，即从当前位置对角线左上，右上，左下和右下。

我们可以通过以下方式计算每次移动中访问的平方数：

Total squares visited in Top Left move = min(r, c) – 1
Total squares visited in Top Right move = min(r, 9 – c) – 1
Total squares visited in Bottom Left move = 8 – max(r, 9 – c)
Total squares visited in Bottom Right move = 8 – max(r, c)
where, r and c are the coordinates of the current position of the Bishop on the chessboard.

下面是上述方法的实现：

C++

// C++ implementation of above approach
#include 
using namespace std;
  
// Function to return the count of
// total positions the Bishop
// can visit in a single move
int countSquares(int row, int column)
{
  
    // Count top left squares
    int topLeft = min(row, column) - 1;
  
    // Count bottom right squares
    int bottomRight = 8 - max(row, column);
  
    // Count top right squares
    int topRight = min(row, 9 - column) - 1;
  
    // Count bottom left squares
    int bottomLeft = 8 - max(row, 9 - column);
  
    // Return total count
    return (topLeft + topRight + bottomRight + bottomLeft);
}
  
// Driver code
int main()
{
  
    // Bishop's Position
    int row = 4, column = 4;
  
    cout << countSquares(row, column);
  
    return 0;
}

Java

// Java implementation of above approach
class GFG {
  
    // Function to return the count of
    // total positions the Bishop
    // can visit in a single move
    static int countSquares(int row, int column)
    {
  
        // Count top left squares
        int topLeft = Math.min(row, column) - 1;
  
        // Count bottom right squares
        int bottomRight = 8 - Math.max(row, column);
  
        // Count top right squares
        int topRight = Math.min(row, 9 - column) - 1;
  
        // Count bottom left squares
        int bottomLeft = 8 - Math.max(row, 9 - column);
  
        // Return total count
        return (topLeft + topRight + bottomRight + bottomLeft);
    }
  
    // Driver code
    public static void main(String[] args)
    {
  
        // Bishop's Position
        int row = 4, column = 4;
  
        System.out.println(countSquares(row, column));
    }
}

C#

// C# implementation of above approach
using System;
class GFG {
  
    // Function to return the count of
    // total positions the Bishop
    // can visit in a single move
    static int countSquares(int row, int column)
    {
  
        // Count top left squares
        int topLeft = Math.Min(row, column) - 1;
  
        // Count bottom right squares
        int bottomRight = 8 - Math.Max(row, column);
  
        // Count top right squares
        int topRight = Math.Min(row, 9 - column) - 1;
  
        // Count bottom left squares
        int bottomLeft = 8 - Math.Max(row, 9 - column);
  
        // Return total count
        return (topLeft + topRight + bottomRight + bottomLeft);
    }
  
    // Driver code
    public static void Main()
    {
  
        // Bishop's Position
        int row = 4, column = 4;
  
        Console.WriteLine(countSquares(row, column));
    }
}

Python3

# Python3 implementation of above approach
  
# Function to return the count of 
# total positions the Bishop 
# can visit in a single move
def countSquares(row, column):
      
    # Count top left squares
    topLeft = min(row, column) - 1
      
    # Count bottom right squares
    bottomRight = 8 - max(row, column)
      
    # Count top right squares
    topRight = min(row, 9-column) -1
      
    # Count bottom left squares
    bottomLeft = 8 - max(row, 9-column) 
  
      
    # Return total count
    return (topLeft + topRight + bottomRight + bottomLeft)
  
# Driver code
  
# Bishop's Position 
row = 4
column = 4
  
print(countSquares(row, column))

输出：

13