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))
PHP
Javascript
输出:
13
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