检查Bishop是否可以删除Pawn

给定Bishop和Pawn在8 * 8棋盘上的位置，任务是检查Bishop是否可以一举将Pawn放下。 Bishop和Pawn的位置用棋盘的行号和列号表示。

例子：

Input: bishopX = 5, bishopY = 5, pawnX = 1, pawnY = 1
Output: Yes

Input: bishopX = 5, bishopY = 5, pawnX = 1, pawnY = 3
Output: No

方法：在象棋游戏中，主教只能沿对角线移动，并且每次移动的距离都没有限制。如果我们以Bishop的位置为原点，可以得出结论，Bishop只能以45度， 135度， 225度和315度角进行攻击。因此，主教只有在处于从其位置获得的倾斜等于45度或135度或225度或315度角的切线的位置时，才可以攻击Pawn。

我们知道两点的斜率方程

还，

对于Bishop的所有迎角(即45°，135°，225°和315°)，坡度值将为1或-1。

tan(45°) = 1
tan(135°) = -1
tan(225°) = 1
tan(315°) = -1

因此，对于45°和225°角，斜率方程为

x2 – x1 = y2 – y1

对于135°和315°角，斜率方程为

-x2 + x1 = y2 – y1

1. 以Bishop和Pawn的位置为二维坐标上的点，请检查两个方程式是否为真
   • x2 – x1 = y2 – y1
   • -x2 + x1 = y2 – y1
2. 否则打印No。

下面是上述方法的实现：

C++

// C++ implementation of above approach
#include 
using namespace std;
  
// Function that return true if
// the Bishop can take down the pawn
bool canTakeDown(int bishopX, int bishopY, int pawnX, int pawnY)
{
  
    // If pawn is at angle
    // 45 or 225 degree from
    // bishop's Position
    if (pawnX - bishopX == pawnY - bishopY)
        return true;
  
    // If pawn is at angle
    // 135 or 315 degree from
    // bishop's Position
    else if (-pawnX + bishopX == pawnY - bishopY)
        return true;
  
    else
        return false;
}
  
// Driver code
int main()
{
    // Bishop's Position
    int bishopX = 5, bishopY = 5;
  
    // Pawn's Position
    int pawnX = 1, pawnY = 1;
  
    if (canTakeDown(bishopX, bishopY, pawnX, pawnY))
        cout << "Yes";
    else
        cout << "No";
}

Java

// Java implementation of above approach
class GFG {
  
    // Function that return true if
    // the Bishop can take down the pawn
    static boolean canTakeDown(int bishopX, int bishopY, int pawnX, int pawnY)
    {
  
        // If pawn is at angle
        // 45 or 225 degree from
        // bishop's Position
        if (pawnX - bishopX == pawnY - bishopY)
            return true;
  
        // If pawn is at angle
        // 135 or 315 degree from
        // bishop's Position
        else if (-pawnX + bishopX == pawnY - bishopY)
            return true;
  
        else
            return false;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        // Bishop's Position
        int bishopX = 5, bishopY = 5;
  
        // Pawn's Position
        int pawnX = 1, pawnY = 1;
  
        if (canTakeDown(bishopX, bishopY, pawnX, pawnY))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

Python3

# Python3 implementation of above approach
  
# Function that return true if 
# the Bishop can take down the pawn
def canTakeDown(bishopX, bishopY, pawnX, pawnY) :
  
  
    # If pawn is at angle 
    # 45 or 225 degree from
    # bishop's Position
    if (pawnX - bishopX == pawnY - bishopY) :
        return True
      
    # If pawn is at angle 
    # 135 or 315 degree from
    # bishop's Position
    elif (-pawnX + bishopX == pawnY - bishopY):
        return True
      
    else:
        return False
  
  
# Driver code
  
# Bishop's Position
bishopX = 5
bishopY = 5
  
# Pawn's Position
pawnX = 1
pawnY = 1
  
if (canTakeDown(bishopX, bishopY, pawnX, pawnY)) :
    print("Yes")
else :
    print("No")

C#

// C# implementation of above approach
using System;
class GFG {
  
    // Function that return true if
    // the Bishop can take down the pawn
    static bool canTakeDown(int bishopX, int bishopY, int pawnX, int pawnY)
    {
  
        // If pawn is at angle
        // 45 or 225 degree from
        // bishop's Position
        if (pawnX - bishopX == pawnY - bishopY)
            return true;
  
        // If pawn is at angle
        // 135 or 315 degree from
        // bishop's Position
        else if (-pawnX + bishopX == pawnY - bishopY)
            return true;
  
        else
            return false;
    }
  
    // Driver code
    public static void Main()
    {
        // Bishop's Position
        int bishopX = 5, bishopY = 5;
  
        // Pawn's Position
        int pawnX = 1, pawnY = 1;
  
        if (canTakeDown(bishopX, bishopY, pawnX, pawnY))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}

输出：

Yes