求曼哈顿距离至少为N的整数(x，y)

给定数字N，任务是找到整数点(x，y)，以使0 <= x，y <= N，并且任意两个点之间的曼哈顿距离将至少为N。

例子：

Input: N = 3
Output: (0, 0) (0, 3) (3, 0) (3, 3)

Input: N = 4
Output: (0, 0) (0, 4) (4, 0) (4, 4) (2, 2)

方法：

• 曼哈顿(x ₁ ，y ₁ )和(x ₂ ，y ₂ )两点之间的距离是：
  | x ₁ – x ₂ | + | y ₁ – y ₂ |
• 对于所有双点，此距离至少为N。
• 由于0 <= x <= N并且0 <= y <= N，所以我们可以想象一个边长为N的正方形，其左下角为(0，0)，右上角为(N，N)。
• 因此，如果我们在此角上放置4个点，则曼哈顿距离将至少为N。
• 现在，由于必须最大化点的数量，因此必须检查正方形内是否有任何可用的点。
• 如果N是偶数，则(N / 2，N / 2)的平方的中点是整数点，否则它将是浮点值，因为当N为奇数时，N / 2不是整数。
• 因此，唯一可用的位置是中间点，只有在N为偶数的情况下，我们才能在该位置放置一个点。
• 因此，如果N为奇数，则点数将为4，如果N为偶数，则点数将为5。

下面是上述方法的实现：

C++

// C++ code to Find the integer points (x, y)
// with Manhattan distance atleast N
  
#include 
using namespace std;
  
// C++ function to find all possible point
vector > FindPoints(int n)
{
  
    vector > v;
  
    // Find all 4 corners of the square
    // whose side length is n
    v.push_back({ 0, 0 });
    v.push_back({ 0, n });
    v.push_back({ n, 0 });
    v.push_back({ n, n });
  
    // If n is even then the middle point
    // of the square will be an integer,
    // so we will take that point
    if (n % 2 == 0)
        v.push_back({ n / 2, n / 2 });
  
    return v;
}
  
// Driver Code
int main()
{
  
    int N = 8;
  
    vector > v
        = FindPoints(N);
  
    // Printing all possible points
    for (auto i : v) {
        cout << "(" << i.first << ", "
             << i.second << ") ";
    }
    return 0;
}

Java

// Java code to Find the integer points (x, y)
// with Manhattan distance atleast N
import java.util.*;
  
class GFG
{
  
static class pair
{ 
    int first, second; 
    public pair(int first, int second) 
    { 
        this.first = first; 
        this.second = second; 
    } 
}
  
// Java function to find all possible point
static Vector FindPoints(int n)
{
    Vector v = new Vector();
  
    // Find all 4 corners of the square
    // whose side length is n
    v.add(new pair( 0, 0 ));
    v.add(new pair( 0, n ));
    v.add(new pair( n, 0 ));
    v.add(new pair( n, n ));
  
    // If n is even then the middle point
    // of the square will be an integer,
    // so we will take that point
    if (n % 2 == 0)
        v.add(new pair( n / 2, n / 2 ));
  
    return v;
}
  
// Driver Code
public static void main(String[] args)
{
    int N = 8;
  
    Vector v = FindPoints(N);
  
    // Printing all possible points
    for (pair i : v) 
    {
        System.out.print("(" + i.first + ", " + 
                               i.second + ") ");
    }
}
}
  
// This code is contributed by PrinciRaj1992

Python3

# Python3 code to Find the integer points (x, y) 
# with Manhattan distance atleast N 
  
# function to find all possible point 
def FindPoints(n) :
  
    v = []; 
  
    # Find all 4 corners of the square 
    # whose side length is n 
    v.append([ 0, 0 ]); 
    v.append([ 0, n ]); 
    v.append([ n, 0 ]); 
    v.append([ n, n ]); 
  
    # If n is even then the middle point 
    # of the square will be an integer, 
    # so we will take that point 
    if (n % 2 == 0) :
        v.append([ n // 2, n // 2 ]); 
  
    return v; 
  
# Driver Code 
if __name__ == "__main__" : 
  
    N = 8; 
  
    v = FindPoints(N); 
  
    # Printing all possible points 
    for element in v :
        print("(", element[0], 
              ",", element[1], ")", end = " "); 
  
# This code is contributed by AnkitRai01

C#

// C# code to Find the integer points (x, y)
// with Manhattan distance atleast N
using System;
using System.Collections.Generic;
  
class GFG
{
  
class pair
{ 
    public int first, second; 
    public pair(int first, int second) 
    { 
        this.first = first; 
        this.second = second; 
    } 
}
  
// Function to find all possible point
static List FindPoints(int n)
{
    List v = new List();
  
    // Find all 4 corners of the square
    // whose side length is n
    v.Add(new pair( 0, 0 ));
    v.Add(new pair( 0, n ));
    v.Add(new pair( n, 0 ));
    v.Add(new pair( n, n ));
  
    // If n is even then the middle point
    // of the square will be an integer,
    // so we will take that point
    if (n % 2 == 0)
        v.Add(new pair( n / 2, n / 2 ));
  
    return v;
}
  
// Driver Code
public static void Main(String[] args)
{
    int N = 8;
  
    List v = FindPoints(N);
  
    // Printing all possible points
    foreach (pair i in v) 
    {
        Console.Write("(" + i.first + ", " + 
                            i.second + ") ");
    }
}
}
  
// This code is contributed by Rajput-Ji

输出：

(0, 0) (0, 8) (8, 0) (8, 8) (4, 4)