最大交点 n 个圆

给定一个数 n，我们需要找到 n 个圆相交的最大次数。
例子：

Input :  n = 2
Output : 2

Input :  n = 3
Output : 6

描述和推导

正如我们在上图中看到的，对于每对圆，最多可以有两个相交点。因此，如果我们有 n 个圆，那么可以有ⁿ C ₂对圆，其中每对圆都有两个交点。因此，由此我们可以得出结论，通过查看所有可能的圆对，可以为 n 个圆的最大交点数得出数学公式为2 * ⁿ C ₂ 。
2 * ⁿ C ₂ = 2 * n * (n – 1)/2 = n * (n-1)

C++

// CPP program to find maximum umber of
// intersections of n circles
#include 
using namespace std;
 
// Returns maximum number of intersections
int intersection(int n)
{
   return n * (n - 1);
}
 
int main()
{
    cout << intersection(3) << endl;
    return 0;
}
// This code is contributed by
// Manish Kumar Rai.

Java

// Java program to find maximum umber of
// intersections of n circles
import java.io.*;
 
public class GFG {
 
    // for the calculation of 2*(nC2)
    static int intersection(int n)
    {
       return n * (n - 1);
    }
 
    public static void main(String[] args) throws IOException
    {
        System.out.println(intersection(3));
    }
}
// This code is contributed by
// Manish Kumar Rai

Python3

# python program to find maximum umber of
# intersections of n circles
# Returns maximum number of intersections
def intersection(n):
     
   return n * (n - 1);
 
# Driver code
print(intersection(3))
 
# This code is contributed by Sam007

C#

// C# program to find maximum umber of
// intersections of n circles
using System;
class GFG {
 
    // for the calculation of 2*(nC2)
    static int intersection(int n)
    {
        return n * (n - 1);
    }
 
// Driver Code
public static void Main()
{
    Console.WriteLine(intersection(3));
}
 
}
 
// This code is contributed by Sam007

Javascript

// Javascript program to find maximum umber of
// intersections of n circles
 
// Returns maximum number of intersections
function intersection(n)
{
   return n * (n - 1);
}
  
// Driver code
document.write(intersection(3));
 
// This code is contributed by _saurabh_jaiswal
 

输出：

6

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。