X 圆和 Y 直线之间可能的最大交点

给定两个整数X和Y ，任务是找到X圆和Y直线之间可能的最大交点数。

例子：

Input: X = 4, Y = 4 
Output: 50 
Explanation: 
4 lines intersect each other at 6 points and 4 circles intersect each other at maximum of 12 points. 
Each line intersects 4 circles at 8 points. 
Hence, 4 lines intersect four circles at a maximum of 32 points. 
Thus, required number of intersections = 6 + 12 + 32 = 50.

Input: X = 3, Y = 4 
Output: 36 
 

方法：
可以观察到，存在三种类型的交叉点：

1. 从 X 个圆中选择一对点的方法数是$\binom{X}{2}$ .每个这样的对最多相交两点。
2. 从 Y 线上选择一对点的方法数是$\binom{Y}{2}$ .每对这样的对至多在一点相交。
3. 从 X 圆和 Y 线中选择一个圆和一条线的方法数是$X*Y$ .每个这样的对最多在两个点上相交。

So, the maximum number of point of intersection can be calculated as: 
=> $2*\binom{X}{2} + \binom{Y}{2} + 2*X*Y$
=> $X*(X - 1) + \frac{Y*(Y - 1)}{2} + 2*X*Y$
 

因此，求X圆与Y直线的最大交点数的公式为：
$\frac{Y*(Y - 1)}{2} + X*(2*Y + X - 1)$

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
int maxPointOfIntersection(int x, int y)
{
    int k = y * (y - 1) / 2;
    k = k + x * (2 * y + x - 1);
    return k;
}
 
// Driver code
int main()
{
     
    // Number of circles
    int x = 3;
     
    // Number of straight lines
    int y = 4;
 
    // Function Call
    cout << (maxPointOfIntersection(x, y));
}
 
// This code is contributed by Ritik Bansal

Java

// Java program to implement
// the above approach
class GFG{
     
static int maxPointOfIntersection(int x, int y)
{
    int k = y * (y - 1) / 2;
    k = k + x * (2 * y + x - 1);
    return k;
}
 
// Driver code
public static void main(String[] args)
{
     
    // Number of circles
    int x = 3;
     
    // Number of straight lines
    int y = 4;
 
    // Function Call
    System.out.print(maxPointOfIntersection(x, y));
}
}
 
// This code is contributed by Princi Singh

Python3

# Python3 program to implement
# the above approach
def maxPointOfIntersection(x, y):
    k = y * ( y - 1 ) // 2
    k = k + x * ( 2 * y + x - 1 )
    return k
 
# Number of circles
x = 3
# Number of straight lines
y = 4
 
# Function Call
print(maxPointOfIntersection(x, y))

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
     
static int maxPointOfIntersection(int x, int y)
{
    int k = y * (y - 1) / 2;
    k = k + x * (2 * y + x - 1);
    return k;
}
 
// Driver code
public static void Main(String[] args)
{
     
    // Number of circles
    int x = 3;
     
    // Number of straight lines
    int y = 4;
 
    // Function Call
    Console.Write(maxPointOfIntersection(x, y));
}
}
 
// This code is contributed by Princi Singh

Javascript

 
// Javascript program to implement
// the above approach
function maxPointOfIntersection(x, y)
{
    let k = y * (y - 1) / 2;
    k = k + x * (2 * y + x - 1);
    return k;
}
 
// Driver code
 
// Number of circles
let x = 3;
   
// Number of straight lines
let y = 4;
 
// Function Call
document.write(maxPointOfIntersection(x, y));
 
// This code is contributed by rameshtravel07
 

输出：

36

时间复杂度： O(1)
辅助空间： O(1)