给定多边形可能的最大和最小子多边形的边总和

给定一个整数A表示正凸多边形的外角（以度为单位），任务是找到形成的最大和最小次多边形的边之和，使得次多边形的每条边都是主多边形的弦多边形。如果不可能形成这样的多边形，则打印“-1” 。

例子：

Input: A = 45
Output: 7
Explanation:
The primary polygon is an Octagon of 8 sides.
Therefore, the smallest secondary polygon consists of 3 edges and the largest secondary polygon consists of 4 edges.

Sum of edges of the smallest secondary polygon + Edges of the largest secondary polygon = 3 + 4 = 7.

Input: A = 60
Output: 6
Explanation: The primary polygon is a Hexagon consisting of 6 sides. Therefore, the smallest secondary polygon consists of 3 edges and the largest secondary polygon consists of 3 edges.

编程需要懂一点英语

方法：这个想法是先找到主多边形的边数，然后检查是否可以制作辅助多边形。请按照以下步骤解决问题：

正多边形的外角和是360度。因此，边数 = 360 / 外角。
最大次多边形的边数是主多边形的边数/2。
多边形是可能的，如果边缘的计数是至少为3，次级多边形是可能的，如果初期多边形≥6的边缘。
最小的多边形总是有3 条边。
打印最大和最小多边形的边数总和。

下面是上述方法的实现：

C++

#include 
using namespace std;
 
// Function to find the sum of largest and
// smallest secondary polygons if possible
void secondary_polygon(int Angle)
{
 
    // Count edges of primary polygon
    int edges_primary = 360/Angle;
 
    if (edges_primary >= 6)
    {
 
        // Calculate edges present in
        // the largest secondary polygon
        int edges_max_secondary = edges_primary / 2;
 
        cout << edges_max_secondary + 3;
    }
    else
        cout << "Not Possible";
}
 
// Driver Code
int main()
{
   
  // Given Exterior Angle
  int Angle = 45;
  secondary_polygon(Angle);
 
  return 0;
}
 
// This code is contributed by mohit kumar 29.

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG
{
 
// Function to find the sum of largest and
// smallest secondary polygons if possible
static void secondary_polygon(int Angle)
{
 
    // Count edges of primary polygon
    int edges_primary = 360/Angle;
 
    if (edges_primary >= 6)
    {
 
        // Calculate edges present in
        // the largest secondary polygon
        int edges_max_secondary = edges_primary / 2;
 
        System.out.println(edges_max_secondary + 3);
    }
    else
        System.out.println("Not Possible");
}
 
 
// Driver Code
public static void main(String[] args)
{
   
    // Given Exterior Angle
  int Angle = 45;
  secondary_polygon(Angle);
}
}
 
// This code is contributed by code_hunt.

Python3

# Python3 program for the above approach:
 
# Function to find the sum of largest and
# smallest secondary polygons if possible
def secondary_polygon(Angle):
 
    # Count edges of primary polygon
    edges_primary = 360//Angle
 
    if edges_primary >= 6:
 
        # Calculate edges present in
        # the largest secondary polygon
        edges_max_secondary = edges_primary // 2
 
        return edges_max_secondary + 3
 
    else:
        return "Not Possible"
 
 
# Driver Code
if __name__ == '__main__':
 
    # Given Exterior Angle
    Angle = 45
    print(secondary_polygon(Angle))

C#

// C# program for the above approach
using System;
class GFG {
 
  // Function to find the sum of largest and
  // smallest secondary polygons if possible
  static void secondary_polygon(int Angle)
  {
 
    // Count edges of primary polygon
    int edges_primary = 360 / Angle;
 
    if (edges_primary >= 6) {
 
      // Calculate edges present in
      // the largest secondary polygon
      int edges_max_secondary = edges_primary / 2;
 
      Console.WriteLine(edges_max_secondary + 3);
    }
    else
      Console.WriteLine("Not Possible");
  }
 
  // Driver Code
  public static void Main(string[] args)
  {
 
    // Given Exterior Angle
    int Angle = 45;
    secondary_polygon(Angle);
  }
}
 
// This code is contributed by ukasp.

Javascript

输出：

时间复杂度： O(1)
辅助空间： O(1)

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