Given an integer A representing the external angle (in degrees) of a regular convex polygon, the task is to find the sum of the sides of the largest and smallest secondary polygons formed such that each edge of the secondary polygon is a chord of the primary polygon. If it is not possible to form such polygons, then print “-1”.
Examples:
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.
Approach: The idea is to first find the number of edges in the primary polygon and then, check whether it is possible to make secondary polygons or not. Follow the steps below to solve the problem:
- The sum of the external angle in a regular polygon is 360 degrees. Therefore, the number of sides = 360 / external angle.
- The number of sides of the maximum secondary polygon is the number of sides of the primary polygon / 2.
- As polygon is possible if the count of edges is at least 3, a secondary polygon is possible if the edges of the initial polygon ? 6.
- The smallest possible polygon has 3 edges always.
- Print the sum of number of sides in both the largest and smallest polygons.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>using namespace std;// Function to find the sum of largest and// smallest secondary polygons if possiblevoid 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 Codeint 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 approachimport java.io.*;import java.util.*;class GFG {// Function to find the sum of largest and// smallest secondary polygons if possiblestatic 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 Codepublic 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 possibledef 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 Codeif __name__ == '__main__': # Given Exterior Angle Angle = 45 print(secondary_polygon(Angle)) |
C#
// C# program for the above approachusing 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
<script>// JavaScript program for the above approach// Function to find the sum of largest and// smallest secondary polygons if possiblefunction secondary_polygon(Angle){ // Count edges of primary polygon var edges_primary = 360/Angle; if (edges_primary >= 6) { // Calculate edges present in // the largest secondary polygon var edges_max_secondary = edges_primary / 2; document.write(edges_max_secondary + 3); } else document.write("Not Possible");}// Driver Code//Given Exterior Anglevar Angle = 45;secondary_polygon(Angle);// This code is contributed by 29AjayKumar </script> |
7
Time Complexity: O(1)
Auxiliary Space: O(1)
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