Given four integers N, A, K, n where N represents the number of sides the polygon, A represents the initial angle of the polygon, K represents the per angle increase, the task is to find the nth angle of the polygon having N sides. If it is not possible then print -1.
Examples:
Input: N = 3, A = 30, K = 30, n = 3
Output: 90
Explanation:
The 3rd angle of the polygon having three sides is 90 degree as the all angles are 30, 60, 90.Input: N = 4, A = 40, K = 30, n = 3
Output: -1
Explanation:
It is not possible to create that polygon whose initial angle is 40 and no. of sides are 4.
Approach: The idea is to observe that the angles of the polygon are increasing in the terms of Arithmetic Progression by a difference of K degree.
- Find out the angular sum of the N sided polygon and the sum of the angles of the given polygon.
- Check if both the values are equal or not. If yes, then the nth angle is possible hence find the nth angle.
- Otherwise, print -1.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <iostream>using namespace std;// Function to check if the angle// is possible or notbool possible(int N, int a, int b, int n){ // Angular sum of a N-sided polygon int sum_of_angle = 180 * (N - 2); // Angular sum of N-sided given polygon int Total_angle = (N * ((2 * a) + (N - 1) * b)) / 2; // Check if both sum are equal if (sum_of_angle != Total_angle) return false; else return true;}// Function to find the nth angleint nth_angle(int N, int a, int b, int n){ int nth = 0; // Calculate nth angle nth = a + (n - 1) * b; // Return the nth angle return nth;}// Driver Codeint main(){ int N = 3, a = 30, b = 30, n = 3; // Checks the possibility of the // polygon exist or not if (possible(N, a, b, n)) // Print nth angle // of the polygon cout << nth_angle(N, a, b, n); else cout << "Not Possible"; return 0;} |
Java
// Java program for the above approach class GFG{ // Function to check if the angle // is possible or not static boolean possible(int N, int a, int b, int n) { // Angular sum of a N-sided polygon int sum_of_angle = 180 * (N - 2); // Angular sum of N-sided given polygon int Total_angle = (N * ((2 * a) + (N - 1) * b)) / 2; // Check if both sum are equal if (sum_of_angle != Total_angle) return false; else return true; } // Function to find the nth angle static int nth_angle(int N, int a, int b, int n) { int nth = 0; // Calculate nth angle nth = a + (n - 1) * b; // Return the nth angle return nth; } // Driver Code public static void main(String[] args) { int N = 3, a = 30, b = 30, n = 3; // Checks the possibility of the // polygon exist or not if (possible(N, a, b, n)) // Print nth angle // of the polygon System.out.print(nth_angle(N, a, b, n)); else System.out.print("Not Possible"); }}// This code is contributed by amal kumar choubey |
Python3
# Python3 program for the above approach# Function to check if the angle# is possible or notdef possible(N, a, b, n): # Angular sum of a N-sided polygon sum_of_angle = 180 * (N - 2) # Angular sum of N-sided given polygon Total_angle = (N * ((2 * a) + (N - 1) * b)) / 2 # Check if both sum are equal if (sum_of_angle != Total_angle): return False else: return True# Function to find the nth angledef nth_angle(N, a, b, n): nth = 0 # Calculate nth angle nth = a + (n - 1) * b # Return the nth angle return nth# Driver Codeif __name__ == '__main__': N = 3 a = 30 b = 30 n = 3 # Checks the possibility of the # polygon exist or not if (possible(N, a, b, n)): # Print nth angle # of the polygon print(nth_angle(N, a, b, n)) else: print("Not Possible")# This code is contributed by Mohit Kumar |
C#
// C# program for the above approach using System;class GFG{ // Function to check if the angle // is possible or not static bool possible(int N, int a, int b, int n) { // Angular sum of a N-sided polygon int sum_of_angle = 180 * (N - 2); // Angular sum of N-sided given polygon int Total_angle = (N * ((2 * a) + (N - 1) * b)) / 2; // Check if both sum are equal if (sum_of_angle != Total_angle) return false; else return true; } // Function to find the nth angle static int nth_angle(int N, int a, int b, int n) { int nth = 0; // Calculate nth angle nth = a + (n - 1) * b; // Return the nth angle return nth; } // Driver Code public static void Main(string[] args) { int N = 3, a = 30, b = 30, n = 3; // Checks the possibility of the // polygon exist or not if (possible(N, a, b, n)) // Print nth angle // of the polygon Console.Write(nth_angle(N, a, b, n)); else Console.Write("Not Possible"); }} // This code is contributed by Ritik Bansal |
Javascript
<script>// JavaScript program for the above approach// Function to check if the angle// is possible or notfunction possible(N, a, b, n){ // Angular sum of a N-sided polygon let sum_of_angle = 180 * (N - 2); // Angular sum of N-sided given polygon let Total_angle = Math.floor((N * ((2 * a) + (N - 1) * b)) / 2); // Check if both sum are equal if (sum_of_angle != Total_angle) return false; else return true;}// Function to find the nth anglefunction nth_angle(N, a, b, n){ let nth = 0; // Calculate nth angle nth = a + (n - 1) * b; // Return the nth angle return nth;}// Driver Code let N = 3, a = 30, b = 30, n = 3; // Checks the possibility of the // polygon exist or not if (possible(N, a, b, n)) // Print nth angle // of the polygon document.write(nth_angle(N, a, b, n)); else document.write("Not Possible");// This code is contributed by Surbhi Tyagi.</script> |
Output:
90
Time Complexity: O(1)
Auxiliary Space: O(1)
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