A circle is given with k equidistant points on its circumference. 2 points A and B are given in the circle. Find the count of all obtuse angles (angles larger than 90 degree) formed from /_ACB, where C can be any point in circle other than A or B.
Note :
A and B are not equal.
A < B.
Points are between 1 and K(both inclusive).
Examples :
Input : K = 6, A = 1, B = 3. Output : 1 Explanation : In the circle with 6 equidistant points, when C = 2 i.e. /_123, we get obtuse angle. Input : K = 6, A = 1, B = 4. Output : 0 Explanation : In this circle, there is no such C that form an obtuse angle.
It can be observed that if A and B have equal elements in between them, there can’t be any C such that ACB is obtuse. Also, the number of possible obtuse angles are the smaller arc between A and B.
Below is the implementation :
C++
// C++ program to count number of obtuse// angles for given two points.#include <bits/stdc++.h>using namespace std;int countObtuseAngles(int a, int b, int k){ // There are two arcs connecting a // and b. Let us count points on // both arcs. int c1 = (b - a) - 1; int c2 = (k - b) + (a - 1); // Both arcs have same number of // points if (c1 == c2) return 0; // Points on smaller arc is answer return min(c1, c2);}// Driver codeint main(){ int k = 6, a = 1, b = 3; cout << countObtuseAngles(a, b, k); return 0;} |
Java
// Java program to count number of obtuse// angles for given two pointsclass GFG { static int countObtuseAngles(int a, int b, int k) { // There are two arcs connecting a // and b. Let us count points on // both arcs. int c1 = (b - a) - 1; int c2 = (k - b) + (a - 1); // Both arcs have same number of // points if (c1 == c2) return 0; // Points on smaller arc is answer return min(c1, c2); } // Driver Program to test above function public static void main(String arg[]) { int k = 6, a = 1, b = 3; System.out.print(countObtuseAngles(a, b, k)); }}// This code is contributed by Anant Agarwal. |
Python
# C++ program to count number of obtuse# angles for given two points.def countObtuseAngles( a, b, k): # There are two arcs connecting a # and b. Let us count points on # both arcs. c1 = (b - a) - 1 c2 = (k - b) + (a - 1) # Both arcs have same number of # points if (c1 == c2): return 0 # Points on smaller arc is answer return min(c1, c2) # Driver codek, a, b = 6, 1, 3print countObtuseAngles(a, b, k)# This code is contributed by Sachin Bisht |
C#
// C# program to count number of obtuse// angles for given two pointsusing System;class GFG { static int countObtuseAngles(int a, int b, int k) { // There are two arcs connecting // a and b. Let us count points // on both arcs. int c1 = (b - a) - 1; int c2 = (k - b) + (a - 1); // Both arcs have same number // of points if (c1 == c2) return 0; // Points on smaller arc is // answer return Math.Min(c1, c2); } // Driver Program to test above // function public static void Main() { int k = 6, a = 1, b = 3; Console.WriteLine( countObtuseAngles(a, b, k)); }}// This code is contributed by vt_m. |
PHP
<?php// PHP program to count number // of obtuse angles for given // two points.function countObtuseAngles($a, $b, $k){ // There are two arcs connecting a // and b. Let us count points on // both arcs. $c1 = ($b - $a) - 1; $c2 = ($k - $b) + ($a - 1); // Both arcs have same number of // points if ($c1 == $c2) return 0; // Points on smaller arc is answer return min($c1, $c2);}// Driver code$k = 6; $a = 1; $b = 3;echo countObtuseAngles($a, $b, $k);// This code is contributed by aj_36?> |
Javascript
<script>// Javascript program to count number of obtuse// angles for given two points function countObtuseAngles(a , b , k) { // There are two arcs connecting a // and b. Let us count points on // both arcs. var c1 = (b - a) - 1; var c2 = (k - b) + (a - 1); // Both arcs have same number of // points if (c1 == c2) return 0; // Points on smaller arc is answer return Math.min(c1, c2); } // Driver Program to test above function var k = 6, a = 1, b = 3; document.write(countObtuseAngles(a, b, k));// This code is contributed by todaysgaurav </script> |
Output :
1
Time Complexity: O(1)
Auxiliary Space: O(1)
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