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++ 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 code int main()
{ int k = 6, a = 1, b = 3;
cout << countObtuseAngles(a, b, k);
return 0;
} |
// Java program to count number of obtuse // angles for given two points 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 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. |
# 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 code k, a, b = 6 , 1 , 3
print countObtuseAngles(a, b, k)
# This code is contributed by Sachin Bisht |
<?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 ?> |
<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)