Given an integer N and a pizza which can be cut into pieces, each cut should be a straight line going from the center of the pizza to its border. Also, the angle between any two cuts must be a positive integer. Two pieces are equal if their appropriate angles are equal. The given pizza can be cut in following three ways:
- Cut the pizza into N equal pieces.
- Cut the pizza into N pieces of any size.
- Cut the pizza into N pieces such that no two of them are equal.
The task is to find if it is possible to cut the pizza in the above ways for a given value of N. Print 1 if possible else 0 for all the cases i.e. print 111 if all the cases are possible.
Examples:
Input: N = 4
Output: 1 1 1
Explanation:
Case 1: All four pieces can have angle = 90
Case 2: Same cut as Case 1
Case 3: 1, 2, 3 and 354 are the respective angles of the four pieces cut.Input: N = 7
Output: 0 1 1
Approach:
- Case 1 will only be possible if 360 is divisible by N.
- For case 2 to be possible, N must be ? 360.
- An ideal solution for case 3 would be to choose pieces in such a way that the angles they form are 1, 2, 3, … respectively. So, in order for this case to be possible, (N * (N + 1)) / 2 must be ? 360.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <iostream> using namespace std;
// Function to check if it is possible // to cut the pizza in the given way void cutPizza( int n)
{ // Case 1
cout << (360 % n == 0) ? "1" : "0" ;
// Case 2
cout << (n <= 360) ? "1" : "0" ;
// Case 3
cout << (((n * (n + 1)) / 2) <= 360) ? "1" : "0" ;
} // Driver code int main()
{ int n = 7;
cutPizza(n);
return 0;
} |
// Java implementation of the approach class GFG
{ // Function to check if it is possible // to cut the pizza in the given way static void cutPizza( int n)
{ // Case 1
System.out.print( ( 360 % n == 0 ) ? "1" : "0" );
// Case 2
System.out.print( (n <= 360 ) ? "1" : "0" );
// Case 3
System.out.print( (((n * (n + 1 )) / 2 ) <= 360 ) ? "1" : "0" );
} // Driver code public static void main(String args[])
{ int n = 7 ;
cutPizza(n);
} } // This code is contributed by Arnab Kundu |
# Python3 implementation of the approach # Function to check if it is possible # to cut the pizza in the given way def cutPizza(n):
# Case 1
if ( 360 % n = = 0 ):
print ( "1" , end = "")
else :
print ( "0" , end = "");
# Case 2
if (n < = 360 ):
print ( "1" , end = "")
else :
print ( "0" , end = "");
# Case 3
if (((n * (n + 1 )) / 2 ) < = 360 ):
print ( "1" , end = "")
else :
print ( "0" , end = "");
# Driver code n = 7 ;
cutPizza(n); # This code is contributed # by Akanksha Rai |
// C# implementation of the approach using System;
class GFG
{ // Function to check if it is possible // to cut the pizza in the given way static void cutPizza( int n)
{ // Case 1
Console.Write((360 % n == 0) ? "1" : "0" );
// Case 2
Console.Write((n <= 360) ? "1" : "0" );
// Case 3
Console.Write((((n * (n + 1)) / 2) <= 360) ? "1" : "0" );
} // Driver code public static void Main(String []args)
{ int n = 7;
cutPizza(n);
} } // This code is contributed by Arnab Kundu |
<?php // PHP implementation of the approach // Function to check if it is possible // to cut the pizza in the given way function cutPizza( $n )
{ // Case 1
echo (360 % $n == 0) ? "1" : "0" ;
// Case 2
echo ( $n <= 360) ? "1" : "0" ;
// Case 3
echo ((( $n * ( $n + 1)) / 2) <= 360) ? "1" : "0" ;
} // Driver code $n = 7;
cutPizza( $n );
// This code is contributed // by Akanksha Rai ?> |
<script> // Javascript implementation of the approach // Function to check if it is possible // to cut the pizza in the given way function cutPizza(n)
{ // Case 1
document.write( (360 % n == 0) ? "1" : "0" );
// Case 2
document.write( (n <= 360) ? "1" : "0" );
// Case 3
document.write( (((n * (n + 1)) / 2) <= 360) ? "1" : "0" );
} // Driver code let n = 7; cutPizza(n); // This code is contributed by avanitrachhadiya2155 </script> |
011
Time Complexity: O(1)
Auxiliary Space: O(1)