Pizza Problem

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: 
 

1. Cut the pizza into N equal pieces.
2. Cut the pizza into N pieces of any size.
3. 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)