# Cake Distribution Problem

Given two integers N and M, where N is the number of friends sitting in a clockwise manner in a circle and M is the number of cakes. The task is to calculate the left number of cakes after distributing i cakes to i’th friend. The distribution of cakes will stop if the count of cakes is less than the required amount.

Examples:

Input: N = 4, M = 11
Output: 0
1st round:
The 1st friend gets 1 cake, 2nd gets 2 cakes,
3rd get 3 and 4th gets 4 cakes.
Remaining cakes = 11 – (1 + 2 + 3 + 4) = 1
2nd round:
This time only 1st friend gets the left 1 cake.
Remaining cakes = 1 – 1 = 0

Input: N = 3, M = 8
Output: 1
1st round:
The 1st friend gets 1 cake, 2nd gets 2 cakes,
and 3rd get 3 cakes.
Remaining cakes = 8 – (1 + 2 + 3) = 2
2nd round:
This time only 1st friend gets the left 1 cake,
and then there is no cake left for 2nd friend.
Remaining cakes = 2 – 1 = 1

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

• Check how many cycles of distribution of cakes are possible from m number of cakes.
• Calculate the number of cakes for 1 cycle which is
`sum = n * (n + 1) / 2`
• Now diving M by sum we get cycle count + some remainder.
• Now check how many remaining cakes are again possible to distribute to x friends.
• The value of x can be easily achieved by solving quadratic equation
`remainder = x * (x + 1) / 2`

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach  ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the  ` `// remaining count of cakes  ` `int` `cntCakes(``int` `n, ``int` `m) ` `{  ` ` `  `    ``// Sum for 1 cycle  ` `    ``int` `sum = (n * (n + 1)) / 2; ` ` `  `    ``// no. of full cycle and remainder  ` `    ``int` `quo = m/sum ; ` `    ``int` `rem = m % sum ; ` `    ``double` `ans = m - quo * sum ; ` ` `  `    ``double` `x = (-1 + ``pow``((8 * rem) + 1, 0.5)) / 2; ` `     `  `    ``ans = ans - x * (x + 1) / 2; ` ` `  `    ``return` `int``(ans); ` `} ` ` `  `// Driver Code ` `int` `main ()  ` `{ ` `    ``int` `n = 3; ` `    ``int` `m = 8;  ` `    ``int` `ans = cntCakes(n, m); ` `    ``cout << (ans); ` `} ` ` `  `// This code is contributed by Surendra_Gangwar `

## Java

 `// Java implementation of the approach  ` `class` `GFG  ` `{ ` `     `  `    ``// Function to return the  ` `    ``// remaining count of cakes  ` `    ``static` `int` `cntCakes(``int` `n, ``int` `m) ` `    ``{  ` `     `  `        ``// Sum for 1 cycle  ` `        ``int` `sum = (n * (n + ``1``)) / ``2``; ` `     `  `        ``// no. of full cycle and remainder  ` `        ``int` `quo = m/sum ; ` `        ``int` `rem = m % sum ; ` `        ``double` `ans = m - quo * sum ; ` `     `  `        ``double` `x = (-``1` `+ Math.pow((``8` `* rem) + ``1``, ``0.5``)) / ``2``; ` `         `  `        ``ans = ans - x * (x + ``1``) / ``2``; ` `     `  `        ``return` `(``int``)ans; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{ ` `        ``int` `n = ``3``; ` `        ``int` `m = ``8``;  ` `        ``int` `ans = cntCakes(n, m); ` `        ``System.out.println(ans); ` `    ``} ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the  ` `# remaining count of cakes ` `def` `cntCakes(n, m): ` ` `  `    ``# Sum for 1 cycle ` `    ``sum` `=` `(n``*``(n ``+` `1``))``/``/``2` ` `  `    ``# no. of full cycle and remainder ` `    ``quo, rem ``=` `m``/``/``sum``, m ``%` `sum` `    ``ans ``=` `m ``-` `quo ``*` `sum` ` `  `    ``x ``=` `int``((``-``1` `+` `(``8` `*` `rem ``+` `1``)``*``*``0.5``)``/``2``) ` `    ``ans ``=` `ans ``-` `x``*``(x ``+` `1``)``/``/``2` ` `  `    ``return` `ans ` ` `  `# Driver code ` `def` `main(): ` `  ``n ``=` `4` `  ``m ``=` `11` `  ``ans ``=` `cntCakes(n, m) ` `  ``print``(ans) ` ` `  `main() `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `    ``// Function to return the  ` `    ``// remaining count of cakes  ` `    ``static` `int` `cntCakes(``int` `n, ``int` `m) ` `    ``{  ` `     `  `        ``// Sum for 1 cycle  ` `        ``int` `sum = (n * (n + 1)) / 2; ` `     `  `        ``// no. of full cycle and remainder  ` `        ``int` `quo = m/sum ; ` `        ``int` `rem = m % sum ; ` `        ``double` `ans = m - quo * sum ; ` `     `  `        ``double` `x = (-1 + Math.Pow((8 * rem) + 1, 0.5)) / 2; ` `         `  `        ``ans = ans - x * (x + 1) / 2; ` `     `  `        ``return` `(``int``)ans; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``static` `public` `void` `Main () ` `    ``{ ` `        ``int` `n = 3; ` `        ``int` `m = 8;  ` `        ``int` `ans = cntCakes(n, m); ` `        ``Console.Write(ans); ` `    ``} ` `} ` ` `  `// This code is contributed by ajit. `

Output:

```0
```

Time Complexity: O(1)

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