# Count of distinct remainders when N is divided by all the numbers from the range [1, N]

Given an integer N, the task is to find the count of total distinct remainders which can be obtained when N is divided by every element from the range [1, N].

Examples:

Input: N = 5
Output: 3
5 % 1 = 0
5 % 2 = 1
5 % 3 = 2
5 % 4 = 1
5 % 5 = 0
The distinct remainders are 0, 1 and 2.

Input: N = 44
Output: 22

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

Approach: It can be easily observed that for even values of N the number of distinct remainders will be N / 2 and for odd values of N it will be 1 + ⌊N / 2⌋.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the count of distinct ` `// remainders that can be obtained when ` `// n is divided by every element ` `// from the range [1, n] ` `int` `distinctRemainders(``int` `n) ` `{ ` ` `  `    ``// If n is even ` `    ``if` `(n % 2 == 0) ` `        ``return` `(n / 2); ` ` `  `    ``// If n is odd ` `    ``return` `(1 + (n / 2)); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 5; ` ` `  `    ``cout << distinctRemainders(n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the above approach  ` `class` `GFG  ` `{  ` ` `  `// Function to return the count of distinct ` `// remainders that can be obtained when ` `// n is divided by every element ` `// from the range [1, n] ` `static` `int` `distinctRemainders(``int` `n) ` `{ ` ` `  `    ``// If n is even ` `    ``if` `(n % ``2` `== ``0``) ` `        ``return` `(n / ``2``); ` ` `  `    ``// If n is odd ` `    ``return` `(``1` `+ (n / ``2``)); ` `} ` ` `  `// Driver code  ` `public` `static` `void` `main(String[] args)  ` `{  ` `    ``int` `n = ``5``; ` `    ``System.out.println(distinctRemainders(n)); ` `}  ` `}  ` ` `  `// This code is contributed by Mohit Kumar `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the count of distinct  ` `# remainders that can be obtained when  ` `# n is divided by every element  ` `# from the range [1, n] ` `def` `distinctRemainders(n): ` `     `  `    ``# If n is even ` `    ``if` `n ``%` `2` `=``=` `0``: ` `        ``return` `n``/``/``2` `     `  `    ``# If n is odd ` `    ``return` `((n``/``/``2``)``+``1``) ` ` `  `# Driver code ` `if` `__name__``=``=``"__main__"``: ` ` `  `    ``n ``=` `5` `    ``print``(distinctRemainders(n)) `

## C#

 `// C# implementation of the above approach  ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `    ``// Function to return the count of distinct ` `    ``// remainders that can be obtained when ` `    ``// n is divided by every element ` `    ``// from the range [1, n] ` `    ``static` `int` `distinctRemainders(``int` `n) ` `    ``{ ` `     `  `        ``// If n is even ` `        ``if` `(n % 2 == 0) ` `            ``return` `(n / 2); ` `     `  `        ``// If n is odd ` `        ``return` `(1 + (n / 2)); ` `    ``} ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main()  ` `    ``{  ` `        ``int` `n = 5; ` `        ``Console.WriteLine(distinctRemainders(n)); ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```3
```

Time Complexity: O(1)

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Improved By : mohit kumar 29, AnkitRai01

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