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

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++

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// C++ implementation of the approach
#include <bits/stdc++.h>
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;
}

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Java

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// 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

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Python3

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# 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))

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C#

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// 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

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Output:

3

Time Complexity: O(1)



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Improved By : mohit kumar 29, AnkitRai01



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