# Count of decreasing pairs formed from numbers 1 to N

Given an integer N, the task is to count of decreasing pairs from numbers 1 to N.

A pair (x, y) is said to be decreasing if x > y

Examples:

Input: N = 8
Output: 3
Explanation:
Decreasing pairs are: (7, 1), (6, 2), (5, 3).

Input: N = 9
Output: 4
Explanation:
Decreasing pairs are: (8, 1), (7, 2), (6, 3), (5, 4).

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

Approach: Consider the below cases:

If N = 1 => Count = 0
If N = 2 => Count = 1 {(2, 1)}
If N = 3 => Count = 1 {(3, 1) or (3, 2)}
If N = 4 => Count = 2 {(4, 3), (2, 1)}
If N = 5 => Count = 2 {(5, 4), (3, 2)}
If N = 6 => Count = 3 {(6, 5), (4, 3), (2, 1)}
.
.
and so on

It can be clearly observed that Below is the implementation of the above approach:

## C++

 // C++ program to count decreasing  // pairs formed from numbers 1 to N     #include  using namespace std;     // Function to count the  // possible number of pairs  int divParts(int N)  {      if (N % 2 == 0)             // if the number is even          // then the answer in (N/2)-1          cout << (N / 2) - 1 << endl;         else         // if the number is odd          // then the answer in N/2          cout << N / 2 << endl;  }     // Driver code  int main()  {      int N = 8;         divParts(N);         return 0;  }

## Java

 // Java program to count decreasing  // pairs formed from numbers 1 to N  import java.util.*;  class GFG{         // Function to count the  // possible number of pairs  static void divParts(int N)  {      if (N % 2 == 0)             // if the number is even          // then the answer in (N/2)-1          System.out.println((N / 2) - 1);         else         // if the number is odd          // then the answer in N/2          System.out.println((N / 2));  }     // Driver code  public static void main(String[] args)  {      int N = 8;         divParts(N);  }  }     // This code is contributed by offbeat

## Python3

 # Python3 program to count decreasing  # pairs formed from numbers 1 to N     # Function to count the  # possible number of pairs  def divParts(N):         if (N % 2 == 0):             # if the number is even          # then the answer in (N/2)-1          print((N / 2) - 1);         else:                     # if the number is odd          # then the answer in N/2          print(N / 2);     # Driver code  N = 8;  divParts(N);     # This code is contributed by Code_Mech

## C#

 // C# program to count decreasing  // pairs formed from numbers 1 to N  using System;  class GFG{         // Function to count the  // possible number of pairs  static void divParts(int N)  {      if (N % 2 == 0)             // if the number is even          // then the answer in (N/2)-1          Console.WriteLine((N / 2) - 1);         else         // if the number is odd          // then the answer in N/2          Console.WriteLine((N / 2));  }     // Driver code  public static void Main()  {      int N = 8;         divParts(N);  }  }     // This code is contributed by Code_Mech

Output:

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Improved By : offbeat, Code_Mech