Count ways to express a number as sum of consecutive numbers

Given a number N, find the number of ways to represent this number as a sum of 2 or more consecutive natural numbers.

Examples:

Input :15 
Output :3
15 can be represented as:
1+2+3+4+5
4+5+6
7+8

Input :10
Output :1
10 can only be represented as:
1+2+3+4



The idea is to represent N as a sequence of length L+1 as:
N = a + (a+1) + (a+2) + .. + (a+L)
=> N = (L+1)*a + (L*(L+1))/2
=> a = (N- L*(L+1)/2)/(L+1)
We substitute the values of L starting from 1 till L*(L+1)/2 < N
If we get 'a' as a natural number then the solution should be counted.

C/C++

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// C++ program to count number of ways to express
// N as sum of consecutive numbers.
#include <bits/stdc++.h>
using namespace std;
  
long int countConsecutive(long int N)
{
    // constraint on values of L gives us the 
    // time Complexity as O(N^0.5)
    long int count = 0;
    for (long int L = 1; L * (L + 1) < 2 * N; L++)
    {
        float a = (1.0 * N-(L * (L + 1)) / 2) / (L + 1);
        if (a-(int)a == 0.0) 
            count++;        
    }
    return count;
}
  
// Driver Code
int main()
{
    long int N = 15;
    cout << countConsecutive(N) << endl;
    N = 10;
    cout << countConsecutive(N) << endl;
    return 0;
}

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Java

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// A Java program to count number of ways 
// to express N as sum of consecutive numbers.
public class SumConsecutiveNumber 
{    
    // Utility method to compute number of ways
    // in which N can be represented as sum of 
    // consecutive number
    static int countConsecutive(int N)
    {
        // constraint on values of L gives us the 
        // time Complexity as O(N^0.5)
        int count = 0;
        for (int L = 1; L * (L + 1) < 2 * N; L++)
        {
            float a = (float) ((1.0 * N-(L * (L + 1)) / 2) / (L + 1));
            if (a-(int)a == 0.0
                count++;        
        }
        return count;
    }
      
    // Driver code to test above function
    public static void main(String[] args) {
        int N = 15;
        System.out.println(countConsecutive(N));
        N = 10;
        System.out.println(countConsecutive(N));
    }
}
// This code is contributed by Sumit Ghosh

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Python

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# Python program to count number of ways to
# express N as sum of consecutive numbers.
  
def countConsecutive(N):
      
    # constraint on values of L gives us the 
    # time Complexity as O(N^0.5)
    count = 0
    L = 1
    while( L * (L + 1) < 2 * N):
        a = (1.0 * N - (L * (L + 1) ) / 2) / (L + 1)
        if (a - int(a) == 0.0):
            count += 1
        L += 1
    return count
  
# Driver code
  
N = 15
print countConsecutive(N)
N = 10
print countConsecutive(N)
  
# This code is contributed by Sachin Bisht

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C#

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// A C# program to count number of
// ways to express N as sum of
// consecutive numbers.
using System;
  
public class GFG {
      
    // Utility method to compute
    // number of ways in which N
    // can be represented as sum
    // of consecutive number
    static int countConsecutive(int N)
    {
          
        // constraint on values of L
        // gives us the time
        // Complexity as O(N^0.5)
        int count = 0;
        for (int L = 1; L * (L + 1)
                         < 2 * N; L++)
        {
            float a = (float) ((1.0 
                    * N-(L * (L + 1))
                     / 2) / (L + 1));
                       
            if (a - (int)a == 0.0) 
                count++;     
        }
          
        return count;
    }
      
    // Driver code to test above
    // function
    public static void Main()
    {
        int N = 15;
        Console.WriteLine(
              countConsecutive(N));
          
        N = 10;
        Console.Write(
              countConsecutive(N));
    }
}
  
// This code is contributed by
// nitin mittal.

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PHP

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<?php
// PHP program to count number 
// of ways to express N as sum 
// of consecutive numbers.
  
function countConsecutive($N)
{
    // constraint on values 
    // of L gives us the 
    // time Complexity as O(N^0.5)
    $count = 0;
    for ($L = 1; 
         $L * ($L + 1) < 2 * $N; $L++)
    {
        $a = (int)(1.0 * $N - ($L
             (int)($L + 1)) / 2) / ($L + 1);
        if ($a - (int)$a == 0.0) 
            $count++; 
    }
    return $count;
}
  
// Driver Code
$N = 15;
echo countConsecutive($N), "\n";
$N = 10;
echo countConsecutive($N), "\n";
  
// This code is contributed by ajit
?>

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Output:

3
1

The Time complexity for this program is O(N^0.5), because of the condition in the for loop.

This article is contributed by Pranav Marathe. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : nitin mittal, jit_t