# 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
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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++

 `// C++ program to count number of ways to express ` `// N as sum of consecutive numbers. ` `#include ` `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; ` `} `

## Java

 `// 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 `

## Python

 `# 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 `

## C#

 `// 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. `

## PHP

 ` `

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.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up

Improved By : nitin mittal, jit_t

Article Tags :
Practice Tags :

19

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.