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++

`// 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;` `}` |

## 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

`<?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` `?>` |

**Output:**

3 1

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

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