# Count ways to express a number as sum of consecutive numbers

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

**Examples:**

Input:N = 15Output:3Explanation:

15 can be represented as:

- 1 + 2 + 3 + 4 + 5
- 4 + 5 + 6
- 7 + 8

Input:N = 10Output:1

**Approach: **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.

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## 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++) {` ` ` `double` `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++) {` ` ` `double` `a = (` `double` `)((` `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` |

## Python3

`# 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++) {` ` ` `double` `a = (` `double` `)((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` `?>` |

## Javascript

`<script>` ` ` `// A Javascript program to count number of` ` ` `// ways to express N as sum of` ` ` `// consecutive numbers.` ` ` ` ` `// Utility method to compute` ` ` `// number of ways in which N` ` ` `// can be represented as sum` ` ` `// of consecutive number` ` ` `function` `countConsecutive(N)` ` ` `{` ` ` ` ` `// constraint on values of L` ` ` `// gives us the time` ` ` `// Complexity as O(N^0.5)` ` ` `let count = 0;` ` ` `for` `(let L = 1; L * (L + 1) < 2 * N; L++)` ` ` `{` ` ` `let a = ((1.0 * N-(L * (L + 1)) / 2) / (L + 1));` ` ` ` ` `if` `(a - parseInt(a, 10) == 0.0)` ` ` `count++; ` ` ` `}` ` ` ` ` `return` `count;` ` ` `}` ` ` ` ` `let N = 15;` ` ` `document.write(countConsecutive(N) + ` `"</br>"` `);` ` ` `N = 10;` ` ` `document.write(countConsecutive(N));` ` ` `</script>` |

**Output:**

3 1

**Time Complexity:** O(N^0.5)

**Auxiliary Space**: O(1)

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