# 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++

`// 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

`// 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

`# 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#

`// 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

`<?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.

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