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# Program to count number of consecutive lists whose sum is n in C++

Suppose we have a number n, we have to find the number of lists of positive consecutive values that sum up to n.

So, if the input is like n = 15, then the output will be 4, as The possible lists are: [1, 2, 3, 4, 5], [4, 5, 6], [7, 8], and [15].

To solve this, we will follow these steps:

- begin := 1, end := 1, x := (n + 1)
- sum := 0
- while end <= x, do:
- sum := sum + end
- while sum >= n, do:
- if sum is same as n, then:
- (increase count by 1)

- sum := sum - begin
- (increase begin by 1)

- if sum is same as n, then:
- (increase end by 1)

- return count + 1

Let us see the following implementation to get better understanding:

## Example

#include using namespace std; int solve(int n) { int begin=1,end=1,x=(n+1)/2,count=0; long int sum=0; while(end <= x){ sum += end; while(sum >= n){ if(sum == n) count++; sum -= begin; begin++; } end++; } return count+1; } main(){ cout << (solve(15)); }

## Input

15

## Output

4

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