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)
- (increase end by 1)
return count + 1

Let us see the following implementation to get better understanding:

Example

Live Demo

#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));
}