Consecutive Numbers Sum in C++

Suppose we have a positive integer N, we have to find how many different ways can we write it as a sum of consecutive positive integers?

So, if the input is like 10, then the output will be 3, this is because we can represent 10 as 5 + 5 and 7 + 3, so there are two different ways.

To solve this, we will follow these steps −

ret := 1
for initialize i := 2, (increase i by 1), do −
- sum := (i * (i + 1)) / 2
- if sum > N, then −
  - Come out from the loop
- rem := N - sum
- ret := ret + (1 when rem mod i is 0, otherwise 0)
return ret

Let us see the following implementation to get better understanding −

Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
   public:
   int consecutiveNumbersSum(int N) {
      int ret = 1;
      for(int i = 2; ; i++){
         int sum = (i * (i + 1)) / 2;
         if(sum > N) break;
         int rem = N - sum; ret += (rem % i == 0);
      }
      return ret;
   }
}; main(){
   Solution ob;cout << (ob.consecutiveNumbersSum(10));
}