Algorithm (Math)

1. Suppose that there exists an $0\leq x<N$ such that $\sum_{j=0}^{k}(x+j)=N.$ Then note that for any $0\leq k<N$, we have that 5 , 5 + 1, 5 + 2, $$\begin{aligned} \sum_{j=0}^{k}(x+j) & =\frac{(2x+k)\cdot(k+1)}{2}=x(k+1)+\frac{1}{2}(k^{2}+k)=N,\end{aligned}$$ which after rearranging yield $$x(k+1)=N-\frac{1}{2}k(k+1).$$
2. So to design an algorithm, we simply need to enumerate $k$ and check if $k+1$ divides $N-\frac{1}{2}k(k+1).$
3. To speed things up, we work out a tighter search range of $k$. Note that we can easily get the following inequalities $$N-\frac{k}{2}(k+1)=x(k+1)>k+1$$ since $x\geq1$, which also implies that $2N>k(k+1)>k^{2};$ so we have $0\,<k\leq\sqrt{2N}.$

Code

class Solution {
 public:
  int consecutiveNumbersSum(int N) {
    return [&](int acc = 0) {
      for (int k = 0; k < sqrt(2 * N); ++k)
        if ((N - k * (k + 1) / 2) >= k + 1 and (N - k * (k + 1) / 2) % (k + 1) == 0) 
            acc++;
      return acc;
    }();
  }
};