<p>Given an integer <code>n</code>, return <em>the number of ways you can write </em><code>n</code><em> as the sum of consecutive positive integers.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 5
<strong>Output:</strong> 2
<strong>Explanation:</strong> 5 = 2 + 3
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 9
<strong>Output:</strong> 3
<strong>Explanation:</strong> 9 = 4 + 5 = 2 + 3 + 4
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 15
<strong>Output:</strong> 4
<strong>Explanation:</strong> 15 = 8 + 7 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def consecutiveNumbersSum(self, n: int) -> int:
        # for k consecutive:
        # n = k * (2v + k - 1) / 2 = (2kv + k^2 - k) / 2 = kv + (k^2 - k)/2
        count = 0
        n2 = 2*n
        for k in range(1, n+1):
            if n2 % k != 0:
                continue
            remaind = n2 // k
            v = remaind + 1 - k
            if v <= 0:
                break
            if v % 2 == 0:
                count += 1
        return count
16200