<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:
        if n <= 2:
            return 1

        # Help us calculate the series sum of a given number; inclusive of 1
        def calcSum(num):
            return ((num+1)*num)//2

        # Return true/false if number can be made with amt of terms
        # F F F ... T T
        def binSearch(numTerms):
            l, r = numTerms, n
            while l < r:
                mid = (l + r)//2
                if (calcSum(mid) - calcSum(mid-numTerms)) >= n:
                    r = mid
                else:
                    l = mid + 1

            return (calcSum(l) - calcSum(l-numTerms)) == n

        numWays = 1
        prefSum = 1
        for i in range(2, n+1):
            prefSum += i
            if prefSum > n:
                break
            if binSearch(i):
                numWays += 1

        return numWays
16400