Given a positive integer N, how many ways can we write it as a sum of consecutive positive integers?

Example 1:

Input: 5
Output: 2
Explanation: 5 = 5 = 2 + 3

Example 2:

Input: 9
Output: 3
Explanation: 9 = 9 = 4 + 5 = 2 + 3 + 4

Example 3:

Input: 15
Output: 4
Explanation: 15 = 15 = 8 + 7 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5

分析：https://leetcode.com/problems/consecutive-numbers-sum/discuss/209317/topic

这道题的要求的另一种说法: 把N表示成一个等差数列（公差为1）的和

我们不妨设这个数列的首项是x，项数为m，则这个数列的和就是[x + (x + (m-1))]m / 2 = mx + m(m-1)/2 = N

接下来，一个很自然的想法就是，枚举m，通过上式判断对于相应的m是否存在合法的x。

x = ((N - m(m-1)/2)) / m

显然枚举的复杂度是O(sqrt(N))。因为m能取到的最大值显然是sqrt(n)数量级的

 1 class Solution {
 2     int consecutiveNumbersSum(int N) {
 3         int ans = 0;
 4         for (int m = 1; ; m++) {
 5             int mx = N - m * (m-1) / 2;
 6             if (mx <= 0)
 7                 break;
 8             if (mx % m == 0)
 9                 ans++;
10         }
11         return ans;
12     }
13 }