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@ -297,5 +297,6 @@ My LeetCode solutions with Chinese explanation. 我的LeetCode中文题解。
| 661 |[Image Smoother](https://leetcode.com/problems/image-smoother)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/661.%20Image%20Smoother.md)|Easy| |
| 665 |[Non-decreasing Array](https://leetcode.com/problems/non-decreasing-array)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/665.%20Non-decreasing%20Array.md)|Easy| |
| 714 |[Best Time to Buy and Sell Stock with Transaction Fee](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/714.%20Best%20Time%20to%20Buy%20and%20Sell%20Stock%20with%20Transaction%20Fee.md)|Medium| |
| 829 |[Consecutive Numbers Sum](https://leetcode.com/problems/consecutive-numbers-sum/)|[C++](solutions/829.%20Consecutive%20Numbers%20Sum.md)|Hard| |
| 905 |[Sort Array By Parity](https://leetcode.com/problems/sort-array-by-parity/)|[C++](solutions/905.%20Sort%20Array%20By%20Parity.md)|Easy| |
| 946 |[Validate Stack Sequences](https://leetcode.com/problems/validate-stack-sequences/)|[C++](solutions/946.%20Validate%20Stack%20Sequences.md)|Medium| |

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solutions/829. Consecutive Numbers Sum.md Normal file
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# [829. Consecutive Numbers Sum](https://leetcode.com/problems/consecutive-numbers-sum/)
# 思路
给定正整数N，问N能写成多少种连续正整数之和，比如9可以写成 4+5，或者2+3+4。
我们假设把N拆分成k个连续的数之和，并设最小的那个数是m，则我们有：
|k| 1 | 2 | 3 | 4 | ... | K |
|---|:-:|:-:|:-:|:-:|:-:|:-:|:-:|
|连续序列|m|m,m+1|m,m+1,m+2|m,m+1,m+2,m+3|...|m,m+1,m+K-1|
|满足关系|N=m|N = 2m+1| N = 3m+3| N = 4m+6|...| N = Km + f(k) |
其中 f(k) = 1 + 2 +...+ k-1。
有了这个规律我们就知道了如果 `(N - f(k)) % k == 0`则说明可以拆成k个连续的数，那我们就可以从 k=1 不断增大k直到不满足`N <= f(k)`即可。
由于 f(k) = 1 + 2 +...+ k-1 = (k*k-1)/2。所以k最大为 sqrt(2N) 向下取整。所以时间复杂度为O(sqrt(N))。
# C++
``` C++
class Solution {
public:
int consecutiveNumbersSum(int N) {
int res = 0, k = 1, fk = 0;
while(N > fk){
if((N - fk) % k == 0) res++;
fk += (k++);
}
return res;
}
};
```