829. Consecutive Numbers Sum

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Description

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
Note: 1 <= N <= 10 ^ 9.
Problem URL

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Solution

给一个数字N，找到它可以有多少种方式，由一串连续的数字加和而成。

If it is a consecutive sequence, we have
$N = (k + 1) + (k + 2) + ... + (k + i)$
$N = \frac{i * (2k + i + 1)}{2})$
$N = k * i + \frac{i * ( i + 1)}{2})$
$N - \frac{i * ( i + 1)}{2})= k * i$
So, for each I, we could judge wether it could be a mutilpe of a constant equal to it. Since I = 1, it is always valid(it is itself), we start iteration from 2.

Code

class Solution {
    public int consecutiveNumbersSum(int N) {
        int count = 1;
        for (int i = 2; i * (i + 1) / 2 <= N; i++){
            if ((N - i * (i + 1) / 2) % i == 0){
                count++;
            }
        }
        return count;
    }
}

Time Complexity: O(N ^ 0.5)
Space Complexity: O(1)

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