829. Consecutive Numbers Sum

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.

Thoughts:

1. If N = i * j, then we want to find i consecutive numbers with average j For example, 15 = 3 * 5, then 4 + 5 + 6, 3 numbers with average 5; and 15 = 5 * 3, then 1 + 2 + 3 + 4 + 5, 5 numbers with average 3 But one constraint is that i must be odd, since if i is even k + 1, k + 2, ... k + i. The average = (ik + (1 + i)*i/2) / i = k + (1 + i) / 2, cannot be an integer. (Original Post)

2. Simple Code, but obscure logic (Original Post)

Code 1

class Solution:
    def consecutiveNumbersSum(self, N: int) -> int:
        if N == 1: 
            return 1
        res = 1
        for i in range(2, int(N ** .5 + 1)):   
            if N % i == 0:
                if i % 2 == 1: # If i is odd, then we can form a sum of length i
                    res += 1
                j = (N // i) # Check the corresponding N // i
                if i != j and j % 2 == 1:
                    res += 1
        if N % 2 == 1: # If N is odd(2k + 1). Then N = k + k + 1, not included above
            res += 1

        return res

Code 2

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