829. 连续整数求和(数论)

​​829. 连续整数求和​​(数论)

class Solution {
public:
    int consecutiveNumbersSum(int n) {
        int ans = 0;
        int bound = 2 * n;
        for (int k = 1; k * (k + 1) <= bound; k++) {
            if (isKConsecutive(n, k)) {
                ans++;
            }
        }
        return ans;
    }
  
    bool isKConsecutive(int n, int k) {
        if (k % 2 == 1) {
            return n % k == 0;
        } else {
            return n % k != 0 && 2 * n % k == 0;
        }
    }
};