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Hard
Math
Enumeration

中文文档

Description

Given an integer n, return the number of ways you can write n as the sum of consecutive positive integers.

 

Example 1:

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

Example 2:

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

Example 3:

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

 

Constraints:

  • 1 <= n <= 109

Solutions

Solution 1

Python3

class Solution:
    def consecutiveNumbersSum(self, n: int) -> int:
        n <<= 1
        ans, k = 0, 1
        while k * (k + 1) <= n:
            if n % k == 0 and (n // k + 1 - k) % 2 == 0:
                ans += 1
            k += 1
        return ans

Java

class Solution {

    public int consecutiveNumbersSum(int n) {
        n <<= 1;
        int ans = 0;
        for (int k = 1; k * (k + 1) <= n; ++k) {
            if (n % k == 0 && (n / k + 1 - k) % 2 == 0) {
                ++ans;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int consecutiveNumbersSum(int n) {
        n <<= 1;
        int ans = 0;
        for (int k = 1; k * (k + 1) <= n; ++k) {
            if (n % k == 0 && (n / k + 1 - k) % 2 == 0) {
                ++ans;
            }
        }
        return ans;
    }
};

Go

func consecutiveNumbersSum(n int) int {
	n <<= 1
	ans := 0
	for k := 1; k*(k+1) <= n; k++ {
		if n%k == 0 && (n/k+1-k)%2 == 0 {
			ans++
		}
	}
	return ans
}