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829. Consecutive Numbers Sum

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

• Java
• C++
• Python
• Go
• TypeScript

• 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;
      }
  }
  

• 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;
      }
  };
  

• 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
  
  

• 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
  }
  

• function consecutiveNumbersSum(n: number): number {
      let ans = 0;
      n <<= 1;
      for (let k = 1; k * (k + 1) <= n; ++k) {
          if (n % k === 0 && (Math.floor(n / k) + 1 - k) % 2 === 0) {
              ++ans;
          }
      }
      return ans;
  }
  
  

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