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Formatted question description: https://leetcode.ca/all/829.html

# 829. Consecutive Numbers Sum (Hard)

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.

Companies:
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Related Topics:
Math

## Solution 1.

The sum of Arithmetic Progression (AP) is:

S = an + n(n-1)d/2


where a is the first element, n is the count of elements, d is the common difference.

In our case, S = N, d = 1 so we have:

a = (N - n(n-1)/2) / n


So we can traverse from n = 1 to n = N to get a, but they need to satisfy the following requirements:

1. a = (N - n(n-1)/2) / n >= 1 so 2N - n^2 + n >= 2n, n(n+1) <= 2N. For n >= 1, n(n+1) > n^2 so n^2 < 2N, n < sqrt(2N). For N >= 1, n < sqrt(2N) is the same or tighter than n <= N, so we can simply use n(n+1) <= 2N as the upper bound.
2. The last element a + n - 1 <= N so
(N - n(n-1)/2)/n + n - 1 <= N
2N - n(n-1) + 2n(n-1) <= 2nN
n(n-1) <= 2(n-1)N
n = 1 or n <= 2N


This is useless.

In sum, we traverse from n = 1 to n(n+1) <= 2N, and if N - n(n-1)/2 can be divided by n, we can increment the count.

// OJ: https://leetcode.com/problems/consecutive-numbers-sum/
// Time: O(sqrt(N))
// Space: O(1)
class Solution {
public:
int ans = 0;
for (int n = 1; n * (n + 1) <= 2 * N; ++n) {
int a = N - n * (n - 1) / 2;
if (a % n == 0) ++ans;
}
return ans;
}
};


## Solution 2.

There are lots of multiplication in Solution 1. Is it possible to reduce them?

If we let n(n+1) as s. In the previous loop, s' = n(n-1), s - s' = 2n. So we can change the multiplication to addition.

Now let s = n(n+1)/2, the upper bound of n will be s <= N. In each step, we do s += n. And the a can be N - s + n now. We can further simply a % n == 0 to (N - s) % n == 0 since if (N - s + n) % n == 0 then (N - s) % n == 0

// OJ: https://leetcode.com/problems/consecutive-numbers-sum/
// Time: O(sqrt(N))
// Space: O(1)
class Solution {
public:
int ans = 0, n = 1, s = 1;
for (; s <= N; s += ++n) {
if ((N - s) % n == 0) ++ans;
}
return ans;
}
};


## Solution 3.

Assume N=(x+1)+(x+2)+...+(x+k), where x >= 0, k >= 1.

We have 2N=k(2x+k+1), which has two factors, k and 2x+k+1, one of which is odd number and another is even. So the question now is how many ways to factor 2N into one even and one odd number.

Assume 2N=2^t * M, where M is odd. If we factor M = a * b, then multiply 2^t to one of them will yield the even number. So the question now becomes how many ways to factor the odd part of N.

If N = 3*3*3*5*5, how many ways to factor N into two numbers?

For one number, we can pick 0 to 3 3s, and 0 to 2 5s, so we have 4 * 3 = 12 options.

Here you can see the answer is multiplication of 1 + {count of a factor}.

To get the count, we can start from d = 3 to d * d <= N with increment 2, and try to divide N with d to count the occurrance of d in N. If the count is c, we multiply 1 + c to answer.

But what if there is a factor greater than sqrt(N)? In this case it have to be a prime number and there will be only one of it. So if eventually N > 1, we multiply 2 to answer.

// OJ: https://leetcode.com/problems/consecutive-numbers-sum/
// Time: O(sqrt(N))
// Space: O(1)
// Ref: https://leetcode.com/problems/consecutive-numbers-sum/solution/
class Solution {
public:
while ((N & 1) == 0) N >>= 1;
int ans = 1, d = 3;
while (d * d <= N) {
int e = 0;
for (; N % d == 0; N /= d, ++e);
ans *= e + 1;
d += 2;
}
if (N > 1) ans <<= 1;
return ans;
}
};

• class Solution {
int max = (int) Math.sqrt(N * 2);
int count = 0;
for (int i = 1; i <= max; i++) {
if (i % 2 == 1) {
if (N % i == 0)
count++;
} else {
if (N % i != 0 && N * 2 % i == 0)
count++;
}
}
return count;
}
}

############

class Solution {
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;
}
}

• // OJ: https://leetcode.com/problems/consecutive-numbers-sum/
// Time: O(sqrt(N))
// Space: O(1)
class Solution {
public:
int ans = 0;
for (int n = 1; n * (n + 1) <= 2 * N; ++n) {
int a = N - n * (n - 1) / 2;
if (a % n == 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
}