Divide an integer into the sum of consecutive positive numbers

Question

Today I am trying to solve an classical problem:

For any $n\in \Bbb{N}^+$, If it can be represent as the sum of consecutive positive numbers, find out them.

For example:

$$15 = 1+2+3+4+5$$ $$15=4+5+6$$ $$15=7+8$$

And I have an ugly method, its time complexity is: $O(n^2)$. I use two for loop to exhaustion all possibilities.

for(int i=1;i<n;i++)
{
    for(int j=i;j<n;j++)
    {
        sum+=j;
        if(sum==n)
        {   //print out the answer
            for(int l=i;l<=j;l++)
            {
                cout << l << "+ ";
            }
            cout << endl;
            break;
        }
    }
    sum = 0;
}

I think there may be exist a more effective solution, But I am failed until now. Please help me.

for i from 1 to n
{
   for j from i to n
   {
       sum <- sum + j
       if sum equal n
       {
          print the result
       }
   }
   sum <- 0
}

Answer 1

Here's an alternative way of viewing D.W.'s hint. Using the formula $\sum_{i=1}^m i = \frac{(m+1)m}{2}$, $$ \sum_{i=a+1}^b i = \sum_{i=1}^b i - \sum_{i=1}^a i = \frac{b^2+b}{2} - \frac{a^2+a}{2} = \frac{(b-a)(b+a+1)}{2}. $$ Given a factorization $2n = xy$, we can solve the system $x = b-a$, $y = b+a+1$. The result is $b = (y+x-1)/2$, $a = (y-x-1)/2$. So we want $x,y$ to have different parities (this is only a restriction if $n$ is even) and $y \geq x+1$. Also, $b \neq a+1$ corresponds to $x \neq 1$.

For example, for $n = 15$ we have $2n = 30 = 2 \cdot 15 = 3 \cdot 10 = 5 \cdot 6$. These correspond to the following pairs $(a,b)$: $$ (6,8),(3,6),(0,5). $$ These, in turn, correspond to the representations $$ 15 = 7+8 = 4+5+6 = 1+2+3+4+5. $$

This leads to an $O(n)$ algorithm (in the appropriate computation model). I'll let you work out the remaining details.

Answer 2

Here's a hint: if $n$ can be represented as the sum of $2k+1$ consecutive integers, and if the middle of those consecutive integers is $m$, then what can you say about the relationship between $n$, $k$, and $m$? Now, given $n$, can you determine whether it is expressible as the sum of an odd number of consecutive integers?

Answer 3

Here's a $O(\sqrt n)$ algorithm. We want to find all length-$k$ expressions for which $$ n=a+(a+1)+(a+2)+\cdots+(a+(k-1)) $$ Rearranging terms, we require $$ n=\sum_{i=0}^{k-1}(a+i) = \sum_{i=0}^{k-1}a+\sum_{i=0}^{k-1}i=ka+\frac{k(k-1)}{2} $$ and so we must have $$ n=k\left(\frac{2a+k-1}{2}\right) $$ for some $a$ and $k$. If $k$ is even we require $$ n=\left(\frac{k}{2}\right)(2a+k-1) $$ and so $n$ must be divisible by $k/2$. If $k$ is odd, we'll have $2a+k-1$ even, so $n$ must be divisible by $k$.

That means that the possible solutions will only be those for which $k$ is odd and $k\mid n$ or $k$ is even and $(k/2)\mid n$. Now that we know that we're looking only for divisors of $n$, we'll have candidate pairs $k$ and $n/k$ for those $k\le \sqrt{n}$ if $k$ is odd and $k\le 2\sqrt{n}$ if $n$ is even. That means we can find all solutions by checking at most only the $k\le2\sqrt{n}$. Since each check involves nothing more than a constant number of steps, we can find all the solutions in time $O(\sqrt{n})$.

Let's look at your example, with $n=15$:

1. ($k=1$). We have $15 = 1\cdot(2a+1-1)/2$ so $a=15$, giving the sum 15=15. The other solution will be when $k=15$, giving an impossible solution.
2. ($k=2$). We have $15=(2/2)(2a+1)$ and so $a=7$, giving the solution 15=7+8. The companion solution, where $k=15/1$ is impossible, again.
3. ($k=3$). We have $15=3\cdot(2a+2)/2$, so $a=4$, giving another solution 15=4+5+6. The companion solution, when $k=15/3=5$ gives $a=1$, which yields the solution 15=1+2+3+4+5.
4. ($k=4$). Since $4/2=2$ doesn't divide 15, there is no possible solution.
5. ($k=5$). Since $5>\sqrt{15}$ we can stop looking at odd candidates.
6. ($k=6$). $6/2$ divides 15, but we've already considered that solution. There are no further even numbers to consider, and so we stop trying, having found all the possible solutions.

Answer 4

You can replace your second "for" loop with binary search, since you can calculate sum of arithmetical progression by formula. I mean traverse all possible values of lengths and try to find starting value by binary search, it replaces second $N$ with $log(N)$

I found my old $O(NlogN)$ code where $N$ is the maximal length of representation $1+2+...+N$. It searches the lenght of this representation.

int ans = 1, n; cin >> n;
    for(int len = 2; len <= 32000; len++)
    {
        int l = 1, r = n/2;
        while (l <= r)
        {
            int mid = (l+r)/2;
            __int64 cur_sum = (2*mid + len - 1) * 1LL * len / 2;
            if (cur_sum == n) { ans = len; break; }
            if (cur_sum < n) l = mid+1;
            else r = mid-1;
        }
    }
    cout << ans;

I'm not sure it works ok on all cases, currently I'm writing binserach differently, but the idea is clear I think.