Question

I just find a problem like this

“Give you N different integers, do you know how many different sequences that the difference between every adjacent pair of numbers is larger than 1?”

And when the integers are “1 2 3”, then the answer is zero, when the integers are “5 3 1”, then the answer is 6, for “1 3 5” “1 5 3” “3 1 5” “3 5 1” “5 1 3” “5 3 1” satisfy the problem, I just tried all I could do but I couldn’t solve it, so my question is, how to write a algorithm to solve it.

Thankyou.

Here is my program

int n;bool vi[30];int a[30];int b[30];int counter = 0;
void dfs(int k)
{
    if ( k == n)
    {
        for (int i = 2; i <= n; i ++)
            if (fabs(b[i] - b[i - 1]) <= 1) return ;
        counter ++;
        return ;
    }
    for (int i = 0; i < n; i ++)
    {
    if (!vi[i])
    {
        b[k + 1] = a[i];
        vi[i] = true;
        dfs(k + 1);
        vi[i] = false;
        }
    }
}
int main (void)
{
        cin >> n;
        for (int i = 0; i < n; i ++)
            cin >> a[i];
        memset(vi, 0, sizeof(vi));
        for (int i = 0; i < n; i ++)
        {
            vi[i] = true;b[1] = a[i];dfs(1);vi[i] = false;
        }
        cout << counter << endl;
    return 0;
}

Answer 1

The answer to the question, “How many different sequences are there…” can be solved without enumerating every combination. Which is a good thing, because 25! is approximately 1.55 x 10^24, or way more than any existing computer is going to enumerate in a second. Or in a year.

This is a math problem, specifically combinatorics. See http://en.wikipedia.org/wiki/Combinatorics and possibly http://en.wikipedia.org/wiki/Combinations for information on how you would go about solving the problem.