Algorithm running time

问题

What would the running time be? I got O(n^2)

`cin >> n;
 min = 2*n;
 max = (n+3)*10;

for(int i=0; i<1000; i++)
    for(int j =0; j<n; j++)
         for(int k = min; k< max;k++)
            p = f+c+m

`

回答1:

No matter what outer loop runs 1000 times..so the complexity ~O(n^2). [innermost runs 10n+30-2n = 8n+30 times and loop with variable j runs n times..]

回答2:

The number of times p is computed is:

   1000 * n * (max - min)
=  1000 * n * ((n + 3)*10 - 2*n)
=  1000 * n * (10*n + 30 - 2*n)
=  1000 * n * (8*n + 30)
=  8000 * n^2 + 30000 * n

Yes, it is O(n^2).

来源：https://stackoverflow.com/questions/33293998/algorithm-running-time