To count number of quadruplets with sum = 0

[Steven Bethard]

n00m wrote:
> http://www.spoj.pl/problems/SUMFOUR/
> 
> 3
> 0 0 0 0
> 0 0 0 0
> -1 -1 1 1
> Answer for this input data is 33.
> 
> My solution for the problem is
> ======================================================================
> 
> import time
> t = time.clock()
> 
> q,w,e,r,sch,h = [],[],[],[],0,{}
> 
> f = open("D:/m4000.txt","rt")
> 
> n = int(f.readline())
> 
> for o in range(n):
>   row = map(long, f.readline().split())
>   q.append(row[0])
>   w.append(row[1])
>   e.append(row[2])
>   r.append(row[3])
> 
> f.close()
> 
> for x in q:
>   for y in w:
>     if h.has_key(x+y):
>       h[x+y] += 1
>     else:
>       h[x+y] = 1

This won't help much, but you can rewrite the above as::

     x_y = x + y
     h[x_y] = h.get(x_y, 1)

Or if you're using Python 2.5, try::

     h = collections.defaultdict(itertools.repeat(0).next)

     ...
     for x in q:
         for y in w:
             h[x + y] += 1
     ...

Not likely to get you an order of magnitude though.

> for x in e:
>   for y in r:
>     sch += h.get(-(x+y),0)

If you use the collections.defaultdict approach above, this becomes::

     for x in e:
         for y in r:
             sch += h[-(x + y)]

Note that you should also probably put all your code into a function -- 
looking up function locals is quicker than looking up module globals.

STeVe