Why not to solve it in O(N) (such that N is the number of the group)??

On May 9, 5:42 pm, Dario <[email protected]> wrote:
> I just made the large input case for problem C by storing 'next
> position' and 'cash' arrays, then just iterating over them R times.
> This seems to be O(N+R). But I've been wondering about possible
> optimizations to this. I can think of an O(NlogR) solution in which I
> nest the 'next position' and 'cash' arrays, only adding up cash/
> changing position when the correct bit of R is on (binary). I've
> pasted my code below (the algorithm described here is really just the
> solve() function and the rest is wrapping). I'd be interested in a
> comparison of different optimizations/algorithms over the problem
> space?
>
> def doProb(fname, ofname):
>     #do problem 3 given a file name
>     f = open(fname, 'r');
>     T = int(f.readline());
>     output = ['Case #' + str(i) + ': '+ \
>                 str(solve([int(j) for j in f.readline().split()],
> [int(j) for j in f.readline().split()])) + \
>                 '\n' for i in range(1,T+1)];
>     f.close();
>     of = open(ofname, 'w');
>     of.writelines(output);
>     of.close();
>
> def solve(x,g):
>     (R,k,N) = x; #unpack the input
>     (np, cash) = getParms(k,g);
>     cur = 0;
>     tot = 0;
>
>     while(R>0):
>         incl = R%2;
>         R = R//2;
>         if(incl==1):
>             tot += cash[cur];
>             cur = np[cur];
>         (np, cash) = nest(np, cash);
>     return tot;
>
> def nest(np, cash):
>     #thinking of np as a permutation, get np^2 and total cash list
>     np2 = [0]*len(np);
>     c2  = [0]*len(np);
>     for i in range(len(np)):
>         np2[i] = np[np[i]];
>         c2[i] = cash[i]+cash[np[i]];
>     return (np2, c2);
>
> def getParms(k,g):
>     N= len(g);
>
>     #special case, all passengers fit in the roller coaster:
>     if(sum(g) <= k):
>         np = [i for i in range(N)];
>         cash = [sum(g)]*N;
>         return (np, cash);
>
>     #other cases
>     np = [];
>     cash = [];
>     for i in range(N):
>         tot = g[i];
>         j = i;
>         while(tot <= k):
>             j = (j+1)%N;
>             tot += g[j];
>         tot -= g[j];
>         np.append(j);
>         cash.append(tot);
>
>     return (np, cash);
>
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