Google code jam exercise -- Your Rank Is Pure

Continue to the Round 1B 2010. The second question does not understand the question. The third question can be understood, but the algorithm has been tossing for a long time. Let's take a look at the third question first.

Given a set S = {2, 3 ,... n}, its subset s' satisfies the nature: n is the k element in s', then k should also be in s', similarly, k is the k' element, so k' should also be in s'. After a finite step, k = 1, not in s', it is called n for s' as a Pure Rank. Given an n, calculate the number of S.

In Content Analysis, the dynamic planning algorithm is excluded. How do I think that many questions are about dynamic planning? I don't know much about state planning. Take a look at the specific algorithm as follows,

If n is a set of S' pure rank, s' = S' ^ {1, 2,... k}, then k is the pure rank of s. The dynamic planning algorithm is to find the method from s' to s.

S 'has k elements, and S ''has k' elements. From S' to S'', it needs to be from {k + 1, k + 2 ,... n-1} select k-k' elements.

Use count [n] [k] to indicate the number of the Set S with k elements and n is pure rank, then there is count [n] [k] = sum count [k] [k'] * C [n-k-1] [k-k'-1] for k' = 1... k-1

Since the final number may be relatively large, we need to take the remainder of 100003 in the process of accumulation. The remainder cannot be replaced by a method equal to or greater than 100003.

The final code is as follows:

#!/usr/bin/python#encoding:UTF-8#Filename:FileFixIt.pyimport sysdef solveN(n):    comb = [[0 for j in xrange(n+1)] for i in xrange(n+1)]    j = 0    for i in xrange(n+1):        comb[i][j] = 1        comb[i][i] = 1    for i in xrange(1,n+1):        for j in xrange(1,i):            comb[i][j] = comb[i-1][j]+comb[i-1][j-1]            if comb[i][j]>=100003:                comb[i][j] = comb[i][j]%100003    count = [[0 for j in xrange(n+1)] for i in xrange(n+1)]    for i in xrange(2,n+1):        count[i][1] = 1    for i in xrange(3,n+1):        for j in xrange(2,i):            count[i][j] = 0            for k in xrange(1,j):                count[i][j] += count[j][k]* comb[i-j-1][j-k-1]                 if count[i][j]>=100003:                    count[i][j] = count[i][j]%100003#    print "count:"            #    for i in xrange(2,n+1):#        for j in xrange(1,i):#            print "count[%d][%d] %d" %(i,j,count[i][j])#    cnt = 0#    for j in xrange(1,n):#        cnt += count[n][j]#    print "cnt:",cnt#    return cnt    return count    inname = "input.txt"outname = "output.txt"if len(sys.argv)>1:    inname = sys.argv[1]    outname = inname.rstrip(".in")    outname = outname + ".out"fin = open(inname,"r")fout = open(outname,"w")testCaseNum = int(fin.readline().rstrip("\n"))caseNum = 0lines = fin.readlines()allCase = [int(line.rstrip("\n")) for line in lines]maxN = max(allCase)count = solveN(maxN)for caseNum,n in enumerate(allCase):    cnt = 0    for j in xrange(1,n):        cnt += count[n][j]        if cnt>=100003:            cnt = cnt%100003    answer = "Case #%d: %d\n" %(caseNum+1,cnt)    fout.write(answer)fin.close()fout.close()

Finally, both the small and large case Tests passed, and large case takes a long time.

Is there a repeat?