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  1. Prepare
  2. Mathematics
  3. Combinatorics
  4. A Chocolate Fiesta
  5. Discussions

A Chocolate Fiesta

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  • walkeazeshoesan1
     + 0 comments

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  • aniketsinharoy
     + 0 comments

    Java 8 solution:

    public static int solve(List<Integer> a) 
    {
        int even_number=0;
        int odd_number=0;
        int i;
        for(i=0;i<a.size();++i)
        {
            if(a.get(i)%2==0)
            {
                ++even_number;
            }
            else
            {
                ++odd_number;
            }
        }
        BigInteger even_comb = BigInteger.valueOf(2).pow(even_number).subtract(BigInteger.valueOf(1)).mod(BigInteger.valueOf(1000000007));
        BigInteger odd_comb =odd_number>0? BigInteger.valueOf(2).pow(odd_number-1).subtract(BigInteger.valueOf(1)).mod(BigInteger.valueOf(1000000007)):BigInteger.valueOf(0);
        BigInteger total_comb = even_comb.multiply(odd_comb).mod(BigInteger.valueOf(1000000007));
        BigInteger ans = even_comb.add(odd_comb).add(total_comb).mod(BigInteger.valueOf(1000000007));
        return ans.intValue();
    }

  • submit_jane_doe
     + 0 comments

    The problem description is exactly why I hate HackerRank.

  • mathemage
     + 0 comments

    FYI nice explanation on the used combinatorial identity: https://math.stackexchange.com/questions/1072881/is-there-a-closed-form-formula-for-sum-of-odd-combinations

  • insoucyant
     + 0 comments

    def solve(a): if len(a) == 0: return 0 evn = [nmbr for nmbr in a if nmbr%2==0] od = [nmbr for nmbr in a if nmbr%2!=0] ven = 2**(len(evn)) - 1 if len(od) > 0: do = 2**(len(od)-1) - 1 else: do = 0 tot = ven+do+do*ven return tot%1000000007