Problem of the Day
A new programming or logic puzzle every Mon-Fri

Comments:

• March boy - 10 years, 4 months ago

  july - 4710 august - 681 september - 9919 october - 71015 november - 81114 december - 8124

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• Anonymous - 10 years, 4 months ago

  #include <iostream>
  #include <cstring>
  
  using namespace std;
  
  const char * MONTHS[] = { "January","February","March","April","May","June","July","August","Septempber","October","November","December"};
  
  /* return the hash of the month */
  unsigned long hash(const char * name) 
  {
  int index;
  unsigned long ret =  strlen(name); /* length of the month */
  
  /* find the ordinal of the month */
  for(index=0; index<12; ++index)
    if(strcmp(MONTHS[index],name) == 0)
       break;
  if (index >= 12) return 0; /* unknown month */
  ret = ret * (index >= 9 ? 100:10) + index + 1;
  /* get the ordinal of the first letter of the month */
  index = tolower(MONTHS[index][0]) + 1 - 'a';
  ret = ret * (index > 9 ? 100:10) + index;
  return ret;
  }
  
  int main() {
  for  (int i=0;i<12;++i)
          cout << MONTHS[i] <<"-" <<hash(MONTHS[i])<<endl;
  return 0;   
  }
  

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• Nick - 10 years, 4 months ago

  Solution in plain English is that the pattern is 3 concatenated numbers, the first number is the number of letters in the month's name, the second number is the month's number, and the third number is the place in the alphabet of the month's first letter.

  Nice one!

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• Patrick - 10 years, 4 months ago

  This is it in Ruby.(Sorry about the indentation)

  @@alphabet = ("A".."Z").to_a
      def self.to_numbers(months_array)
        result = {}
        months_array.map.with_index do |m, i|
          number = "#{ m.to_s.length }" << "#{ i+1 }" << "#{ @@alphabet.index(m[0])+1 }"
          result[m.to_sym] = number.to_i
        end
        result
      end
  

  {:January=>7110, :February=>826, :March=>5313, :April=>541, :May=>3513, :June=>4610, :July=>4710, :August=>681, :October=>7915, :November=>81014, :December=>8114}

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