CPMagical CP Library

Problem solution · Turing

CCC 2001 S1 - Keeping Score

CCC 2001 S1 - Keeping Score: a Turing solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to CCCSolutions.

Technique
Direct simulation
Source
CCCSolutions
Length
91 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For CCC 2001 S1 - Keeping Score, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 91 lines of Turing from the credited upstream file ccc01s1.t.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from CCCSolutions by CCCSolutions contributors · Milliken Mills High School and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCCC 2001 S1 - Keeping Score · TuringTuring
Use this to learn the idea, then write your own version.
% CCC 2001% Problem J3S1: Keeping Score % determine the points of a Bridge hand.% ace = 4, king = 3, queen = 2 and jack = 1% void = 3, singleton = 2 and doubleton = 1 % Input: cards come in as a single string, eg:"C258JKD69QAHSTJA"% output is a list of cards, by suit, points per suit and total points  % Calculate the points for the first suit in a given string % The points are returned as well as the shortened string% (The first suit is removed form the string.procedure Points (var s : string, var p : int)    var i : int    i := 1    p := 0    loop        exit when i > length (s) or s (i) = "C" or s (i) = "D" or s (i) =            "H" or s (i) = "S"        put s (i), " " ..        if s (i) = "J" then            p := p + 1        elsif s (i) = "Q" then            p := p + 2        elsif s (i) = "K" then            p := p + 3        elsif s (i) = "A" then            p := p + 4        end if        i := i + 1    end loop    if i = 1 then        p := p + 3    elsif i = 2 then        p := p + 2    elsif i = 3 then        p := p + 1    end if    s := s (i .. *)end Points  % Output spacing was arrived at by considering the possibility that% one suit might contain all 13 cards! var p, tp : intvar s : string put "Enter cards:"get s tp := 0put ""put "Cards Dealt                       Points" % for each suit, strip off the "C", "D", "H", or "S" and send to % Points method above.s := s (2 .. *)put "Clubs " ..Points (s, p)Text.Locate (Text.WhatRow, 37)put p : 2tp := tp + p s := s (2 .. *)put "Diamonds " ..Points (s, p)Text.Locate (Text.WhatRow, 37)put p : 2tp := tp + p s := s (2 .. *)put "Hearts " ..Points (s, p)Text.Locate (Text.WhatRow, 37)put p : 2tp := tp + p s := s (2 .. *)put "Spades " ..Points (s, p)Text.Locate (Text.WhatRow, 37)put p : 2tp := tp + p put "                              Total ", tp : 2

Keep practising

Did this explanation save you time? I'm a Grade 11 student building this free library to make difficult algorithms easier to understand.

Buy me a coffee ↗