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Contest format 5 hours, around 8-12 problems One computer running (likely)Linux, plus printer 3 people on one machine No cell phones, calculators, USB.

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Presentation on theme: "Contest format 5 hours, around 8-12 problems One computer running (likely)Linux, plus printer 3 people on one machine No cell phones, calculators, USB."— Presentation transcript:

1 Contest format 5 hours, around 8-12 problems One computer running (likely)Linux, plus printer 3 people on one machine No cell phones, calculators, USB drives, Internet (C++ STL available, and java api’s are available) All the paper resources you want, including books, print outs of code, anything! – If it fits in the van, bring it!

2 Scoring Sorted first by number of problems solved Then sorted by time – If you submit a problem at :30 and one at 1:30, your total time is 120 minutes – Find the easy problems and do them first!!! – Watch the standings and see what other teams are doing! – 20 minute penalty for wrong answer on a single submission, but it’s only counted if you eventually solve that problem

3 Submitting You will receive one of several responses: – Format error – Time limit reached – Runtime error (division by 0, out of memory, exception thrown, etc.) – Compile error – Wrong answer – Correct! The judges only give you one at a time – If you have two or more problems, you’ll usually only get the more embarrassing of them

4 Always have someone typing Typing and compiling is time intensive, and there’s only one keyboard – If your program isn’t working, print it and debug it by hand – Let someone else sit and type! If you’re waiting for the computer, write some code out by hand or ask a neighbor for their opinion on your algorithm If it has you read until end of input, use: while (cin >> x)

5 Questions You can submit questions to the judges about problems Updates will be given to everyone if there is a typo or other error You will get one of two responses: – A clarification – No answer (i.e. read the problem more closely)

6 Test the judge’s input They give you 1 or 2 sample inputs and solutions; test them! – There will normally be simple cases. Make sure your format exactly matches the judge’s sample output! – They use a file compare (via a script) so it must be very close

7 End cases The judges are very tricky with their tests If the problem says inputs will be between A and B, you can almost bet that inputs of size A and B will be tested Be wary of carefully worded questions!

8 Tree or not a tree A tree is a well-known data structure that is either empty (null, void, nothing) or is a set of one or more nodes connected by directed edges between nodes satisfying the following properties: – There is exactly one node, called the root, to which no directed edges point. – Every node except the root has exactly one edge pointing to it. – There is a unique sequence of directed edges from the root to each node.

9 Counting characters in a range Input will consist of two integers, 0 < N < 100 For each of the numbers in between these two numbers (inclusive), count the occurrences of each digit Example: 17 21 – 17 18 19 20 21 – 0=>17=>1 1=>48=>1 2=>29=>1 cin >> a >> b; for (i = a; i <= b; ++i) { ++arr[i / 10]; ++arr[i % 10]; } 17 21 ? 21 17 ?

10 Be generous with your memory! Make your arrays twice as big as necessary – Off by one error are difficult to find! Use the STL (strings, vectors, everything!) Use long long instead of int Use double instead of float

11 Code quickly at the cost of efficiency The faster you type, the faster you submit! Use the STL if it makes it easier for you – If you can’t remember how to use the STL sort, write a simple (bubble?)sort. Who cares! Generally, if you get a “time limit reached”, your algorithm needs to be changed, not just little things in your code

12 Helpful suggestion Bring printed code, such as the algorithms we’ll talk about. You won’t have to remember them and know you have a working/correct version too. – If someone is not typing in an answer, type in the algorithm so the template is ready to use. – Also data structures you may want to use (trees for example). – Including a “read a file” code. You know it works, then one least thing to think about.

13 math Number theory – Very popular in the program contests – For ICPC, you need a rather small but useful set Prime table generation, primality testing, greatest common divisor, modular arithmetic and congruence (solving linear congruences), and Euler’s – A Note, Java’s BigInteger class has a number of number-theoretic functions, like gcd, modular exponentiation, primality testing, etc.

14 String manipulation There have been a number of string manipulation questions over the years. Learn the string library – At the least substring, replace, find etc. – Regex maybe really helpful.

15 algorithms Brute force algorithms – From nested loop algorithms to backtracking (easier with recursion). Breath first search. Depth first search is recursive and has nice bracktracking features. Dynamic Programming – Recursive algorithm that is composed of subproblems Coin flipping and fibonacci are simple examples Longest Common Subsequence (LCS), Longest Increasing Subsequence (LIS), Optimal Binary Search tree (OBST), 0-1 knapsack, edit distance, Matrix Chain Product are increasing harder examples.

16 algorithms Trees and priority queues, not necessary an algorithms, but can speed things up. Graph theory – How to represent things and then use BFS and DFS, and topological sorting. Does the graph have cycles?

17 Classic Problems algorithms Shortest paths (Dijkstra for example) Spanning trees (Prim or Kruskal) Eulerain paths and circuits Matchings in bipartite graphs Network flow (max flow, min cost flows) Geometry.

18 STL: Deque #include deque x; x.push_back(20); x.pop_back(); x.back(); x.push_front(20); x.pop_front(); x.front(); x.resize(100); x[10] OR; x.clear();

19 STL: Strings #include string str; string str(“foo”); string str(10, ‘c’); str += “bar”; Find – str.find(“aaba”); str.rfind(“aaba”); – str.find_first_of(“AEIOU”); – str.find_last_not_of(“AEIOU”, 5); – Returns an int, or string::npos if none found str.substr(int position, int length)

20 STL: Algorithms #include swap(a, b); // Any type that has = can go here! reverse(arr, arr + 10); reverse(deq.begin(), deq.end()); Sorting – sort(arr, arr + 10); sort(deq.begin(), deq.end()); – sort(arr, arr + 10, lessThanFunction); bool lessThanFunction(const Type& t1, const Type& t2) { if (t1 < t2) return true; return false; }

21 STL: Algorithms #include Permutations int x[] = {3, 5, 4, 1, 2}; sort(x, x + 5); do { // stuff } while (next_permutation(x, x + 5));

22 STL: formatting #include double d = 12345.6789; cout << d << endl; cout << setprecision(3) << d << endl; cout << setprecision(3) << fixed << d << endl; cout << setprecision(1) << fixed << 0.55 << endl; int i = 42; cout << hex << i << endl; cout << hex << uppercase << i << endl; cout << i << endl; cout << dec << i << endl; 12345.7 1.23e+04 12345.679 0.6 2a 2A 42

23 Algorithms Brush up on – depth-first search, breadth-first search (or just use iterative deepening DFS) N-Trees, but lots of other uses as well. minimum spanning trees ing_tree – Lots of varying algorithms listed at the bottom of the page

24 Algorithms (2) shortest path, like Dijkstra’s algorithm –’s_algorithm – em (Max) flow problems – Good demo of max flow and min cut algorithms. Also links to some other versions of spanning tree algorithms.

25 Algorithms (3) Greatest common divisor is a fun one to remember too – And remember, if gcd(a, b) == 1, then a and b are relatively prime!

26 Dynamic programming/memoization Recursive algorithm that is composed of subproblems – You keep recomputing the subproblems! – Save them in an array and look them up – Start with the recursive version first, then modify it to save work Examples – Fibonacci – Coin problem

27 Geometric algorithms

28 Geometric algorithms Intersection – Four points: a1, a2, b1, b2 – Compute: dir1 = direction(b1, b2, a1) dir2 = direction(b1, b2, a2) dir3 = direction(a1, a2, b1) dir4 = direction(a1, a2, b2) – If dir1/dir2 are opposite signs, and dir3/dir4 are opposite signs, they intersect

29 a1 a2 b1 b2 dir1 dir2 dir3 dir4 a1 a2 b1 b2 dir1 dir2 dir3 dir4


31 1 4 6 5 2 3

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