Online Round 2 March 8, 2006 Match summary This round of the TCO goes to tomek, thanks to a solid time on all three tasks. A successful challenge phase helped ante to jump to the second place, ploh finished third. The problem set was reasonably easy, more than 20 coders were able to correctly solve the whole set. The only glitch in this round were the broadcasts concerning the 550point problem. On one hand, it was a bit of a shame, as the problem really needed a correct formal problem statement. On the other hand, the examples allowed only the interpretation given in the broadcasts. Thus there was no reason to cancel the match, and the results are valid. Congratulations to all advancers! The ProblemsPowerDigitUsed as: Division One  Level One:
As turuthok's submission shows, Java coders had a distinct advantage in this problem, thanks to the bigInteger class. In other programming languages the coders had to be a bit more careful. The hardest part of this problem is determining when to return 1. The rest is simple: If the number x^{y} has the kth digit (as defined in the problem statement), we can compute it as floor( x^{y} / 10^{k} ) mod 10. Probably the simplest way to solve both parts at once: Sequentially compute powers of x modulo 100,000. Keep a flag whether the result is already at least equal to 100,000. At the end, you'll have the value x^{y} mod 100,000, and you will know whether this is the actual value. This is enough information to determine the result. C++ code follows. int digitK(int x, int y, int k) { bool large = false; // compute: // tmp = (x^y) mod 100000 // large = (x^y >= 100000) long long tmp = 1; while (y) { tmp *= x; if (tmp >= 100000) large = true; tmp %= 100000; } // convert the number into a vector of digits vector<int> digits; while (tmp) { digits.push_back(tmp%10); tmp /= 10; } // BE CAREFUL! A SPECIAL CASE! if (x==0) digits.push_back(0); // fill in the missing digits correctly if (large) { while (digits.size() <= k) digits.push_back(0); } else { while (digits.size() <= k) digits.push_back(1); } return digits[k]; }InverseCollatz Used as: Division One  Level Two:
Given the string S and a valid result R, we can easily "invert" the result to get the initial number – we can go backwards and compute all the intermediate values. For example, if the string S="EOE" and the result was 14, we can deduce the whole sequence as follows: In the last step, the evenrule was applied, thus the previous number was twice as big, i.e., 28. The previous step had to be 9 → 28, and the first step was 18 → 9. Thus we started with the number 18. Note that each of the two operations preserves order, in the sense that if we have a larger (valid) input, we will get a larger output. This also means that the inverse operation described above has to preserve order in the same way: if we take a larger (valid) result and "invert" it, we will get a larger starting number. This can bring us to the idea to try to compute the solution to the given task by going backwards in the given string. A brute force solution could work as follows: Consider all possible result values: 1, 2, 3, ... For each of them, verify whether it is valid by going backwards, and if it is valid, note the starting number. As soon as we have found two valid starting numbers P, Q (P<Q), we are done. From the observation about preserving order we know that these two numbers are the smallest two. Thus the sequence we seek has to be "(QP)k + P". The large example in the problem statement shows that this approach, while correct, will be too slow to solve the large cases. Luckily, we can easily fix this. We won't be going backwards through the whole string. Instead, will compute the solution one step at a time. For example, suppose that S = 'O' + T, and that we already know that for the input T the solution is "8k+6". If we want to determine the solution for the whole sequence S, we have to determine which of these values are reachable after an 'O'step, and what are the corresponding input numbers. In this case, the first few elements of the solution for T are { 6, 14, 22, 30, 38, 46, 54 ... } Let's consider them one after the other
We can stop here. The first two reachable values are 22 and 46, the smallest two original values are 7 and 15. This means that the answer is "8k+7". The case when the first step is an 'O'step can be solved as shown in the example. The 'E'case is even simpler, all values can be reached, the answer changes from "Pk+Q" to "(2*P)k+(2*Q)". Still, we can use the same approach here. While it is possible to compute exact formulas giving the changed sequence in both cases, I consider this approach to be more robust, less errorprone, and it still leads to a pretty short piece of code. First, we will write a function to make one step backwards, or return zero if the step is impossible: long long inverseStep(char ch, long long x) { if (ch=='E') return 2*x; if (x%3 != 1) return 0; // no valid original x /= 3; if (x%2 == 0) return 0; // the original is NOT odd, we wouldn't take an 'O'step! return x; } Now we can write the main function as follows: string getForm(string s) { long long a=1, b=0; for (int d=s.size()1; d>=0; d) { vector<long long> originals; for (long long k=0; ; k++) { long long x = inverseStep( s[d], a*k+b ); if (x) originals.push_back(x); if (originals.size() == 2) break; } a = originals[1]  originals[0]; b = originals[0]; // one more small fix necessary: // the correct form of a sequence {a, 2a, 3a, ...} is "ak+0", not "ak+a" if (b==a) b=0; } stringstream SS; SS << a << "k+" << b; return SS.str(); } Homework for advancers #1. To gain a deeper understanding of this problem, try to answer the following questions:
Used as: Division One  Level Three:
The immediate question after reading the problem statement should've been: How do I build a graph corresponding to this problem? What we really want is to make sure that we reached the exit. What we initially know is that we can be at any of the free squares. This is a piece of information we can update after we move: For each of our possible positions, try to make a move, and note the position where it ended. Also the small size of the maze (at most 20 squares) suggests that a state space search is the way to go. We have at most 21 possible positions: at most 20 squares, and the exit. Our current state at any moment of the game can be uniquely described by giving the set of positions where we can be at that moment. Thus we have 2^{21} theoretically possible states. Now consider a graph where states are vertices, and each edge corresponds to us making one move in the game. The starting vertex will be the state containing all free squares, the destination vertex is the state containing only the exit. The solution of our problem corresponds to the lexicographically smallest of the shortest paths between these two vertices. (Some coders decided to use only 2^{20} states, with the state "you can't be on any of the squares" corresponding to the destination vertex. In this situation, both approaches are correct. However, if the problem statement would ask for the length of the path, this approach would have to handle a maze full of 'X's separately.) The canonical approach to finding shortest paths (in unweighted graphs) is breadthfirst search. If we want a lexicographically smallest shortest path, we have to be a bit more careful. There are two possible approaches: The first approach is to search backwards. If you find several ways how to reach a vertex in an optimal number of steps, consider all possible predecessors and keep only that one that corresponds to the lexicographically smallest first step. (This can be done on the fly, for each vertex you keep its predecessor, you only have to adjust the code that handles a situation when you process an edge that leads to an already visited vertex.) The second approach is simpler to code but harder to believe. Run a normal BFS. Each time you process a vertex, process the outgoing edges in lexicographical order. For each vertex, keep the first found predecessor. (The easiest way to show why this works is to split the vertices into "layers" according to their distance from the starting vertex. Then we can prove that the vertices in each layer will be processed in lexicograpical order (i.e., the path into a later discovered vertex will be lexicographically greater). krijgertje's solution is a pretty clean implementation of the second approach. Homework for advancers #2.

