Monday, June 19, 2006 Match summary The 1,000point problem turned out to be easy in this match, with 40 coders successfully solving it. Weiqi took the lead after the coding phase but, thanks to two +75s in challenge phase, Burunduk1 got up to the first place and won his third HS SRM in a row. Weiqi finished second, third place went to Loner, and fourth and fifth went to newcomers rlp and ilyakor respectively. The ProblemsBestSellerUsed as: Level One:
This problem asked for a value that occurs most times in the given array. Constraints were so low that it was not neccessary to use any sophisticated data structure; you could just iterate values and, for each value, count the amount of times it occurs in the array, then choose the best according to the criteria described in the problem. Java code follows: public String findBestSeller(String[] items) { int n, i, j, best, cur; String ret; n = items.length; best = 1; ret = ""; for (i = 0; i < n; i++) { cur = 0; for (j = 0; j < n; j++) if (items[j].compareTo(items[i]) == 0) cur++; if (cur > best  (cur == best && items[i].compareTo(ret) < 0)) { best = cur; ret = items[i]; } } return ret; }KidsWordGame Used as: Level Two:
First of all let's write a function which will be able to say whether the word b might be obtained from word a without violating the rules (i.e. without cheating). Java code of this function follows: boolean ok(String a, String b) { if (b.length() == a.length() + 1 && b.indexOf(a) != 1) return true; else return false; }Now we can write the solution itself: public int getCheater(String[] first, String[] second, String[] third) { int n, i, j; String s; s = first[0]; // The original word n = Math.max(first.length, Math.max(second.length, third.length)); for (i = 0; i < n; i++) { if (i == first.length) break; // Notice that there is no previous move for the first move in the game if (i == 0  ok(s, first[i])) s = first[i]; else return 3; if (i == second.length) break; if (ok(s, second[i])) s = second[i]; else return 1; if (i == third.length) break; if (ok(s, third[i])) s = third[i]; else return 2; } return 1; } If some player saw a word that could not be obtained from the previous word, it would indicate that the previous palyer is a cheater. BestDecompositionUsed as: Level Three:
Here you were asked to find such f_{1}, f_{2},
..., f_{m}, that they sum up to n and their product is maximal.
Let's forget that all f's must be integers. Suppose that they may be float,
but don't forget that they must be nonnegative.
Now let me prove, that all f's must be equal in order to maximize their
product. Let's take some two of these numbers, say f_{i} = a and
_{j} = b. Then: struct BestDecomposition { int maxProduct(int n) { int i, j, k; k = 1; while (n > 4) { k *= 3; k %= 10007; n = 3; } k *= n; k %= 10007; return k; } }; 
