USACOW:

1. USACOW: The Aftermath

2. PotW Solution Scanner s = newScanner(System.in); intn = s.nextInt(), k = s.nextInt(), x = 0; for(int i = 0; i < n; i++) x ^= s.nextInt() % (k + 1); System.out.println(x == 0 ? "John" : "Bessie");

3. USACO 2011November Contest • Officially over now (hopefully everyone took it) • Preliminary results can be found at: http://usaco.org/index.php?page=nov11results • Yes, we can see everyone's scores! • Problems in all divisions and their test data are also available. • Yay, more practice problems to do!

4. Bronze Problem: digits • A number, N, is converted to base 2 and base 3, but one digit is written incorrectly in each case. • Given these two (incorrect) results, find N. • N < 1,000,000,000 • Test all possible changes to the base 2 number and see if resulting number can be reached with one digit change to the base 3 number • If difference between resulting number and base 3 input is 1 or 2 times power of 3

5. Bronze Solution BufferedReader f = newBufferedReader(newFileReader("digits.in")); PrintWriter out = newPrintWriter(newBufferedWriter(newFileWriter("digits.out"))); String base2Str = f.readLine(); String base3Str = f.readLine(); intbase2 = 0, base3 = 0; for(inti = base2Str.length() - 1, digit = 1; i >= 0; digit *= 2, i--) base2 += digit * (base2Str.charAt(i) - '0'); for(inti = base3Str.length() - 1, digit = 1; i >= 0; digit *= 3, i--) base3 += digit * (base3Str.charAt(i) - '0'); for(intpowerOf2 = 1, digitIndex = base2Str.length() - 1; digitIndex>= 0; powerOf2 *= 2, digitIndex--) { intpossibleN = (base2Str.charAt(digitIndex) == '1') ? base2 - powerOf2 : base2 + powerOf2; intdiff = Math.abs(possibleN - base3); if(isPowerOf3(diff)) { out.println(possibleN); out.close(); System.exit(0); } } privatestaticbooleanisPowerOf3(intnum) { while(num % 3 == 0) num/= 3; returnnum == 1 || num == 2; }

6. Silver Problem: lineup • N cows are standing at various positions along a line. • Each cow has an integer position and an integer “breed ID” • Given each cow’s position and breed number, determine the minimum size (range of positions) of a photograph capturing at least one of each breed of cow.

7. Silver Problem (cont.) class Pair implements Comparable<Pair> {intx, id;      Pair(int x_,int id_){x = x_; id = id_;}publicintcompareTo(Pair o) {if(x != o.x)return (x < o.x) ? -1 : 1;return (id < o.id) ? -1 : 1;      }} • Utility class for sorting pairs of numbers • Implement comparable • Sort by first, then by second • Very often useful

8. Silver Solution Scanner s = new Scanner(System.in);intnum = s.nextInt(), idc = 0; // idc is number of unique idsArrayList<Pair> list = newArrayList<Pair>();HashMap<Integer,Integer> map = newHashMap<Integer,Integer>();for(inti = 0; i < num; i++) {intx = s.nextInt(), id = s.nextInt(); // scan in x coordinates and ids if(!map.containsKey(id)) // compress ids using HashSet map.put(id,idc++); list.add(newPair(x,map.get(id))); }Collections.sort(list); // sort cows by x coordinateint[] cnt = newint[idc]; // bins cows by idintexistc = 0, left = 0, right = 0, best = 1100000000;while(left < num && right < num) {while(right < num && existc < idc) { // until all ids are present intcur = list.get(right++).id; // sweep right, adding cows if(cnt[cur]++ == 0) existc++; } if(existc== idc) { // if this is valid, modify best accordingly intd = list.get(right-1).x - list.get(left).x; if(d < best) best = d; } intcur = list.get(left++).id; // increment left, erase cow if(cnt[cur]-- == 1) existc--; }System.out.println(best);

9. Gold Problem: median Problem: Given n≤100,000 numbers, count how many continuous subsequences have median≥k, where the median's index is rounded up if the subsequence has even length. • Obsrv. #1: O(n2) is too slow for full credit • Obsrv. #2: This problem is equivalent to counting how many subsequences have at least half their numbers greater than k. • Obsrv. #3: Note that if you replace k with 0, you can then replace every number in the list ≥k with 1, and <k with -1.

10. Gold Problem (cont.) • Obsrv. #4: Take prefix sums of the modified list - e.g. {-1,1,-1,-1,1} turns into {-1,0,-1,-2,-1} • Obsrv. #5: Now you just have to determine, for each number in the prefix sum list, how many numbers to the left of it are ≤ it. • The problem has been greatly simplified. • Note that this answer is essentially the reverse of the number of inversions of the prefix sum list • From this point on, use either: • Mergesort - used to count # of inversions • Binary Indexed or Fenwick Tree - take advantage of each prefix sum lying within [-n,n]

11. Gold Solution • bit: stores the Binary Indexed Tree • j & -j isolates the last significant digit - this operation forms the core of BITs • sum: prefix sum - artificially shifted by num + 1 to become nonnegative • sum++ or sum-- depending on s.nextInt() • ans: stores final answer - queries from the BIT are added to it Scanner s = new Scanner(System.in);intnum = s.nextInt(), thresh = s.nextInt(), sum = num + 1;int[] bit = newint[2 * num + 2];longans = 0;for(inti = 0; i < num; i++) {for (int j = sum; j < bit.length; j += (j & -j))    bit[j]++;if (s.nextInt() >= thresh)    sum++;else    sum--;for (int j = sum; j > 0; j -= (j & -j))ans += bit[j];}System.out.println(ans);

12. PotW: Cow Whitewashing • One day, at school, Bessie wrote a string of matching '('s and ')'s. As soon as she got home, however, she noticed that she had definitely made some mistakes! • She plans on covering this up by simply erasing some of the parentheses on the left and right, while aiming for the longest substring. • The substring she obtains must be a valid sequence of parentheses - each '(' must match to a ')' to its right, and vice versa. Tell Bessie the length of the longest valid substring. • Sample Input - strings of parentheses())                         ()()                )(()))(()()) • Sample Output - length of longest valid substring2                           4                  6 • For 15 pts: length < 100 • For 25 pts: length < 1,000 • For 35 pts: length < 1,000,000