2016 Google code jam 大赛
二，RoundC
import java.io.BufferedReader; import java.io.FileInputStream; import java.io.FileNotFoundException; import java.io.FileOutputStream; import java.io.InputStreamReader; import java.io.PrintStream; import java.util.ArrayList; import java.util.HashMap; import java.util.Scanner; public class Main { public static void main(String[] args) throws FileNotFoundException { FileInputStream fis = new FileInputStream("A-large.in"); PrintStream out = new PrintStream(new FileOutputStream("A-large.out")); System.setIn(fis); System.setOut(out); Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in))); // System.out.println((char)(65)); int T = in.nextInt(); for (int t = 1; t <= T; ++t) { int n = in.nextInt(); HashMap<Character, Long> map = new HashMap(); int start = 65; for (int i = 0; i < n; ++i) { map.put((char)(start++), in.nextLong()); } // 用贪心每次都找最多的。 // System.out.println(hash); ArrayList<String> res = new ArrayList(); // 特判三人的时候 String now = new String(); while(!map.isEmpty()) { Long max = (long) 0; char cur = ' '; for (char c : map.keySet()) { if (map.get(c) > max) { cur = c; max = map.get(c); } } if (map.get(cur) == 1) { map.remove(cur); } else { map.put(cur, map.get(cur) - 1); } now += cur; if (now.length() == 2 || map.isEmpty()) { res.add(now); now = ""; } } if (res.size() != 1 && res.get(res.size() - 1).length() == 1) { // swap String a = res.get(res.size() - 1); String b = res.get(res.size() - 2); res.set(res.size() - 1, b); res.set(res.size() - 2, a); } System.out.print("Case #" + t + ":" ); for (String s : res) { System.out.print(" " + s); } System.out.println(); } } }
一，资格赛：（100分，拿了65分。最后有事没做第四题。————2016年-04月-09日）
1，Problem A. Counting Sheep
Problem A. Counting Sheep Confused? Read the quick-start guide. Small input 7 points You have solved this input set. Large input 8 points You have already tried this input set. (Judged at the end of the contest.) Problem Bleatrix Trotter the sheep has devised a strategy that helps her fall asleep faster. First, she picks a number N. Then she starts naming N, 2 × N, 3 × N, and so on. Whenever she names a number, she thinks about all of the digits in that number. She keeps track of which digits (0, 1, 2, 3, 4, 5, 6, 7, 8, and 9) she has seen at least once so far as part of any number she has named. Once she has seen each of the ten digits at least once, she will fall asleep. Bleatrix must start with N and must always name (i + 1) × N directly after i × N. For example, suppose that Bleatrix picks N = 1692. She would count as follows: N = 1692. Now she has seen the digits 1, 2, 6, and 9. 2N = 3384. Now she has seen the digits 1, 2, 3, 4, 6, 8, and 9. 3N = 5076. Now she has seen all ten digits, and falls asleep. What is the last number that she will name before falling asleep? If she will count forever, print INSOMNIA instead. Input The first line of the input gives the number of test cases, T. T test cases follow. Each consists of one line with a single integer N, the number Bleatrix has chosen. Output For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the last number that Bleatrix will name before falling asleep, according to the rules described in the statement. Limits 1 ≤ T ≤ 100. Small dataset 0 ≤ N ≤ 200. Large dataset 0 ≤ N ≤ 106. Sample Input Output 5 0 1 2 11 1692 Case #1: INSOMNIA Case #2: 10 Case #3: 90 Case #4: 110 Case #5: 5076 In Case #1, since 2 × 0 = 0, 3 × 0 = 0, and so on, Bleatrix will never see any digit other than 0, and so she will count forever and never fall asleep. Poor sheep! In Case #2, Bleatrix will name 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. The 0 will be the last digit needed, and so she will fall asleep after 10. In Case #3, Bleatrix will name 2, 4, 6... and so on. She will not see the digit 9 in any number until 90, at which point she will fall asleep. By that point, she will have already seen the digits 0, 1, 2, 3, 4, 5, 6, 7, and 8, which will have appeared for the first time in the numbers 10, 10, 2, 30, 4, 50, 6, 70, and 8, respectively. In Case #4, Bleatrix will name 11, 22, 33, 44, 55, 66, 77, 88, 99, 110 and then fall asleep. Case #5 is the one described in the problem statement. Note that it would only show up in the Large dataset, and not in the Small dataset.
答案：
import java.util.*; import java.io.*; public class Main { public static void main(String[] args) throws FileNotFoundException { FileInputStream fis = new FileInputStream("A-large.in"); PrintStream out = new PrintStream(new FileOutputStream("A-large.out")); System.setIn(fis); System.setOut(out); Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in))); int t = in.nextInt(); // Scanner has functions to read ints, longs, strings, chars, etc. for (int i = 1; i <= t; ++i) { int n = in.nextInt(); HashSet<Integer> hash = new HashSet(); if (n == 0) { System.out.println("Case #" + i + ": INSOMNIA"); } else { int k = 1; while (hash.size() != 10) { hash.addAll(getNum(n * k)); k++; // System.out.println(hash.toString()); } System.out.println("Case #" + i + ": " + n * (k - 1)); } } } private static HashSet<Integer> getNum (int n) { HashSet<Integer> hash = new HashSet(); while (n != 0) { hash.add(n % 10); n = n / 10; } return hash; } }
2，Problem B. Revenge of the Pancakes
Problem B. Revenge of the Pancakes This contest is open for practice. You can try every problem as many times as you like, though we won't keep track of which problems you solve. Read the Quick-Start Guide to get started. Small input 10 points Solve B-small Large input 10 points Solve B-large Problem The Infinite House of Pancakes has just introduced a new kind of pancake! It has a happy face made of chocolate chips on one side (the "happy side"), and nothing on the other side (the "blank side"). You are the head waiter on duty, and the kitchen has just given you a stack of pancakes to serve to a customer. Like any good pancake server, you have X-ray pancake vision, and you can see whether each pancake in the stack has the happy side up or the blank side up. You think the customer will be happiest if every pancake is happy side up when you serve them. You know the following maneuver: carefully lift up some number of pancakes (possibly all of them) from the top of the stack, flip that entire group over, and then put the group back down on top of any pancakes that you did not lift up. When flipping a group of pancakes, you flip the entire group in one motion; you do not individually flip each pancake. Formally: if we number the pancakes 1, 2, ..., N from top to bottom, you choose the top i pancakes to flip. Then, after the flip, the stack is i, i-1, ..., 2, 1, i+1, i+2, ..., N. Pancakes 1, 2, ..., i now have the opposite side up, whereas pancakes i+1, i+2, ..., N have the same side up that they had up before. For example, let's denote the happy side as + and the blank side as -. Suppose that the stack, starting from the top, is --+-. One valid way to execute the maneuver would be to pick up the top three, flip the entire group, and put them back down on the remaining fourth pancake (which would stay where it is and remain unchanged). The new state of the stack would then be -++-. The other valid ways would be to pick up and flip the top one, the top two, or all four. It would not be valid to choose and flip the middle two or the bottom one, for example; you can only take some number off the top. You will not serve the customer until every pancake is happy side up, but you don't want the pancakes to get cold, so you have to act fast! What is the smallest number of times you will need to execute the maneuver to get all the pancakes happy side up, if you make optimal choices? Input The first line of the input gives the number of test cases, T. T test cases follow. Each consists of one line with a string S, each character of which is either + (which represents a pancake that is initially happy side up) or - (which represents a pancake that is initially blank side up). The string, when read left to right, represents the stack when viewed from top to bottom. Output For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the minimum number of times you will need to execute the maneuver to get all the pancakes happy side up. Limits 1 ≤ T ≤ 100. Every character in S is either + or -. Small dataset 1 ≤ length of S ≤ 10. Large dataset 1 ≤ length of S ≤ 100. Sample Input Output 5 - -+ +- +++ --+- Case #1: 1 Case #2: 1 Case #3: 2 Case #4: 0 Case #5: 3 In Case #1, you only need to execute the maneuver once, flipping the first (and only) pancake. In Case #2, you only need to execute the maneuver once, flipping only the first pancake. In Case #3, you must execute the maneuver twice. One optimal solution is to flip only the first pancake, changing the stack to --, and then flip both pancakes, changing the stack to ++. Notice that you cannot just flip the bottom pancake individually to get a one-move solution; every time you execute the maneuver, you must select a stack starting from the top. In Case #4, all of the pancakes are already happy side up, so there is no need to do anything. In Case #5, one valid solution is to first flip the entire stack of pancakes to get +-++, then flip the top pancake to get --++, then finally flip the top two pancakes to get ++++.
答案：
import java.io.BufferedReader; import java.io.FileInputStream; import java.io.FileNotFoundException; import java.io.FileOutputStream; import java.io.InputStreamReader; import java.io.PrintStream; import java.util.Scanner; public class Main { public static void main(String[] args) throws FileNotFoundException { FileInputStream fis = new FileInputStream("B-large.in"); PrintStream out = new PrintStream(new FileOutputStream("B-large.out")); System.setIn(fis); System.setOut(out); Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in))); int t = in.nextInt(); // Scanner has functions to read ints, longs, strings, chars, etc. for (int i = 1; i <= t; ++i) { String str = in.next(); boolean[] b = new boolean[str.length()]; for (int j = 0; j < str.length(); ++j) { if (str.charAt(j) == '+') { b[j] = true; } else { b[j] = false; } } long result = 0; int end = str.length() - 1; while (end >= 0) { while (end >= 0 && b[end]) { end--; } if (end < 0) { break; } int start = 0; if (!b[0]) { flip(b, 0, end); result++; } else { while (start <= end && b[start]) { start++; } flip(b, 0, start - 1); flip(b, 0, end); result = result + 2; } } System.out.println("Case #" + i + ": " + result); } } private static void flip(boolean[] b, int start, int end) { while (start <= end) { if (start == end) { b[start] = !b[start]; break; } boolean temp = !b[start]; b[start] = !b[end]; b[end] = temp; start++; end--; } } }
3，Problem C. Coin Jam
Problem C. Coin Jam This contest is open for practice. You can try every problem as many times as you like, though we won't keep track of which problems you solve. Read the Quick-Start Guide to get started. Small input 10 points Solve C-small Large input 20 points Solve C-large Problem A jamcoin is a string of N ≥ 2 digits with the following properties: Every digit is either 0 or 1. The first digit is 1 and the last digit is 1. If you interpret the string in any base between 2 and 10, inclusive, the resulting number is not prime. Not every string of 0s and 1s is a jamcoin. For example, 101 is not a jamcoin; its interpretation in base 2 is 5, which is prime. But the string 1001 is a jamcoin: in bases 2 through 10, its interpretation is 9, 28, 65, 126, 217, 344, 513, 730, and 1001, respectively, and none of those is prime. We hear that there may be communities that use jamcoins as a form of currency. When sending someone a jamcoin, it is polite to prove that the jamcoin is legitimate by including a nontrivial divisor of that jamcoin's interpretation in each base from 2 to 10. (A nontrivial divisor for a positive integer K is some positive integer other than 1 or K that evenly divides K.) For convenience, these divisors must be expressed in base 10. For example, for the jamcoin 1001 mentioned above, a possible set of nontrivial divisors for the base 2 through 10 interpretations of the jamcoin would be: 3, 7, 5, 6, 31, 8, 27, 5, and 77, respectively. Can you produce J different jamcoins of length N, along with proof that they are legitimate? Input The first line of the input gives the number of test cases, T. T test cases follow; each consists of one line with two integers N and J. Output For each test case, output J+1 lines. The first line must consist of only Case #x:, where x is the test case number (starting from 1). Each of the last J lines must consist of a jamcoin of length N followed by nine integers. The i-th of those nine integers (counting starting from 1) must be a nontrivial divisor of the jamcoin when the jamcoin is interpreted in base i+1. All of these jamcoins must be different. You cannot submit the same jamcoin in two different lines, even if you use a different set of divisors each time. Limits T = 1. (There will be only one test case.) It is guaranteed that at least J distinct jamcoins of length N exist. Small dataset N = 16. J = 50. Large dataset N = 32. J = 500. Note that, unusually for a Code Jam problem, you already know the exact contents of each input file. For example, the Small dataset's input file will always be exactly these two lines: 1 16 50 So, you can consider doing some computation before actually downloading an input file and starting the clock. Sample Input Output 1 6 3 Case #1: 100011 5 13 147 31 43 1121 73 77 629 111111 21 26 105 1302 217 1032 513 13286 10101 111001 3 88 5 1938 7 208 3 20 11 In this sample case, we have used very small values of N and J for ease of explanation. Note that this sample case would not appear in either the Small or Large datasets. This is only one of multiple valid solutions. Other sets of jamcoins could have been used, and there are many other possible sets of nontrivial base 10 divisors. Some notes: 110111 could not have been included in the output because, for example, it is 337 if interpreted in base 3 (1*243 + 1*81 + 0*27 + 1*9 + 1*3 + 1*1), and 337 is prime. 010101 could not have been included in the output even though 10101 is a jamcoin, because jamcoins begin with 1. 101010 could not have been included in the output, because jamcoins end with 1. 110011 is another jamcoin that could have also been used in the output, but could not have been added to the end of this output, since the output must contain exactly J examples. For the first jamcoin in the sample output, the first number after 100011 could not have been either 1 or 35, because those are trivial divisors of 35 (100011 in base 2).
答案：
small
import java.io.BufferedReader; import java.io.FileInputStream; import java.io.FileNotFoundException; import java.io.FileOutputStream; import java.io.InputStreamReader; import java.io.PrintStream; import java.util.ArrayList; import java.util.Arrays; import java.util.Scanner; public class Main { public static void main(String[] args) throws FileNotFoundException { // FileInputStream fis = new FileInputStream("C-small-attempt1.in"); // PrintStream out = new PrintStream(new FileOutputStream("C-small-attempt1.out")); // System.setIn(fis); // System.setOut(out); Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in))); int t = in.nextInt(); // Scanner has functions to read longs, longs, strings, chars, etc. for (int i = 1; i <= t; ++i) { int n = in.nextInt(); int J = in.nextInt(); long[] num = new long[n]; num[0] = 1; num[n - 1] = 1; long jNum = 0; System.out.println("Case #" + i + ": "); for (long k = 1; k < (long)Math.pow(2, n); k++) { ArrayList<Long> list = new ArrayList(); for (long base = 2; base <= 10; base++) { long value = getBase(num, base); // System.out.prlongln(value); if (!isPrime(value)) { list.add(value); } } if (list.size() == 9) { jNum++; for (int p = 0; p < n; p++) { System.out.print(num[p]); } for (long li : list) { System.out.print(" " + getDivisor(li) + " "); } System.out.println(); if (jNum == J) { break; } } binarySum(num); } } } private static long getDivisor(long value) { long ans = 0; for (long i = 2; i < value; i++) { if (value % i == 0) { ans = i; break; } } return ans; } private static void binarySum(long[] num) { long digit = 1; for (int i = num.length - 2; i >=1; i--) { if (num[i] + digit != 2) { num[i] = num[i] + digit; break; } else { num[i] = 0; digit = 1; } } } private static long getBase(long[] num, long base) { long result = 0; for (int i = 0; i < num.length; i++) { result = result * base + num[i]; } return result; } private static boolean isPrime(long n) { if (n <= 1) { return false; } long k = (long) Math.sqrt(n); for (long i = 2; i <= k; i++) { if(n % i == 0) { return false; } } return true; } }
大数据，因为数据巨大。所以，用了BigInteger。
import java.io.BufferedReader; import java.io.FileInputStream; import java.io.FileNotFoundException; import java.io.FileOutputStream; import java.io.InputStreamReader; import java.io.PrintStream; import java.math.BigInteger; import java.util.ArrayList; import java.util.Scanner; public class Main { public static void main(String[] args) throws FileNotFoundException { FileInputStream fis = new FileInputStream("C-large.in"); PrintStream out = new PrintStream(new FileOutputStream("C-large.out")); System.setIn(fis); System.setOut(out); Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in))); int t = in.nextInt(); // Scanner has functions to read longs, longs, strings, chars, etc. for (int i = 1; i <= t; ++i) { int n = in.nextInt(); int J = in.nextInt(); long[] num = new long[n]; num[0] = 1; num[n - 1] = 1; long jNum = 0; System.out.println("Case #" + i + ": "); for (long k = 1; k < (long)Math.pow(2, n); k++) { ArrayList<BigInteger> list = new ArrayList(); for (long base = 2; base <= 10; base++) { BigInteger value = getBase(num, base); // System.out.prlongln(value); if (!isPrime(value)) { list.add(value); } } if (list.size() == 9) { jNum++; for (int p = 0; p < n; p++) { System.out.print(num[p]); } for (BigInteger li : list) { System.out.print(" " + getDivisor(li) + " "); } System.out.println(); if (jNum == J) { break; } } // System.out.println("here" + jNum); binarySum(num); } } } private static BigInteger getDivisor(BigInteger value) { BigInteger ans = new BigInteger("0"); for (BigInteger i = new BigInteger("2"); i.compareTo(value) < 0; i = i.add(new BigInteger("1"))) { if ((value.remainder(i)).equals(new BigInteger("0"))) { ans = i; break; } // System.out.println("get"); } return ans; } private static void binarySum(long[] num) { long digit = 1; for (int i = num.length - 2; i >=1; i--) { if (num[i] + digit != 2) { num[i] = num[i] + digit; break; } else { num[i] = 0; digit = 1; } } } private static BigInteger getBase(long[] num, long base) { BigInteger result = new BigInteger("0"); for (int i = 0; i < num.length; i++) { result = result.multiply(new BigInteger(String.valueOf(base))); result = result.add(new BigInteger(String.valueOf(num[i]))); // System.out.println("base"); } return result; } private static boolean isPrime(BigInteger n) { int[] temp = {2, 3, 5, 7, 11,3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101}; for (int a : temp) { if((n.remainder(new BigInteger(a + ""))).equals(new BigInteger("0"))) { return false; } // System.out.println("prime"); } return true; } }
4，
Problem D. Fractiles This contest is open for practice. You can try every problem as many times as you like, though we won't keep track of which problems you solve. Read the Quick-Start Guide to get started. Small input 10 points Solve D-small Large input 25 points Solve D-large Problem Long ago, the Fractal civilization created artwork consisting of linear rows of tiles. They had two types of tile that they could use: gold (G) and lead (L). Each piece of Fractal artwork is based on two parameters: an original sequence of K tiles, and a complexity C. For a given original sequence, the artwork with complexity 1 is just that original sequence, and the artwork with complexity X+1 consists of the artwork with complexity X, transformed as follows: replace each L tile in the complexity X artwork with another copy of the original sequence replace each G tile in the complexity X artwork with K G tiles For example, for an original sequence of LGL, the pieces of artwork with complexity 1 through 3 are: C = 1: LGL (which is just the original sequence) C = 2: LGLGGGLGL C = 3: LGLGGGLGLGGGGGGGGGLGLGGGLGL Here's an illustration of how the artwork with complexity 2 is generated from the artwork with complexity 1: You have just discovered a piece of Fractal artwork, but the tiles are too dirty for you to tell what they are made of. Because you are an expert archaeologist familiar with the local Fractal culture, you know the values of K and C for the artwork, but you do not know the original sequence. Since gold is exciting, you would like to know whether there is at least one G tile in the artwork. Your budget allows you to hire S graduate students, each of whom can clean one tile of your choice (out of the KC tiles in the artwork) to see whether the tile is G or L. Is it possible for you to choose a set of no more than S specific tiles to clean, such that no matter what the original pattern was, you will be able to know for sure whether at least one G tile is present in the artwork? If so, which tiles should you clean? Input The first line of the input gives the number of test cases, T. T test cases follow. Each consists of one line with three integers: K, C, and S. Output For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is either IMPOSSIBLE if no set of tiles will answer your question, or a list of between 1 and S positive integers, which are the positions of the tiles that will answer your question. The tile positions are numbered from 1 for the leftmost tile to KC for the rightmost tile. Your chosen positions may be in any order, but they must all be different. If there are multiple valid sets of tiles, you may output any of them. Remember that once you submit a Small and it is accepted, you will not be able to download and submit another Small input. See the FAQ for a more thorough explanation. This reminder won't appear in problems in later rounds. Limits 1 ≤ T ≤ 100. 1 ≤ K ≤ 100. 1 ≤ C ≤ 100. KC ≤ 1018. Small dataset S = K. Large dataset 1 ≤ S ≤ K. Sample Input Output 5 2 3 2 1 1 1 2 1 1 2 1 2 3 2 3 Case #1: 2 Case #2: 1 Case #3: IMPOSSIBLE Case #4: 1 2 Case #5: 2 6 Note: for some of these sample cases, other valid solutions exist. In sample case #1, there are four possible original sequences: GG, GL, LG, and LL. They would produce the following artwork, respectively: Original sequence GG: GGGGGGGG Original sequence GL: GGGGGGGL Original sequence LG: LGGGGGGG Original sequence LL: LLLLLLLL One valid solution is to just look at tile #2. If tile #2 turns out to be G, then you will know for sure the artwork contains at least one G. (You will not know whether the original sequence is GG, GL, or LG, but that doesn't matter.) If tile #2 turns out to be L, then you will know that the original sequence must be LL, so there are no Gs in the artwork. So 2 is a valid solution. On the other hand, it would not be valid to just look at tile #1. If it turns out to be L, you will only know that the original sequence could have been either LG or LL. If the original sequence is LG, there is at least one G in the artwork, but if the original sequence is LL, there are no Gs. So 1 would not be a valid solution. Note that 1 2 is also a valid solution, because tile #2 already provides all the information you need. 1 2 3 is not a valid solution, because it uses too many tiles. In sample case #2, the artwork must consist of only one tile: either G or L. Looking at that tile will trivially tell you whether or not the artwork has a G in it. In sample case #3, which would not appear in the Small dataset, the artwork must be either GG, GL, LG, or LL. You can only look at one tile, and neither of them on its own is enough to answer the question. If you see L for tile #1, you will not know whether the artwork is LG or LL, so you will not know whether any Gs are present. If you see L for tile #2, you will not know whether the artwork is GL or LL, so you will not know whether any Gs are present. Sample case #4 is like sample case #3, but with access to one more tile. Now you can just look at the entire artwork. In sample case #5, there are eight possible original sequences, and they would produce the following artwork: Original sequence GGG: GGGGGGGGG Original sequence GGL: GGGGGGGGL Original sequence GLG: GGGGLGGGG Original sequence GLL: GGGGLLGLL Original sequence LGG: LGGGGGGGG Original sequence LGL: LGLGGGLGL Original sequence LLG: LLGLLGGGG Original sequence LLL: LLLLLLLLL One valid solution is to look at tiles #2 and #6. If they both turn out to be Ls, the artwork must be all Ls. Otherwise, there must at least one G. Note that 1 2 would not be a valid solution, because even if those tiles both turn out to be Ls, that does not rule out an original sequence of LLG. 6 2 would be a valid solution, since the order of the positions in your solution does not matter.
