UVa11127 - Triple-Free Binary Strings

Problem J
Triple-Free Binary Strings
Input: Standard Input

Output: Standard Output

 

A binary string consists of ones and zeros. Given a binary string T, if  there is no binary string S such that SSS (concatenate three copies of S together) is a substring of T, we say T is triple-free.

A pattern consists of ones, zeros and asterisks, where an asterisk(*) can be replaced by either one or zero. For example, the pattern 0**1 contains strings 0001, 0011, 0101, 0111, but not 1001 or 0000.

Given a pattern P, how many triple-free binary strings does it contain?

InputEach line of the input represents a test case, which contains the length of pattern, n(0<n<31), and the pattern P. There can be maximum 35 test cases. 

The input terminates when n=0.

OutputFor each test case, print the case number and the answer, shown below. 

 

Sample Input                       Output for Sample Input

4 0**1

5 *****

10 **01**01**

0

Case 1: 2

Case 2: 16

Case 3: 9

import java.io.IOException;import java.io.FileReader;import java.io.BufferedReader;import java.io.InputStreamReader;import java.util.Scanner;import java.util.TreeSet;class Main{public static final boolean DEBUG = false;public static final int N = 40;public static char[] buf = new char[N];public static char[] b = new char[N];public static TreeSet<Integer> ts = new TreeSet<Integer>();public static int ans, n;public static void dfs(int cur){if (cur >= 3) {int t = cur / 3 * 3;for (int i = 3; i <= t; i += 3) {int temp = 0;for (int j = cur - i; j < cur; j++) {temp <<= 1;temp |= b[j] - '0';}if (ts.contains(temp)) return;}}if (cur == n) {ans++;return;}if (buf[cur] != '*') {b[cur] = buf[cur];dfs(cur + 1);} else {b[cur] = '0';dfs(cur + 1);b[cur] = '1';dfs(cur + 1);}}public static void main(String[] args) throws IOException{Scanner cin;String s;int t = 1;if (DEBUG) {cin = new Scanner(new FileReader("d:\\OJ\\uva_in.txt"));} else {cin = new Scanner(new InputStreamReader(System.in));}for (int i = 1; i <= 10; i++) {for (int j = 0; j < (1 << i); j++) {ts.add(j | (j << i) | (j << (2 * i)));}}while (cin.hasNext()) {n = cin.nextInt();if (n == 0) break;s = cin.next();buf = s.toCharArray();ans = 0;dfs(0);System.out.println("Case " + (t++) + ": " + ans);}}}