hdu 4810 Wall Painting 位操作

来源：互联网发布：知乎英文名编辑：程序博客网时间：2024/05/16 06:04

本题题意甚是费解。
找到合适的位操作，再运用组合数，是关键。

/** * 2015年7月15日 上午11:21:15 * PrjName:hdu4825 * @ Semprathlon */import java.io.*;class Trie {    private final int maxd = 33;    private long data;    private Trie ch0, ch1;    void insert(long n) {        Trie rt = this;        for (int i = maxd - 1; i >= 0; i--) {            if ((n & (1L << i)) == 0L) {// 0                if (rt.ch0 == null)                    rt.ch0 = new Trie();                rt = rt.ch0;            } else {// 1                if (rt.ch1 == null)                    rt.ch1 = new Trie();                rt = rt.ch1;            }            if (i == 0)                rt.data = n;        }    }    long query(long n) {        Trie rt = this;        for (int i = maxd - 1; i >= 0; i--) {            if ((n & (1L << i)) > 0L && rt.ch0 != null || rt.ch1 == null)// 0                rt = rt.ch0;            else if (rt.ch1 != null)// 1                rt = rt.ch1;        }        return rt.data;    }}public class Main {    public static void main(String[] args) throws IOException {        StreamTokenizer in = new StreamTokenizer(new BufferedReader(                new InputStreamReader(System.in)));        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));        int cas = 0;        in.nextToken();        int T = (int) in.nval;        while (T-- > 0) {            Trie tr = new Trie();            in.nextToken();            int n = (int) in.nval;            in.nextToken();            int m = (int) in.nval;            for (int i = 1; i <= n; i++) {                in.nextToken();                tr.insert((long) in.nval);            }            out.println("Case #" + (++cas) + ":");            for (int i = 1; i <= m; i++) {                in.nextToken();                out.println(tr.query((long) in.nval));            }        }        out.flush();        out.close();    }}

0 0