hdu 4759 Poker Shuffle

读通题意应该要往二进制的方向去想,因为牌的个数是(1<<n)个的,

尝试发现每一次操作都是对数字的循环右移一位,如果是把奇数放到前面则还要异或(1<<n-1);

发现这个规律后,我们可以选循环右移再异或(而且可以异或上任何值,我们可以假设我们已经事先循环右移了n次);

枚举循环右移的次数,然后异或上一个值使它变成我们想要的值,因为  a ^ c = b - > a ^ b = c

这样我们这样比较所得到的c是否相同就可以了;

 1 import java.util.*;
 2 import java.math.*;
 3 import java.io.*;
 4 
 5 public class Main {
 6     public static void main(String[] arg) {
 7         Solve t = new Solve();
 8         t.main();
 9     }
10 }
11 class Solve{
12     int n;
13     BigInteger right(BigInteger x) {
14         BigInteger tp = x.and(BigInteger.valueOf(1));
15         return x.shiftRight(1).or(tp.shiftLeft(n-1));
16     }
17     public void main(){
18         Scanner cin = new Scanner(System.in);
19         int T = cin.nextInt();
20         int cas = 0;
21         while (T-- != 0) {
22             n = cin.nextInt();
23             BigInteger a = cin.nextBigInteger();
24             BigInteger x = cin.nextBigInteger();
25             BigInteger b = cin.nextBigInteger();
26             BigInteger y = cin.nextBigInteger();
27             BigInteger tp = BigInteger.valueOf(-1);
28             a = a.add(tp);
29             x = x.add(tp);
30             b = b.add(tp);
31             y = y.add(tp);
32             int fg = 0;
33             for (int i = 0; i < n; i++) {
34                 x = right(x);
35                 y = right(y);
36                 BigInteger ta = a.xor(x);
37                 BigInteger tb = b.xor(y);
38                 if (ta.compareTo(tb) == 0) {
39                     fg = 1;
40                     break;
41                 }
42             }
43             cas++;
44             System.out.print("Case "+cas+": ");
45             if (fg == 1) System.out.println("Yes");
46             else System.out.println("No");
47         }
48         
49     }
50     
51 }
View Code

 

 

posted @ 2013-11-03 19:54  Rabbit_hair  阅读(404)  评论(0编辑  收藏  举报