HDU 4759 Poker Shuffle

Poker Shuffle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 226    Accepted Submission(s): 78

Problem Description

Jason is not only an ACMer, but also a poker nerd. He is able to do a perfect shuffle. In a perfect shuffle, the deck containing K cards, where K is an even number, is split into equal halves of K/2 cards which are then pushed together in a certain way so as to make them perfectly interweave. Suppose the order of the cards is (1, 2, 3, 4, …, K-3, K-2, K-1, K). After a perfect shuffle, the order of the cards will be (1, 3, …, K-3, K-1, 2, 4, …, K-2, K) or (2, 4, …, K-2, K, 1, 3, …, K-3, K-1). 
Suppose K=2^N and the order of the cards is (1, 2, 3, …, K-2, K-1, K) in the beginning, is it possible that the A-th card is X and the B-th card is Y after several perfect shuffles?

 

 

Input

Input to this problem will begin with a line containing a single integer T indicating the number of datasets.
Each case contains five integer, N, A, X, B, Y. 1 <= N <= 1000, 1 <= A, B, X, Y <= 2^N.

 

 

Output

For each input case, output 「Yes」 if it is possible that the A-th card is X and the B-th card is Y after several perfect shuffles, otherwise 「No」.

 

 

Sample Input

3 1 1 1 2 2 2 1 2 4 3 2 1 1 4 2

 

 

Sample Output

Case 1: Yes Case 2: Yes Case 3: No

 

 

Source

2013 ACM/ICPC Asia Regional Changchun Online

 

 

Recommend

liuyiding

 

舉個例子，當n=3，每張牌的編號從0開始時：

 每次洗牌的時候，奇數在後偶數在前時，只需循環右移一下，以下：

0(000) -->0(000) -->0(000) -->0(000)

1(001) -->4(100) -->2(010) -->1(001)

2(010) -->1(001) -->4(100) -->2(010)

3(011) -->5(101) -->6(110) -->3(011)

4(100) -->2(010) -->1(001) -->4(100)

5(101) -->6(110) -->3(011) -->5(101)

6(110) -->3(011) -->5(101) -->6(110)

7(111) -->7(111) -->7(111) -->7(111)

 

奇數在前偶數在後時，只需右移一下，高位亦或1，以下（從右向左看）：

000(0) -> 100(4) -> 110(6) -> 111(7) -> 011(3) -> 001(1) -> 000(0)

001(1) -> 000(0) -> 100(4) -> 110(6) -> 111(7) -> 011(3) -> 001(1)

010(2) -> 101(5) -> 010(2) -> 101(5) -> 010(2) -> 101(5) -> 010(2)

011(3) -> 001(1) -> 000(0) -> 100(4) -> 110(6) -> 111(7) -> 011(3)

100(4) -> 110(6) -> 111(7) -> 011(3) -> 001(1) -> 000(0) -> 100(4)

101(5) -> 010(2) -> 101(5) -> 010(2) -> 101(5) -> 010(2) -> 101(5)

110(6) -> 111(7) -> 011(3) -> 001(1) -> 000(0) -> 100(4) -> 110(6)

111(7) -> 011(3) -> 001(1) -> 000(0) -> 100(4) -> 110(6) -> 111(7)

 

從上面兩個表能夠看出，每層的任意2個數異或的結果相同，且都爲n。

 

eg：

000(0) -> 100(4) -> 110(6) -> 111(7) -> 011(3) -> 001(1) -> 000(0)

001(1) -> 000(0) -> 100(4) -> 110(6) -> 111(7) -> 011(3) -> 001(1)

010(2) -> 101(5) -> 010(2) -> 101(5) -> 010(2) -> 101(5) -> 010(2)

每行的紅色數字異或，粉紅色數字異或，藍色數字異或------>都爲7，因此，你懂得

 

import java.math.*;
import java.util.*;

public class Main{
    static int n;
    static Scanner cin = new Scanner(System.in);
    
    static BigInteger rot(BigInteger x){
        BigInteger tmp = x.and(BigInteger.valueOf(1));
        return x.shiftRight(1).or(tmp.shiftLeft(n-1));
    }
    
    public static void main(String[] args){
        int t = cin.nextInt(),    cases=0;
        while(++cases <= t){
            n = cin.nextInt();
            BigInteger A = cin.nextBigInteger(),    X = cin.nextBigInteger();
            BigInteger B = cin.nextBigInteger(),    Y = cin.nextBigInteger();
            A = A.add(BigInteger.valueOf(-1));
            B = B.add(BigInteger.valueOf(-1));
            X = X.add(BigInteger.valueOf(-1));
            Y = Y.add(BigInteger.valueOf(-1));
            String ans = "No";
            for(int i=0; i <= n; i++){
                A = rot(A);
                B = rot(B);
                BigInteger tmpx = A.xor(X);
                BigInteger tmpy = B.xor(Y);
                if(tmpx.compareTo(tmpy)==0){
                    ans = "Yes";
                    break;
                }
            }
            System.out.println("Case " + cases + ": " + ans);
        }
    }
}