HDU 4759 Poker Shuffle 二进制分析(2013 ACM/ICPC Asia Regional Changchun Online A)

你为何这么叼！

Poker Shuffle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 214    Accepted Submission(s): 70

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

 

思路：主要是二进制的运用。---by（随心所欲）

为了方便从0开始，首先看下右移一下，高位异或1的规律：(可以从右往左一列一列看)

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)

每次洗牌的时候，奇数在后偶数在前时，只需右移一下；奇数在前偶数在后时，只需右移一下，高位亦或1.

而且每层的任意2个数异或的结果相同。

对于给定的n,a,b,x,y;只需判断a^b==x^y(此处是异或运算符)既可。

 

import java.math.*;
import java.util.*;
public class Main {
    public static void main(String arg[]){
        BigInteger a,b,c,x,y;
        Scanner cin=new Scanner(System.in);
        int t=1,n,tt;
        c=BigInteger.ONE;
        tt=cin.nextInt();
        while(tt-->0){
            n=cin.nextInt();
            a=cin.nextBigInteger();a=a.subtract(c);
            x=cin.nextBigInteger();x=x.subtract(c);
            b=cin.nextBigInteger();b=b.subtract(c);
            y=cin.nextBigInteger();y=y.subtract(c);
            a=a.xor(b);
            x=x.xor(y);
            boolean flag=false;
            for(int i=0;i<n;i++){
                if(a.equals(x)){
                    flag=true;
                    break;
                }
                if(a.testBit(0)){
                    a=a.shiftRight(1);
                    a=a.setBit(n-1);
                }else a=a.shiftRight(1);
            }
            if(flag) System.out.println("Case "+t+": Yes");
            else System.out.println("Case "+t+": No");
            t++;
        }
    }
}