HDU 4759 Poker Shuffle(2013长春网络赛1001题)

Poker Shuffle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 95    Accepted Submission(s): 24


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
 

 

Recommend
liuyiding
 

 

 

题目意思很简单。

就是洗牌,抽出奇数和偶数,要么奇数放前面,要么偶数放前面。

 

总共2^N张牌。

需要问的是,给了A X B Y  问经过若干洗牌后,第A个位置是X,第B个位置是Y 是不是可能的。

 

题目给的牌编号是1开始的,先转换成0开始。

一开始位置是0~2^N-1.  对应的牌是0~2^N-1

首先来看每次洗牌的过程。

对于第一种洗牌:将奇数放前面,偶数放后面。其实每个位置数的变化就是相当于循环右移一位,然后高位异或1.

对于第二种洗牌:讲偶数放前面,奇数放后面。其实每个位置数的变化就是相当于循环右移一位,然后高位异或0.

 

所以经过若干次洗牌,可以看成是循环右移了K位,然后异或上一个数。

 

所以对于题目的查询:

首先将A X B Y都减一。  然后枚举X,Y循环右移了K位以后,能不能同时异或上相同的数得到A,B

 

需要大数,然后转化成二进制就可以解决了。

循环右移X,Y,然后判断A ^ X 是不是等于 B ^ Y

 

 

  1 /* ***********************************************
  2 Author        :kuangbin
  3 Created Time  :2013/9/28 星期六 11:57:13
  4 File Name     :2013长春网络赛\1001.cpp
  5 ************************************************ */
  6 
  7 #pragma comment(linker, "/STACK:1024000000,1024000000")
  8 #include <stdio.h>
  9 #include <string.h>
 10 #include <iostream>
 11 #include <algorithm>
 12 #include <vector>
 13 #include <queue>
 14 #include <set>
 15 #include <map>
 16 #include <string>
 17 #include <math.h>
 18 #include <stdlib.h>
 19 #include <time.h>
 20 using namespace std;
 21 
 22 /*
 23  * 完全大数模板
 24  * 输出cin>>a
 25  * 输出a.print();
 26  * 注意这个输入不能自动去掉前导0的,可以先读入到char数组,去掉前导0,再用构造函数。
 27  */
 28 #define MAXN 9999
 29 #define MAXSIZE 1010
 30 #define DLEN 4
 31 
 32 class BigNum
 33 {
 34 public:
 35     int a[500];  //可以控制大数的位数
 36     int len;
 37 public:
 38     BigNum(){len=1;memset(a,0,sizeof(a));}  //构造函数
 39     BigNum(const int);     //将一个int类型的变量转化成大数
 40     BigNum(const char*);   //将一个字符串类型的变量转化为大数
 41     BigNum(const BigNum &); //拷贝构造函数
 42     BigNum &operator=(const BigNum &); //重载赋值运算符,大数之间进行赋值运算
 43     friend istream& operator>>(istream&,BigNum&); //重载输入运算符
 44     friend ostream& operator<<(ostream&,BigNum&); //重载输出运算符
 45 
 46     BigNum operator+(const BigNum &)const;  //重载加法运算符,两个大数之间的相加运算
 47     BigNum operator-(const BigNum &)const;  //重载减法运算符,两个大数之间的相减运算
 48     BigNum operator*(const BigNum &)const;  //重载乘法运算符,两个大数之间的相乘运算
 49     BigNum operator/(const int &)const;     //重载除法运算符,大数对一个整数进行相除运算
 50 
 51     BigNum operator^(const int &)const;     //大数的n次方运算
 52     int operator%(const int &)const;        //大数对一个int类型的变量进行取模运算
 53     bool operator>(const BigNum &T)const;   //大数和另一个大数的大小比较
 54     bool operator>(const int &t)const;      //大数和一个int类型的变量的大小比较
 55 
 56     void print();        //输出大数
 57 };
 58 BigNum::BigNum(const int b)   //将一个int类型的变量转化为大数
 59 {
 60     int c,d=b;
 61     len=0;
 62     memset(a,0,sizeof(a));
 63     while(d>MAXN)
 64     {
 65         c=d-(d/(MAXN+1))*(MAXN+1);
 66         d=d/(MAXN+1);
 67         a[len++]=c;
 68     }
 69     a[len++]=d;
 70 }
 71 BigNum::BigNum(const char *s)  //将一个字符串类型的变量转化为大数
 72 {
 73     int t,k,index,L,i;
 74     memset(a,0,sizeof(a));
 75     L=strlen(s);
 76     len=L/DLEN;
 77     if(L%DLEN)len++;
 78     index=0;
 79     for(i=L-1;i>=0;i-=DLEN)
 80     {
 81         t=0;
 82         k=i-DLEN+1;
 83         if(k<0)k=0;
 84         for(int j=k;j<=i;j++)
 85             t=t*10+s[j]-'0';
 86         a[index++]=t;
 87     }
 88 }
 89 BigNum::BigNum(const BigNum &T):len(T.len)  //拷贝构造函数
 90 {
 91     int i;
 92     memset(a,0,sizeof(a));
 93     for(i=0;i<len;i++)
 94         a[i]=T.a[i];
 95 }
 96 BigNum & BigNum::operator=(const BigNum &n)  //重载赋值运算符,大数之间赋值运算
 97 {
 98     int i;
 99     len=n.len;
100     memset(a,0,sizeof(a));
101     for(i=0;i<len;i++)
102         a[i]=n.a[i];
103     return *this;
104 }
105 istream& operator>>(istream &in,BigNum &b)
106 {
107     char ch[MAXSIZE*4];
108     int i=-1;
109     in>>ch;
110     int L=strlen(ch);
111     int count=0,sum=0;
112     for(i=L-1;i>=0;)
113     {
114         sum=0;
115         int t=1;
116         for(int j=0;j<4&&i>=0;j++,i--,t*=10)
117         {
118             sum+=(ch[i]-'0')*t;
119         }
120         b.a[count]=sum;
121         count++;
122     }
123     b.len=count++;
124     return in;
125 }
126 ostream& operator<<(ostream& out,BigNum& b)  //重载输出运算符
127 {
128     int i;
129     cout<<b.a[b.len-1];
130     for(i=b.len-2;i>=0;i--)
131     {
132         printf("%04d",b.a[i]);
133     }
134     return out;
135 }
136 BigNum BigNum::operator+(const BigNum &T)const   //两个大数之间的相加运算
137 {
138     BigNum t(*this);
139     int i,big;
140     big=T.len>len?T.len:len;
141     for(i=0;i<big;i++)
142     {
143         t.a[i]+=T.a[i];
144         if(t.a[i]>MAXN)
145         {
146             t.a[i+1]++;
147             t.a[i]-=MAXN+1;
148         }
149     }
150     if(t.a[big]!=0)
151        t.len=big+1;
152     else t.len=big;
153     return t;
154 }
155 BigNum BigNum::operator-(const BigNum &T)const  //两个大数之间的相减运算
156 {
157     int i,j,big;
158     bool flag;
159     BigNum t1,t2;
160     if(*this>T)
161     {
162         t1=*this;
163         t2=T;
164         flag=0;
165     }
166     else
167     {
168         t1=T;
169         t2=*this;
170         flag=1;
171     }
172     big=t1.len;
173     for(i=0;i<big;i++)
174     {
175         if(t1.a[i]<t2.a[i])
176         {
177             j=i+1;
178             while(t1.a[j]==0)
179                 j++;
180             t1.a[j--]--;
181             while(j>i)
182                 t1.a[j--]+=MAXN;
183             t1.a[i]+=MAXN+1-t2.a[i];
184         }
185         else t1.a[i]-=t2.a[i];
186     }
187     t1.len=big;
188     while(t1.a[len-1]==0 && t1.len>1)
189     {
190         t1.len--;
191         big--;
192     }
193     if(flag)
194         t1.a[big-1]=0-t1.a[big-1];
195     return t1;
196 }
197 BigNum BigNum::operator*(const BigNum &T)const  //两个大数之间的相乘
198 {
199     BigNum ret;
200     int i,j,up;
201     int temp,temp1;
202     for(i=0;i<len;i++)
203     {
204         up=0;
205         for(j=0;j<T.len;j++)
206         {
207             temp=a[i]*T.a[j]+ret.a[i+j]+up;
208             if(temp>MAXN)
209             {
210                 temp1=temp-temp/(MAXN+1)*(MAXN+1);
211                 up=temp/(MAXN+1);
212                 ret.a[i+j]=temp1;
213             }
214             else
215             {
216                 up=0;
217                 ret.a[i+j]=temp;
218             }
219         }
220         if(up!=0)
221            ret.a[i+j]=up;
222     }
223     ret.len=i+j;
224     while(ret.a[ret.len-1]==0 && ret.len>1)ret.len--;
225     return ret;
226 }
227 BigNum BigNum::operator/(const int &b)const  //大数对一个整数进行相除运算
228 {
229     BigNum ret;
230     int i,down=0;
231     for(i=len-1;i>=0;i--)
232     {
233         ret.a[i]=(a[i]+down*(MAXN+1))/b;
234         down=a[i]+down*(MAXN+1)-ret.a[i]*b;
235     }
236     ret.len=len;
237     while(ret.a[ret.len-1]==0 && ret.len>1)
238         ret.len--;
239     return ret;
240 }
241 int BigNum::operator%(const int &b)const   //大数对一个 int类型的变量进行取模
242 {
243     int i,d=0;
244     for(i=len-1;i>=0;i--)
245         d=((d*(MAXN+1))%b+a[i])%b;
246     return d;
247 }
248 BigNum BigNum::operator^(const int &n)const  //大数的n次方运算
249 {
250     BigNum t,ret(1);
251     int i;
252     if(n<0)exit(-1);
253     if(n==0)return 1;
254     if(n==1)return *this;
255     int m=n;
256     while(m>1)
257     {
258         t=*this;
259         for(i=1;(i<<1)<=m;i<<=1)
260            t=t*t;
261         m-=i;
262         ret=ret*t;
263         if(m==1)ret=ret*(*this);
264     }
265     return ret;
266 }
267 bool BigNum::operator>(const BigNum &T)const    //大数和另一个大数的大小比较
268 {
269     int ln;
270     if(len>T.len)return true;
271     else if(len==T.len)
272     {
273         ln=len-1;
274         while(a[ln]==T.a[ln]&&ln>=0)
275           ln--;
276         if(ln>=0 && a[ln]>T.a[ln])
277            return true;
278         else
279            return false;
280     }
281     else
282        return false;
283 }
284 bool BigNum::operator>(const int &t)const  //大数和一个int类型的变量的大小比较
285 {
286     BigNum b(t);
287     return *this>b;
288 }
289 void BigNum::print()   //输出大数
290 {
291     int i;
292     printf("%d",a[len-1]);
293     for(i=len-2;i>=0;i--)
294       printf("%04d",a[i]);
295     printf("\n");
296 }
297 bool ONE(BigNum a)
298 {
299     if(a.len == 1 && a.a[0] == 1)return true;
300     else return false;
301 }
302 BigNum A,B,X,Y;
303 char str1[10010],str2[10010],str3[10010],str4[10010];
304 
305 
306 int a[1010],b[1010],x[1010],y[1010];
307 int c[1010];
308 int main()
309 {
310     //freopen("in.txt","r",stdin);
311     //freopen("out.txt","w",stdout);
312     int T;
313     int n;
314     int iCase = 0;
315     scanf("%d",&T);
316     while(T--)
317     {
318         iCase++;
319         scanf("%d",&n);
320         cin>>A>>X>>B>>Y;
321         printf("Case %d: ",iCase) ;
322         A = A-1;
323         X = X-1;
324         B = B-1;
325         Y = Y-1;
326         for(int i = 0;i < n;i++)
327         {
328             if(A.a[0]%2 == 0)a[i] = 0;
329             else a[i] = 1;
330             if(B.a[0]%2 == 0)b[i] = 0;
331             else b[i] = 1;
332             if(X.a[0]%2 == 0)x[i] = 0;
333             else x[i] = 1;
334             if(Y.a[0]%2 == 0)y[i] = 0;
335             else y[i] = 1;
336             A = A/2;
337             B = B/2;
338             X = X/2;
339             Y = Y/2;
340         }
341         bool flag = false;
342         for(int k = 0;k <= n;k++)
343         {
344             x[n] = x[0];
345             y[n] = y[0];
346             for(int i = 0;i < n;i++)
347             {
348                 x[i] = x[i+1];
349                 y[i] = y[i+1];
350             }
351             for(int i = 0;i < n;i++)
352             {
353                 if(a[i] == x[i])c[i] = 0;
354                 else c[i] = 1;
355             }
356             bool fff = true;
357             for(int i = 0;i < n;i++)
358                 if(b[i]^c[i] != y[i])
359                 {
360                     fff = false;
361                     break;
362                 }
363             if(fff)flag = true;
364             if(flag)break;
365 
366         }
367         if(flag)printf("Yes\n");
368         else printf("No\n");
369     }
370     return 0;
371 }

 

 

 

 

 

 再贴一下JAVA的程序。

 

JAVA写大数很方便啊。。。。

 

 1 import java.io.*;
 2 import java.util.*;
 3 import java.math.*;
 4 public class Main {
 5     
 6     public static void main(String args[])
 7     {
 8         int T;
 9         int iCase = 0;
10         int n;
11         BigInteger A,X,B,Y;
12         int []a = new int[1010];
13         int []x = new int[1010];
14         int []b = new int[1010];
15         int []y = new int[1010];
16         Scanner cin = new Scanner(System.in);
17         T = cin.nextInt();
18         while(T > 0)
19         {
20             iCase++;
21             n = cin.nextInt();
22             A = cin.nextBigInteger();
23             X = cin.nextBigInteger();
24             B = cin.nextBigInteger();
25             Y = cin.nextBigInteger();
26             A = A.subtract(BigInteger.ONE);
27             X = X.subtract(BigInteger.ONE);
28             B = B.subtract(BigInteger.ONE);
29             Y = Y.subtract(BigInteger.ONE);
30             for(int i = 0;i < n;i++)
31             {
32                 a[i] = A.mod(BigInteger.valueOf(2)).intValue();
33                 b[i] = B.mod(BigInteger.valueOf(2)).intValue();
34                 x[i] = X.mod(BigInteger.valueOf(2)).intValue();
35                 y[i] = Y.mod(BigInteger.valueOf(2)).intValue();
36                 A = A.divide(BigInteger.valueOf(2));
37                 B = B.divide(BigInteger.valueOf(2));
38                 X = X.divide(BigInteger.valueOf(2));
39                 Y = Y.divide(BigInteger.valueOf(2));
40                 //System.out.println(a[i]+" "+ x[i]+" "+ b[i]+" "+y[i]);
41             }
42             boolean flag = false;
43             for(int k = 0;k <= n;k++)
44             {
45                 x[n] = x[0];
46                 for(int i = 0;i < n;i++)x[i] = x[i+1];
47                 y[n] = y[0];
48                 for(int i = 0;i < n;i++)y[i] = y[i+1];
49                 boolean ff = true;
50                 for(int i = 0;i < n;i++)
51                     if((x[i]^a[i]) != (y[i]^b[i]))
52                     {
53                         ff = false;
54                         break;
55                     }
56                 if(ff)
57                 {
58                     flag = true;
59                     break;
60                 }
61             }
62             if(flag)System.out.println("Case "+iCase+": Yes");
63             else System.out.println("Case "+iCase+": No");
64             T--;
65         }
66     }
67 
68 }

 

 

 

 

 

 

 

posted on 2013-09-29 16:56 kuangbin 阅读(...) 评论(...) 编辑 收藏

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