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

2013 ACM/ICPC Asia Regional Changchun Online

 

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

 /* ***********************************************
 Author        :kuangbin
 Created Time  :2013/9/28 星期六 11:57:13
 File Name     :2013长春网络赛\1001.cpp
 ************************************************ */

 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <set>
 #include <map>
 #include <string>
 #include <math.h>
 #include <stdlib.h>
 #include <time.h>
 using namespace std;

 /*
  * 完全大数模板
  * 输出cin>>a
  * 输出a.print();
  * 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。
  */
 #define MAXN 9999
 #define MAXSIZE 1010
 #define DLEN 4

 class BigNum
 {
 public:
     int a[];  //可以控制大数的位数
     int len;
 public:
     BigNum(){len=;memset(a,,sizeof(a));}  //构造函数
     BigNum(const int);     //将一个int类型的变量转化成大数
     BigNum(const char*);   //将一个字符串类型的变量转化为大数
     BigNum(const BigNum &); //拷贝构造函数
     BigNum &operator=(const BigNum &); //重载赋值运算符，大数之间进行赋值运算
     friend istream& operator>>(istream&,BigNum&); //重载输入运算符
     friend ostream& operator<<(ostream&,BigNum&); //重载输出运算符

     BigNum operator+(const BigNum &)const;  //重载加法运算符，两个大数之间的相加运算
     BigNum operator-(const BigNum &)const;  //重载减法运算符，两个大数之间的相减运算
     BigNum operator*(const BigNum &)const;  //重载乘法运算符，两个大数之间的相乘运算
     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

     BigNum operator^(const int &)const;     //大数的n次方运算
     int operator%(const int &)const;        //大数对一个int类型的变量进行取模运算
     bool operator>(const BigNum &T)const;   //大数和另一个大数的大小比较
     bool operator>(const int &t)const;      //大数和一个int类型的变量的大小比较

     void print();        //输出大数
 };
 BigNum::BigNum(const int b)   //将一个int类型的变量转化为大数
 {
     int c,d=b;
     len=;
     memset(a,,sizeof(a));
     while(d>MAXN)
     {
         c=d-(d/(MAXN+))*(MAXN+);
         d=d/(MAXN+);
         a[len++]=c;
     }
     a[len++]=d;
 }
 BigNum::BigNum(const char *s)  //将一个字符串类型的变量转化为大数
 {
     int t,k,index,L,i;
     memset(a,,sizeof(a));
     L=strlen(s);
     len=L/DLEN;
     if(L%DLEN)len++;
     index=;
     for(i=L-;i>=;i-=DLEN)
     {
         t=;
         k=i-DLEN+;
         if(k<)k=;
         for(int j=k;j<=i;j++)
             t=t*+s[j]-'';
         a[index++]=t;
     }
 }
 BigNum::BigNum(const BigNum &T):len(T.len)  //拷贝构造函数
 {
     int i;
     memset(a,,sizeof(a));
     for(i=;i<len;i++)
         a[i]=T.a[i];
 }
 BigNum & BigNum::operator=(const BigNum &n)  //重载赋值运算符，大数之间赋值运算
 {
     int i;
     len=n.len;
     memset(a,,sizeof(a));
     for(i=;i<len;i++)
         a[i]=n.a[i];
     return *this;
 }
 istream& operator>>(istream &in,BigNum &b)
 {
     char ch[MAXSIZE*];
     int i=-;
     in>>ch;
     int L=strlen(ch);
     int count=,sum=;
     for(i=L-;i>=;)
     {
         sum=;
         int t=;
         for(int j=;j<&&i>=;j++,i--,t*=)
         {
             sum+=(ch[i]-'')*t;
         }
         b.a[count]=sum;
         count++;
     }
     b.len=count++;
     return in;
 }
 ostream& operator<<(ostream& out,BigNum& b)  //重载输出运算符
 {
     int i;
     cout<<b.a[b.len-];
     for(i=b.len-;i>=;i--)
     {
         printf("%04d",b.a[i]);
     }
     return out;
 }
 BigNum BigNum::operator+(const BigNum &T)const   //两个大数之间的相加运算
 {
     BigNum t(*this);
     int i,big;
     big=T.len>len?T.len:len;
     for(i=;i<big;i++)
     {
         t.a[i]+=T.a[i];
         if(t.a[i]>MAXN)
         {
             t.a[i+]++;
             t.a[i]-=MAXN+;
         }
     }
     if(t.a[big]!=)
        t.len=big+;
     else t.len=big;
     return t;
 }
 BigNum BigNum::operator-(const BigNum &T)const  //两个大数之间的相减运算
 {
     int i,j,big;
     bool flag;
     BigNum t1,t2;
     if(*this>T)
     {
         t1=*this;
         t2=T;
         flag=;
     }
     else
     {
         t1=T;
         t2=*this;
         flag=;
     }
     big=t1.len;
     for(i=;i<big;i++)
     {
         if(t1.a[i]<t2.a[i])
         {
             j=i+;
             while(t1.a[j]==)
                 j++;
             t1.a[j--]--;
             while(j>i)
                 t1.a[j--]+=MAXN;
             t1.a[i]+=MAXN+-t2.a[i];
         }
         else t1.a[i]-=t2.a[i];
     }
     t1.len=big;
     while(t1.a[len-]== && t1.len>)
     {
         t1.len--;
         big--;
     }
     if(flag)
         t1.a[big-]=-t1.a[big-];
     return t1;
 }
 BigNum BigNum::operator*(const BigNum &T)const  //两个大数之间的相乘
 {
     BigNum ret;
     int i,j,up;
     int temp,temp1;
     for(i=;i<len;i++)
     {
         up=;
         for(j=;j<T.len;j++)
         {
             temp=a[i]*T.a[j]+ret.a[i+j]+up;
             if(temp>MAXN)
             {
                 temp1=temp-temp/(MAXN+)*(MAXN+);
                 up=temp/(MAXN+);
                 ret.a[i+j]=temp1;
             }
             else
             {
                 up=;
                 ret.a[i+j]=temp;
             }
         }
         if(up!=)
            ret.a[i+j]=up;
     }
     ret.len=i+j;
     while(ret.a[ret.len-]== && ret.len>)ret.len--;
     return ret;
 }
 BigNum BigNum::operator/(const int &b)const  //大数对一个整数进行相除运算
 {
     BigNum ret;
     int i,down=;
     for(i=len-;i>=;i--)
     {
         ret.a[i]=(a[i]+down*(MAXN+))/b;
         down=a[i]+down*(MAXN+)-ret.a[i]*b;
     }
     ret.len=len;
     while(ret.a[ret.len-]== && ret.len>)
         ret.len--;
     return ret;
 }
 int BigNum::operator%(const int &b)const   //大数对一个 int类型的变量进行取模
 {
     int i,d=;
     for(i=len-;i>=;i--)
         d=((d*(MAXN+))%b+a[i])%b;
     return d;
 }
 BigNum BigNum::operator^(const int &n)const  //大数的n次方运算
 {
     BigNum t,ret();
     int i;
     if(n<)exit(-);
     if(n==)return ;
     if(n==)return *this;
     int m=n;
     while(m>)
     {
         t=*this;
         for(i=;(i<<)<=m;i<<=)
            t=t*t;
         m-=i;
         ret=ret*t;
         if(m==)ret=ret*(*this);
     }
     return ret;
 }
 bool BigNum::operator>(const BigNum &T)const    //大数和另一个大数的大小比较
 {
     int ln;
     if(len>T.len)return true;
     else if(len==T.len)
     {
         ln=len-;
         while(a[ln]==T.a[ln]&&ln>=)
           ln--;
         if(ln>= && a[ln]>T.a[ln])
            return true;
         else
            return false;
     }
     else
        return false;
 }
 bool BigNum::operator>(const int &t)const  //大数和一个int类型的变量的大小比较
 {
     BigNum b(t);
     return *this>b;
 }
 void BigNum::print()   //输出大数
 {
     int i;
     printf("%d",a[len-]);
     for(i=len-;i>=;i--)
       printf("%04d",a[i]);
     printf("\n");
 }
 bool ONE(BigNum a)
 {
     if(a.len ==  && a.a[] == )return true;
     else return false;
 }
 BigNum A,B,X,Y;
 char str1[],str2[],str3[],str4[];

 int a[],b[],x[],y[];
 int c[];
 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     int T;
     int n;
     int iCase = ;
     scanf("%d",&T);
     while(T--)
     {
         iCase++;
         scanf("%d",&n);
         cin>>A>>X>>B>>Y;
         printf("Case %d: ",iCase) ;
         A = A-;
         X = X-;
         B = B-;
         Y = Y-;
         for(int i = ;i < n;i++)
         {
             if(A.a[]% == )a[i] = ;
             else a[i] = ;
             if(B.a[]% == )b[i] = ;
             else b[i] = ;
             if(X.a[]% == )x[i] = ;
             else x[i] = ;
             if(Y.a[]% == )y[i] = ;
             else y[i] = ;
             A = A/;
             B = B/;
             X = X/;
             Y = Y/;
         }
         bool flag = false;
         for(int k = ;k <= n;k++)
         {
             x[n] = x[];
             y[n] = y[];
             for(int i = ;i < n;i++)
             {
                 x[i] = x[i+];
                 y[i] = y[i+];
             }
             for(int i = ;i < n;i++)
             {
                 if(a[i] == x[i])c[i] = ;
                 else c[i] = ;
             }
             bool fff = true;
             for(int i = ;i < n;i++)
                 if(b[i]^c[i] != y[i])
                 {
                     fff = false;
                     break;
                 }
             if(fff)flag = true;
             if(flag)break;

         }
         if(flag)printf("Yes\n");
         else printf("No\n");
     }
     return ;
 }

再贴一下JAVA的程序。

JAVA写大数很方便啊。。。。

 import java.io.*;
 import java.util.*;
 import java.math.*;
 public class Main {

     public static void main(String args[])
     {
         int T;
         int iCase = 0;
         int n;
         BigInteger A,X,B,Y;
         int []a = new int[1010];
         int []x = new int[1010];
         int []b = new int[1010];
         int []y = new int[1010];
         Scanner cin = new Scanner(System.in);
         T = cin.nextInt();
         while(T > 0)
         {
             iCase++;
             n = cin.nextInt();
             A = cin.nextBigInteger();
             X = cin.nextBigInteger();
             B = cin.nextBigInteger();
             Y = cin.nextBigInteger();
             A = A.subtract(BigInteger.ONE);
             X = X.subtract(BigInteger.ONE);
             B = B.subtract(BigInteger.ONE);
             Y = Y.subtract(BigInteger.ONE);
             for(int i = 0;i < n;i++)
             {
                 a[i] = A.mod(BigInteger.valueOf(2)).intValue();
                 b[i] = B.mod(BigInteger.valueOf(2)).intValue();
                 x[i] = X.mod(BigInteger.valueOf(2)).intValue();
                 y[i] = Y.mod(BigInteger.valueOf(2)).intValue();
                 A = A.divide(BigInteger.valueOf(2));
                 B = B.divide(BigInteger.valueOf(2));
                 X = X.divide(BigInteger.valueOf(2));
                 Y = Y.divide(BigInteger.valueOf(2));
                 //System.out.println(a[i]+" "+ x[i]+" "+ b[i]+" "+y[i]);
             }
             boolean flag = false;
             for(int k = 0;k <= n;k++)
             {
                 x[n] = x[0];
                 for(int i = 0;i < n;i++)x[i] = x[i+1];
                 y[n] = y[0];
                 for(int i = 0;i < n;i++)y[i] = y[i+1];
                 boolean ff = true;
                 for(int i = 0;i < n;i++)
                     if((x[i]^a[i]) != (y[i]^b[i]))
                     {
                         ff = false;
                         break;
                     }
                 if(ff)
                 {
                     flag = true;
                     break;
                 }
             }
             if(flag)System.out.println("Case "+iCase+": Yes");
             else System.out.println("Case "+iCase+": No");
             T--;
         }
     }

 }