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


[align=left]Problem Description[/align]
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?

[align=left]Input[/align]
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.

[align=left]Output[/align]
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”.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

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

[align=left]Source[/align]
2013 ACM/ICPC Asia Regional Changchun Online

[align=left]Recommend[/align]
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[500];  //可以控制大数的位数
int len;
public:
BigNum(){len=1;memset(a,0,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=0;
memset(a,0,sizeof(a));
while(d>MAXN)
{
c=d-(d/(MAXN+1))*(MAXN+1);
d=d/(MAXN+1);
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,0,sizeof(a));
L=strlen(s);
len=L/DLEN;
if(L%DLEN)len++;
index=0;
for(i=L-1;i>=0;i-=DLEN)
{
t=0;
k=i-DLEN+1;
if(k<0)k=0;
for(int j=k;j<=i;j++)
t=t*10+s[j]-'0';
a[index++]=t;
}
}
BigNum::BigNum(const BigNum &T):len(T.len)  //拷贝构造函数
{
int i;
memset(a,0,sizeof(a));
for(i=0;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,0,sizeof(a));
for(i=0;i<len;i++)
a[i]=n.a[i];
return *this;
}
istream& operator>>(istream &in,BigNum &b)
{
char ch[MAXSIZE*4];
int i=-1;
in>>ch;
int L=strlen(ch);
int count=0,sum=0;
for(i=L-1;i>=0;)
{
sum=0;
int t=1;
for(int j=0;j<4&&i>=0;j++,i--,t*=10)
{
sum+=(ch[i]-'0')*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-1];
for(i=b.len-2;i>=0;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=0;i<big;i++)
{
t.a[i]+=T.a[i];
if(t.a[i]>MAXN)
{
t.a[i+1]++;
t.a[i]-=MAXN+1;
}
}
if(t.a[big]!=0)
t.len=big+1;
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=0;
}
else
{
t1=T;
t2=*this;
flag=1;
}
big=t1.len;
for(i=0;i<big;i++)
{
if(t1.a[i]<t2.a[i])
{
j=i+1;
while(t1.a[j]==0)
j++;
t1.a[j--]--;
while(j>i)
t1.a[j--]+=MAXN;
t1.a[i]+=MAXN+1-t2.a[i];
}
else t1.a[i]-=t2.a[i];
}
t1.len=big;
while(t1.a[len-1]==0 && t1.len>1)
{
t1.len--;
big--;
}
if(flag)
t1.a[big-1]=0-t1.a[big-1];
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=0;i<len;i++)
{
up=0;
for(j=0;j<T.len;j++)
{
temp=a[i]*T.a[j]+ret.a[i+j]+up;
if(temp>MAXN)
{
temp1=temp-temp/(MAXN+1)*(MAXN+1);
up=temp/(MAXN+1);
ret.a[i+j]=temp1;
}
else
{
up=0;
ret.a[i+j]=temp;
}
}
if(up!=0)
ret.a[i+j]=up;
}
ret.len=i+j;
while(ret.a[ret.len-1]==0 && ret.len>1)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=0;
for(i=len-1;i>=0;i--)
{
ret.a[i]=(a[i]+down*(MAXN+1))/b;
down=a[i]+down*(MAXN+1)-ret.a[i]*b;
}
ret.len=len;
while(ret.a[ret.len-1]==0 && ret.len>1)
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=0;
for(i=len-1;i>=0;i--)
d=((d*(MAXN+1))%b+a[i])%b;
return d;
}
BigNum BigNum::operator^(const int &n)const  //大数的n次方运算
{
BigNum t,ret(1);
int i;
if(n<0)exit(-1);
if(n==0)return 1;
if(n==1)return *this;
int m=n;
while(m>1)
{
t=*this;
for(i=1;(i<<1)<=m;i<<=1)
t=t*t;
m-=i;
ret=ret*t;
if(m==1)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-1;
while(a[ln]==T.a[ln]&&ln>=0)
ln--;
if(ln>=0 && 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-1]);
for(i=len-2;i>=0;i--)
printf("%04d",a[i]);
printf("\n");
}
bool ONE(BigNum a)
{
if(a.len == 1 && a.a[0] == 1)return true;
else return false;
}
BigNum A,B,X,Y;
char str1[10010],str2[10010],str3[10010],str4[10010];

int a[1010],b[1010],x[1010],y[1010];
int c[1010];
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int T;
int n;
int iCase = 0;
scanf("%d",&T);
while(T--)
{
iCase++;
scanf("%d",&n);
cin>>A>>X>>B>>Y;
printf("Case %d: ",iCase) ;
A = A-1;
X = X-1;
B = B-1;
Y = Y-1;
for(int i = 0;i < n;i++)
{
if(A.a[0]%2 == 0)a[i] = 0;
else a[i] = 1;
if(B.a[0]%2 == 0)b[i] = 0;
else b[i] = 1;
if(X.a[0]%2 == 0)x[i] = 0;
else x[i] = 1;
if(Y.a[0]%2 == 0)y[i] = 0;
else y[i] = 1;
A = A/2;
B = B/2;
X = X/2;
Y = Y/2;
}
bool flag = false;
for(int k = 0;k <= n;k++)
{
x
= x[0];
y
= y[0];
for(int i = 0;i < n;i++)
{
x[i] = x[i+1];
y[i] = y[i+1];
}
for(int i = 0;i < n;i++)
{
if(a[i] == x[i])c[i] = 0;
else c[i] = 1;
}
bool fff = true;
for(int i = 0;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 0;
}

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
= x[0];
for(int i = 0;i < n;i++)x[i] = x[i+1];
y
= 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--;
}
}

}