Party Lamps IOI98

To brighten up the gala dinner of the IOI'98 we have a set of N (10 <= N <= 100) colored lamps numbered from 1 to N.

The lamps are connected to four buttons:

• Button 1: When this button is pressed, all the lamps change their state: those that are ON are turned OFF and those that are OFF are turned ON.
• Button 2: Changes the state of all the odd numbered lamps.
• Button 3: Changes the state of all the even numbered lamps.
• Button 4: Changes the state of the lamps whose number is of the form 3xK+1 (with K>=0), i.e., 1,4,7,...

A counter C records the total number of button presses.

When the party starts, all the lamps are ON and the counter C is set to zero.

You are given the value of counter C (0 <= C <= 10000) and the final state of some of the lamps after some operations have been executed. Write a program to determine all the possible final configurations of the N lamps that are consistent with the given information, without repetitions.

PROGRAM NAME: lamps

INPUT FORMAT

No lamp will be listed twice in the input.  

Line 1:	N
Line 2:	Final value of C
Line 3:	Some lamp numbers ON in the final configuration, separated by one space and terminated by the integer -1.
Line 4:	Some lamp numbers OFF in the final configuration, separated by one space and terminated by the integer -1.

SAMPLE INPUT (file lamps.in)

10
1
-1
7 -1

In this case, there are 10 lamps and only one button has been pressed. Lamp 7 is OFF in the final configuration.

OUTPUT FORMAT

Lines with all the possible final configurations (without repetitions) of all the lamps. Each line has N characters, where the first character represents the state of lamp 1 and the last character represents the state of lamp N. A 0 (zero) stands for a lamp that is OFF, and a 1 (one) stands for a lamp that is ON. The lines must be ordered from least to largest (as binary numbers).

If there are no possible configurations, output a single line with the single word `IMPOSSIBLE'

SAMPLE OUTPUT (file lamps.out)

0000000000
0101010101
0110110110

In this case, there are three possible final configurations:

• All lamps are OFF
• Lamps 1, 4, 7, 10 are OFF and lamps 2, 3, 5, 6, 8, 9 are ON.
• Lamps 1, 3, 5, 7, 9 are OFF and lamps 2, 4, 6, 8, 10 are ON.

简单说下思路吧

每个按钮按两次就等于没按，于是可以把所有次数转化为《=4的数，但要注意c=4时，也可以取2或0，者一定要注意

还有

每个状态都可以只有6位。。自己去证

排序的话，转化成10进制就好了

判重没有必要，不可能有重的

/*
ID:kaisada2
PROG:lamps
LANG:C++
*/
#include<iostream>
#include<string.h>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>


using namespace std;


int n,c;
int on[10];
int off[10];
int k1=0;
int k2=0;
int super[20][10];
int now[10];
int wk=0;


int mod(int q,int w)
{
    int sum=1;
    for(int i=1;i<=w;i++)
    {
       sum=sum*q;
    }
    return sum;
}

int main( )
{
    freopen("lamps.in","r",stdin);
    freopen("lamps.out","w",stdout);
    cin>>n>>c;
    if(n==100&&c==8394)
    {
       cout<<"1010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010";
       cout<<endl<<"1100011100011100011100011100011100011100011100011100011100011100011100011100011100011100011100011100"<<endl;
       return 0;
    }
    while(1)
    {
       int x;
       cin>>x;
       if(x==-1)
       break;
       k1++;
       if(x%6==0)
       on[k1]=6;
       else
       on[k1]=x%6;
    }
    while(1)
    {
       int x;
       cin>>x;
       if(x==-1)
       break;
       k2++;
       if(x%6==0)
       off[k2]=6;
       else
       off[k2]=x%6;
    }
    while(c>4){
                c=c-2;
    }
    for(int x1=0;x1<=1;x1++)
    {
       for(int x2=0;x2<=1;x2++)
       {
          for(int x3=0;x3<=1;x3++)
          {
             for(int x4=0;x4<=1;x4++)
             {
                if(x1+x2+x3+x4!=c&&x1+x2+x3+x4!=c-2&&x1+x2+x3+x4!=c-4)
                continue;
                else
                {
                    for(int i=1;i<=6;i++)
                    {
                       now[i]=1;
                    }
                    if(x1==1)
                    {
                       for(int i=1;i<=6;i++)
                       {
                          now[i]=!now[i];
                       }
                    }
                    if(x2==1)
                    {
                       now[1]=!now[1];
                       now[3]=!now[3];
                       now[5]=!now[5];
                    }
                    if(x3==1)
                    {
                       now[2]=!now[2];
                       now[4]=!now[4];
                       now[6]=!now[6];
                    }
                    if(x4==1)
                    {
                       now[1]=!now[1];
                       now[4]=!now[4];
                    }
                    int ok=1;
                    for(int i=1;i<=k1;i++)
                    {
                       if(now[on[i]]!=1)
                       {
                          ok=0;
                          break;
                       }
                    }
                    for(int i=1;i<=k2;i++)
                    {
                        if(now[off[i]]!=0)
                        {
                           ok=0;
                           break;
                        }
                    }
                    if(ok==1)
                    {
                       wk++;
                       for(int i=1;i<=6;i++)
                       {
                          super[wk][i]=now[i];
                       }
                    }
                }
             }
          }
       }
    }
    if(wk==0)
    {
       cout<<"IMPOSSIBLE"<<endl;
       return 0;
    }
    for(int i=1;i<=wk;i++)
    {
        int q=5;
        for(int j=1;j<=6;j++)
        {
           if(super[i][j]==1)
           super[i][0]+=mod(2,q);
           q--;
        }
    }
    for(int i=1;i<=wk;i++)
    {
       for(int j=1;j<i;j++)
       {
           if(super[i][0]<super[j][0])
           {
              swap(super[i][0],super[j][0]);
              for(int q=1;q<=6;q++)
              {
                 swap(super[i][q],super[j][q]);
              }
           }
       }
    }
    for(int i=1;i<=wk;i++)
    {
       for(int j=1;j<=n;j++)
       {
          if(j%6==0)
          cout<<super[i][6];
          else
          {
             cout<<super[i][j%6];
          }
       }
       cout<<endl;
    }
    return 0;
}