Usaco 2.2.4 party lamps solution report

Party lamps
IOI 98

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

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., 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 executions of the N lamps that are consistent with the given information, without repetitions.

Program name: lampsinput 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)

101-17 -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 invocations (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 distributions, output a single line with the single word 'impossible'

Sample output (File lamps. out)

000000000001010101010110110110

In this case, there are three possible final invocations:

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

The question is to give the number of lights, and the operating rules of the lights, and after the C-step operation, part of the lights are on and part of the lights are off, to find the final possibility of the lights, if 'impossible 'is not output'

• Button 1: When you press this button, all the lights will be changed: the lights that were originally on will go off and the lights that were originally turned off will be lit.
• Button 2: When you press this button, all the lights with odd numbers are changed.
• Button 3: When you press this button, all lights with even numbers are changed.
• Button 4: When you press this button, all the lights whose serial numbers are 3 * k + 1 (k> = 0) will be changed. Example:, 7...

Note that when each operation is performed twice, the effect is the same as that of no implementation, so c greater than 4 can be reduced to 3 or 4;

Then, you can start to search for any possible situations after one-step operations without a lamp. You can see that step C has been performed. If the problem conditions are met, save the data and search for the next group.

/* ID:shiryuw1 PROG:lamps LANG:C++ */ #include<iostream> #include<cstdlib> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; int op[105]; int cl[105]; int n; int c; int clk=0; int opk=0; bool found=false; struct prog{     int a1;     int a2;     int a3;     int a4;     bool str[105]; }; prog ans[10000]; int p=0; bool istrue(bool* lap) {     int i;     for(i=0;i<opk;i++)     {         if(lap[op[i]-1]!=true)             return false;     }     for(i=0;i<clk;i++)     {         if(lap[cl[i]-1]!=false)             return false;     }     return true; } void change(bool* lap,int k) {     int st=0;     int ad=k;     if(k==4)     {         st=1;         ad=2;     }     int i;     for(i=st;i<n;i+=ad)         lap[i]=!lap[i]; } void DFS(bool *lap,int k) {     int i,j;     if(k/2==c/2)     {         if(istrue(lap))         {                         for(j=0;j<n;j++)                 ans[p].str[j]=lap[j];             for(j=0;j<25;j++)             {                 ans[p].a1=ans[p].a1*2+ans[p].str[j];                            }            for(j=25;j<50;j++)             {                ans[p].a2=ans[p].a2*2+ans[p].str[j];                             }            for(j=50;j<75;j++)             {                 ans[p].a3=ans[p].a3*2+ans[p].str[j];                             }            for(j=75;j<100;j++)             {                 ans[p].a4=ans[p].a4*2+ans[p].str[j];             }             p++;             found=true;         }         return ;     }     for(i=1;i<=4;i++)     {             bool tmp[105];        for(j=0;j<n;j++)         {             tmp[j]=lap[j];         }         change(tmp,i);         DFS(tmp,k+1);     } } int cmp(const void *a,const void *b) {     if((*(prog *)a).a1!=(*(prog *)b).a1)         return (*(prog *)a).a1-(*(prog *)b).a1;     if((*(prog *)a).a2!=(*(prog *)b).a2)         return (*(prog *)a).a2-(*(prog *)b).a2;     if((*(prog *)a).a3!=(*(prog *)b).a3)         return (*(prog *)a).a3-(*(prog *)b).a3;     return (*(prog *)a).a4-(*(prog *)b).a4;} int main() {    memset(ans,0,sizeof(ans));     bool lap[105];     freopen("lamps.in","r",stdin);     freopen("lamps.out","w",stdout);     memset(lap,true,sizeof(lap));     cin>>n;     cin>>c;     while(1)     {         cin>>op[opk];                 if(op[opk]==-1)             break;         opk++;     }     while(1)     {         cin>>cl[clk];                 if(cl[clk]==-1)             break;         clk++;     }     if(c>4)     {         if(c%2)             c=3;         else             c=4;     }     DFS(lap,0);     int i,j;     if(!found)         cout<<"IMPOSSIBLE"<<endl;     else     {         qsort(ans,p,sizeof(ans[0]),cmp);         for(i=0;i<p;i++)         {             if(i&&ans[i].a1==ans[i-1].a1&&ans[i].a2==ans[i-1].a2&&ans[i].a3==ans[i-1].a3&&ans[i].a4==ans[i-1].a4)                 continue;             for(j=0;j<n;j++)             {                                 cout<<ans[i].str[j];             }             cout<<endl;         }     }     return 0; }