USACO Party Lamps

题目大意:一排灯有n个,有４种开关,每种开关能改变一些灯现在的状态(亮的变暗,暗的变亮)现在已知一些灯的亮暗情况,问所以可能的情况是哪些

思路:同一种开关开两次显然是没效果的,那么枚举每个开关是否开就好了，还是暴力大法好

 /*{
 ID:a4298442
 PROB:lamps
 LANG:C++
 }
 */
 #include<iostream>
 #include<fstream>
 #include<cstring>
 #include<algorithm>
 #define maxn 109
 using namespace std;
 ifstream fin("lamps.in");
 ofstream fout("lamps.out");
 //#define fin cin
 //#define fout cout
 int lamp[maxn],temp[maxn],n1,n2,a[maxn],b[maxn];
 struct T
 {
     char ch[maxn];
 }ans[maxn];
 void but1(int n)
 {
     for(int i=;i<=n;i++)lamp[i]^=;
 }
 void but2(int n)
 {
     for(int i=;i<=n;i+=)lamp[i]^=;
 }
 void but3(int n)
 {
     for(int i=;i<=n;i+=)lamp[i]^=;
 }
 void but4(int n)
 {
     for(int i=;i*+<=n;i++)lamp[i*+]^=;
 }
 int check(int n,int c,int count_now)
 {
     for(int i=;i<n1;i++)if(lamp[a[i]]==)return ;
     for(int i=;i<n2;i++)if(lamp[b[i]]==)return ;
     if(count_now>c)return ;
     int u=c-count_now;
     if(u&)return ;
     return ;
 }
 int cmp(T x, T y)
 {
     return strcmp(x.ch+,y.ch+)>=?false:true;
 }
 int main()
 {
     int n,c;
     fin>>n>>c;
         while(fin>>a[++n1]&&a[n1]!=-);
     while(fin>>b[++n2]&&b[n2]!=-);
     for(int i=;i<=n;i++)temp[i]=;
     int h=;
     for(int i=;i<=(1LL<<)-;i++)
     {
         int count_now=;
         memcpy(lamp,temp,sizeof(lamp));
         for(int j=,idx=;j<=i;j<<=,idx++)if((i&j)!= )
         {
             count_now++;
             if(idx==)but1(n);
             if(idx==)but2(n);
             if(idx==)but3(n);
             if(idx==)but4(n);
         }
         if(check(n,c,count_now))
         {
             h++;
             for(int i=;i<=n;i++)
             {
                 ans[h].ch[i]=lamp[i]+'';
             }
         }
     }
     if(h==)fout<<"IMPOSSIBLE"<<endl;else
     {
         sort(ans+,ans++h,cmp);
         for(int i=;i<=h;i++)fout<<ans[i].ch+<<endl;
     }
     return ;
 }