//派对灯 //每个开关按偶数次的时候，相当于没按 #include<bits/stdc++.h> using namespace std; int flag[7],n,c,OK; int state[9][7]= { 0,0,0,0,0,0,0,//已排好序 1,0,0,0,0,0,0, 2,0,0,1,1,1,0, 1,0,1,0,1,0,1, 1,0,1,1,0,1,1, 2,1,0,0,1,0,0, 1,1,0,1,0,1,0, 2,1,1,0,0,0,1, 0,1,1,1,1,1,1, };//循环节是6，并且能达到的只有8种形式,第一列保存达到该状态需要的步数 int Loc(int x)//映射为6个一循环的位置 { return (x%6==0?6:x%6); } int Check(int s) { if (state[s][0]>c) return 0;//步数超了 if (c==2 && s==4) return 0;//2步，达不到011011状态 if (c==3 && ((s==2)||(s==5)||(s==7))) return 0;//3步，达不到001110/100100/110001 if (c==1 && s==8) return 0;//1步，达不到111111状态 for (int i=1; i<=6; i++) { if (flag[i]==-1) continue;//状态没有变，忽略 if (state[s][i]!=flag[i]) return 0; } return 1; } int main() { freopen("lamps.in","r",stdin); freopen("lamps.out","w",stdout); memset(flag,-1,sizeof(flag)); scanf("%d%d",&n,&c); for(int k; scanf("%d",&k) && k!=-1; flag[Loc(k)]=1); for(int k; scanf("%d",&k) && k!=-1; flag[Loc(k)]=0); for (int i=1; i<=8; i++)//穷举8种状态 if (Check(i)) { for (int j=1; j<=n; j++) printf("%d",state[i][Loc(j)]); printf("\n"); OK=1; } if (!OK) printf("IMPOSSIBLE\n"); return 0; }