I - Infinite Improbability Drive

I - Infinite Improbability Drive
http://codeforces.com/gym/241750/problem/I
不断构造，先填n-1个0，然后能放1就放1，最后这个序列的长度就是(1<<n)+n-1，也就是每添加1位就要匹配出来一个。

 #include<iostream>
 #include<cstdio>
 #include<queue>
 #include<algorithm>
 #include<cmath>
 #include<ctime>
 #include<set>
 #include<map>
 #include<stack>
 #include<cstring>
 #define inf 2147483647
 #define ls rt<<1
 #define rs rt<<1|1
 #define lson ls,nl,mid,l,r
 #define rson rs,mid+1,nr,l,r
 #define N 100010
 #define For(i,a,b) for(int i=a;i<=b;i++)
 #define p(a) putchar(a)
 #define g() getchar()

 using namespace std;
 int n;
 int a[];
 bool vis[];
 int ans,t,fin;

 void in(int &x){
     int y=;
     char c=g();x=;
     while(c<''||c>''){
         if(c=='-')y=-;
         c=g();
     }
     while(c<=''&&c>=''){
         x=(x<<)+(x<<)+c-'';c=g();
     }
     x*=y;
 }
 void o(int x){
     if(x<){
         p('-');
         x=-x;
     }
     if(x>)o(x/);
     p(x%+'');
 }
 int main(){
     freopen("infinite.in","r",stdin);
     freopen("infinite.out","w",stdout);
     in(n);
     fin=(<<n)+n-;
     a[n]=;
     vis[]=;
     For(i,n+,fin){
         a[i]=;
         t=;
         ans=;
         for(int j=i;j>=i-n+;j--){
             ans+=t*a[j];
             t<<=;
         }
         if(vis[ans]){
             vis[ans-]=;
             a[i]=;
         }
         else
             vis[ans]=;
     }
     For(i,,fin)
         o(a[i]);
     return ;

 }