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 ;

 }