I

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[10000000];
bool vis[10000000];
int ans,t,fin;

void in(int &x){
    int y=1;
    char c=g();x=0;
    while(c<'0'||c>'9'){
        if(c=='-')y=-1;
        c=g();
    }
    while(c<='9'&&c>='0'){
        x=(x<<1)+(x<<3)+c-'0';c=g();
    }
    x*=y;
}
void o(int x){
    if(x<0){
        p('-');
        x=-x;
    }
    if(x>9)o(x/10);
    p(x%10+'0');
}
int main(){
    freopen("infinite.in","r",stdin);
    freopen("infinite.out","w",stdout);
    in(n);
    fin=(1<<n)+n-1;
    a[n]=1;
    vis[1]=1;
    For(i,n+1,fin){
        a[i]=1;
        t=1;
        ans=0;
        for(int j=i;j>=i-n+1;j--){
            ans+=t*a[j];
            t<<=1;
        }
        if(vis[ans]){
            vis[ans-1]=1;
            a[i]=0;
        }
        else
            vis[ans]=1;
    }
    For(i,1,fin)
        o(a[i]);
    return 0;
        
}