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。
1 #include<iostream> 2 #include<cstdio> 3 #include<queue> 4 #include<algorithm> 5 #include<cmath> 6 #include<ctime> 7 #include<set> 8 #include<map> 9 #include<stack> 10 #include<cstring> 11 #define inf 2147483647 12 #define ls rt<<1 13 #define rs rt<<1|1 14 #define lson ls,nl,mid,l,r 15 #define rson rs,mid+1,nr,l,r 16 #define N 100010 17 #define For(i,a,b) for(int i=a;i<=b;i++) 18 #define p(a) putchar(a) 19 #define g() getchar() 20 21 using namespace std; 22 int n; 23 int a[10000000]; 24 bool vis[10000000]; 25 int ans,t,fin; 26 27 void in(int &x){ 28 int y=1; 29 char c=g();x=0; 30 while(c<'0'||c>'9'){ 31 if(c=='-')y=-1; 32 c=g(); 33 } 34 while(c<='9'&&c>='0'){ 35 x=(x<<1)+(x<<3)+c-'0';c=g(); 36 } 37 x*=y; 38 } 39 void o(int x){ 40 if(x<0){ 41 p('-'); 42 x=-x; 43 } 44 if(x>9)o(x/10); 45 p(x%10+'0'); 46 } 47 int main(){ 48 freopen("infinite.in","r",stdin); 49 freopen("infinite.out","w",stdout); 50 in(n); 51 fin=(1<<n)+n-1; 52 a[n]=1; 53 vis[1]=1; 54 For(i,n+1,fin){ 55 a[i]=1; 56 t=1; 57 ans=0; 58 for(int j=i;j>=i-n+1;j--){ 59 ans+=t*a[j]; 60 t<<=1; 61 } 62 if(vis[ans]){ 63 vis[ans-1]=1; 64 a[i]=0; 65 } 66 else 67 vis[ans]=1; 68 } 69 For(i,1,fin) 70 o(a[i]); 71 return 0; 72 73 }
