相关链接
题目传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=3033
神犇题解：http://blog.csdn.net/clover_hxy/article/details/56003661
解题报告
这个东西，看一看样例，感觉答案是$2^k$，然后写一发无脑贪心就过了
不过这题的证明非常妙妙啊：
考虑将每一个$0 \sim 2^{k+1}-1$的数看成一个点，标号为这个数
把在这个数末尾或者开头加$0/1$看作边
那么每个点度数恰好为$4$，并且整个图显然连通
那么这图显然存在欧拉回路，那么就能够一笔画
于是在这个图上贪心走就可以了
Code
#include<bits/stdc++.h>
#define LL long long
using namespace std;
const int N = 3000;
int n,k,arr[N];
set<int> S;
inline int read() {
char c=getchar(); int f=1,ret=0;
while (c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();}
while (c<='9'&&c>='0') {ret=ret*10+c-'0';c=getchar();}
return ret * f;
}
inline int get(int p) {
int ret = 0;
for (int i=0;i<k;i++) ret += (1 << i) * arr[p + i];
return ret;
}
int main() {
k = read(); n = 1 << k;
for (int i=1;i<=k;i++) arr[i] = 0; S.insert(0);
for (int i=n;i>n-k;i--) arr[i] = 1, S.insert(get(i));
for (int i=2;i<=n-k;i++) {
if (S.count(get(i))) arr[i+k-1] = 1;
S.insert(get(i));
}
printf("%d ",n);
for (int i=1;i<=n;i++) printf("%d",arr[i]);
return 0;
}
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