poj 1392 Ouroboros Snake

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1120		Accepted: 587

Description
Ouroboros is a mythical snake from ancient Egypt. It has its tail in its mouth and continously devours itself.

The Ouroboros numbers are binary numbers of 2^n bits that have the property of "generating" the whole set of numbers from 0 to 2^n - 1. The generation works as follows: given an Ouroboros number, we place its 2^n bits wrapped in a circle. Then, we can take
2^n groups of n bits starting each time with the next bit in the circle. Such circles are called Ouroboros circles for the number n. We will work only with the smallest

Ouroboros number for each n.

Example: for n = 2, there are only four Ouroboros numbers. These are 0011;0110;1100; and 1001. In this case, the smallest one is 0011. Here is the Ouroboros circle for 0011:



The table describes the function o(n;k) which calculates the k-th number in the Ouroboros circle of the smallest Ouroboros number of size n. This function is what your program should compute.

Input
The input consists of several test cases. For each test case, there will be a line containing two integers n and k (1<=n<=15; 0<=k<2^n). The end of the input file is indicated by a line containing two zeros. Don抰 process that line.
Output
For each test case, output o(n;k) on a line by itself.
Sample Input

2 0
2 1
2 2
2 3
0 0

Sample Output

0
1
3
2

Source
Mid-Central European Regional Contest 2000

题意：让你找个字典序最小的序列使得排成环旋转后取n个，最后可以取到[0-2^n)的所有数。

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int path[2<<16],h,n,k,vis[2<<16],ans[2<<16];
void dfs(int x)
{
int t1=(x<<1)&((1<<n)-1),t2=t1+1;//t1为x去掉最高位，末尾加0；t2为x去掉最高位，末尾加1；
if(!vis[t1])//注意2个if的先后顺序。
{
vis[t1]=1;
dfs(t1);
path[h++]=0;//记录字典序最小的01序列，
}
if(!vis[t2])
{
vis[t2]=1;
dfs(t2);
path[h++]=1;
}
}
int main()
{
int i,j,sum;
while(~scanf("%d %d",&n,&k))
{
if(n==0&&k==0)break;
h=0;
for(i=0;i<n-1;i++)//进行初始化
path[h++]=0,ans[i]=0;
memset(vis,0,sizeof(vis));
dfs(0);
j=n-1;
for(i=h-1;i>=n-1;i--)//注意倒序一下。
ans[j++]=path[i];
sum=0;
for(i=k;i<k+n;i++)
sum=sum*2+ans[i];
cout<<sum<<endl;
}
return 0;
}