S-Trees 
A Strange Tree (S-tree) over the variable set  is a binary tree representing a Boolean function. Each path of the S-tree begins at the root node and consists of n+1 nodes. Each of the S-tree's nodes has adepth, which is the amount of nodes between itself and the root (so the root has depth 0). The nodes with depth less than n are callednon-terminal nodes. All non-terminal nodes have two children: the right child and the left child. Each non-terminal node is marked with some variable xi from the variable set Xn. All non-terminal nodes with the same depth are marked with the same variable, and non-terminal nodes with different depth are marked with different variables. So, there is a unique variable xi1 corresponding to the root, a unique variable xi2 corresponding to the nodes with depth 1, and so on. The sequence of the variables  is called the variable ordering. The nodes having depth n are called terminal nodes. They have no children and are marked with either 0 or 1. Note that the variable ordering and the distribution of 0's and 1's on terminal nodes are sufficient to completely describe an S-tree.
As stated earlier, each S-tree represents a Boolean function f. If you have an S-tree and values for the variables , then it is quite simple to find out what  is: start with the root. Now repeat the following: if the node you are at is labelled with a variable xi, then depending on whether the value of the variable is 1 or 0, you go its right or left child, respectively. Once you reach a terminal node, its label gives the value of the function.
Figure 1: S-trees for the function 
On the picture, two S-trees representing the same Boolean function, , are shown. For the left tree, the variable ordering is x1, x2, x3, and for the right tree it is x3, x1, x2.
The values of the variables , are given as a Variable Values Assignment (VVA) 
with 
. For instance, ( 
x
일
 = 1, 
x
이
 = 1 
x
삼
 = 0) would be a valid VVA for 
n
 = 3, resulting for the sample function above in the value 
. The corresponding paths are shown bold in the picture.
Your task is to write a program which takes an S-tree and some VVAs and computes  as described above.

Input 

The input file contains the description of several S-trees with associated VVAs which you have to process. Each description begins with a line containing a single integer 
n
, 
, the depth of the S-tree. This is followed by a line describing the variable ordering of the S-tree. The format of that line is  x
i
일
  x
i
이
 ... x
i
n
. (There will be exactly 
n
 different space-separated strings). So, for 
n
 = 3 and the variable ordering 
x
삼
, 
x
일
, 
x
이
, this line would look as follows:
x3 x1 x2
In the next line the distribution of 0's and 1's over the terminal nodes is given. There will be exactly 2n characters (each of which can be 0 or 1), followed by the new-line character. The characters are given in the order in which they appear in the S-tree, the first character corresponds to the leftmost terminal node of the S-tree, the last one to its rightmost terminal node.
The next line contains a single integer m, the number of VVAs, followed by m lines describing them. Each of the m lines contains exactly n characters (each of which can be 0 or 1), followed by a new-line character. Regardless of the variable ordering of the S-tree, the first character always describes the value of x1, the second character describes the value of x2, and so on. So, the line
110
corresponds to the VVA ( x1 = 1, x2 = 1, x3 = 0).
The input is terminated by a test case starting with n = 0. This test case should not be processed.

Output 

For each S-tree, output the line `` S-Tree #
j :
", where 
j
 is the number of the S-tree. Then print a line that contains the value of
 for each of the given 
m
 VVAs, where 
f
 is the function defined by the S-tree.
Output a blank line after each test case.

Sample Input 

3
x1 x2 x3
00000111
4
000
010
111
110
3
x3 x1 x2
00010011
4
000
010
111
110
0

Sample Output 

S-Tree #1:
0011

S-Tree #2:
0011

제목 대의: 하나의 서열 집합 {x1,x2,x3,...,xn}, 집합 중 0이 아니면 1이다. 그리고 n층의 나무가 하나 있다. 집합의 서열은 나무의 충전 방향을 나타낸다. (뒤에 주어진 지령에 대응) 그 결점에서 출발하여 그 결점이 0이면 왼쪽 아들 방향으로 가고, 1이면 오른쪽 아들 방향으로 간다.마지막으로 마지막 층의 잎사귀 결점에 떨어져서 이 숫자를 출력합니다
문제 풀이 사고방식: 세우지 않고 0에 닿으면 *2, 1에 닿으면 *2+1.

#include<stdio.h>
#include<string.h>
#include<math.h>
#define N 1000
#define M 10

int main(){
	int t = 1, n, m;
	char str[N], tem[M], q[M];
	int num[M];

	while (scanf("%d", &n), n){
		// Init.
		memset(str, 0, sizeof(str));
		memset(tem, 0, sizeof(tem));
		memset(q, 0, sizeof(q));
		memset(num, 0, sizeof(num));

		// Read.
		for (int i = 0; i < n; i++){
			scanf("%s", tem);
			num[i] = tem[1] - '0';
		}
		scanf("%s", str);
		scanf("%d", &m);

		// Handle.
		int f = pow(2, n);
		for (int i = 0; i < m; i++){
			scanf("%s", tem);
			int k = 1;
			for (int j = 0; j < n; j++){
				if (tem[num[j] - 1] == '1')
					k = k * 2 + 1;
				else
					k = k * 2;
			}
			q[i] = str[k - f];
		}
		q[m] = '\0';
		printf("S-Tree #%d:
%s

", t++, q);
	}
	return 0;}