uva 712 - S-Trees

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 a depth, 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 called non-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, (

x1 = 1, x2 = 1 x3 = 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 xi1 xi2 ...xin. (There will be exactly n different space-separated strings).
So, for n = 3 and the variable ordering

x3, x1, x2, 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

#include <iostream>
#include <stack>
#include <cstring>
#include <cstdio>
#include <string>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
#include <fstream>
#include <stack>
#include <list>
#include <sstream>
#include <cmath>

using namespace std;
/*

*/
#define ms(arr, val) memset(arr, val, sizeof(arr))
#define N 10
#define INF 0x3fffffff
#define vint vector<int>
#define setint set<int>
#define mint map<int, int>
#define lint list<int>
#define sch stack<char>
#define qch queue<char>
#define sint stack<int>
#define qint queue<int>

int stree
;
char term[1000];
char vva
;
/*
看懂了题目还是比较简单的。
给一个vva，找对应的叶子节点的值。
vva中0表示往左子树走，1表示在右子树找
例如：
题目中的right tree it is x3, x1, x2；
给的vva是1 1 0，也就是x3=0,x1=1,x2=2
那么找的话，就是先左再右再右就是答案1了
*/
int expo(int a, int n)//a^n
{
int ans = 1;
while (n)
{
if (n & 1)
{
ans *= a;
}
a *= a;
n >>= 1;
}
return ans;
}
void bin_search(int i, int j, int p)//二分搜索
{
if (i >= j)
{
putchar(term[j]);
return;
}
if (vva[stree[p]] == '0')
bin_search(i, (i + j) / 2, p + 1);
else
bin_search((i + j) / 2 + 1, j, p + 1);
}
int main()
{
int n, len, i, m, cases = 1;
string t;
while (scanf("%d", &n), n)
{
i = 1;
cin>>t;
len = t.length() - 1;
stree[i] = t[len] - '0';
for (i++; i <= n; i++)
{
cin>>t;
stree[i] = t[len] - '0';
}
len = expo(2, n);
scanf("%s", term + 1);
scanf("%d", &m);
printf("S-Tree #%d:\n", cases++);
while (m--)
{
scanf("%s", vva + 1);
bin_search(1, len, 1);
}
printf("\n\n");
}
return 0;
}