UVA 712 (13.08.23)

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, andso 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 variablexi, 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 isx3, x1, x2.
The values of the variables ,are given as a Variable Values Assignment (VVA)
with .For instance, ( x 1 = 1, x 2 = 1 x 3 = 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 computesas 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 1 x i 2 ... x i n. (There will be exactly n different space-separated strings).So, for n = 3 and the variable ordering x 3, x 1, x 2, 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 thefunction 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 및 방법:
변형된 반복 문제는 먼저 층수 n을 입력한 다음에 n개의 순서를 입력한다.
다음은 마지막 잎결점(잎결점의 개수, 사실은 2의 n차방)
그리고 m를 입력하면 m를 반복합니다.
뒤이어 m행은 반복 지령, 0은 왼쪽, 1은 오른쪽~
m번 반복해서 얻은 결과를 저장하여 한꺼번에 출력합니다!
요점:
수조로 나무를 저장하면 결점 번호는 k이고 왼쪽 아들 번호는 2*k이며 오른쪽 아들 번호는 2*k+1이다.
나는 처음부터 k를 1로 설정하여 마지막 잎사귀 결점의 번호를 구했다. 주의했다. 이때의 번호 k는 수조의 하표가 아니라 위의 결점을 빼야 한다
AC 코드:
 

#include<stdio.h>

#include<string.h>



char order[10][5];

char node[512];

char ans[10000];



int main() {

    int n;

    int cas = 0;

    while(scanf("%d", &n) != EOF && n) {

        int i, j;

        int presum = 1;

        for(i = 0; i < n; i++) {

            scanf("%s", order[i]);

            presum = presum * 2;

        }



        getchar();

        gets(node);



        int m;

        scanf("%d", &m);

        getchar();



        int pos = 0;

        char tmp[10];

        for(i = 0; i < m; i++) {

            gets(tmp);

            int len = strlen(tmp);

            int k = 1;

            for(j = 0; j < len; j++) {

                if(tmp[j] == '0')

                    k = 2*k;

                else

                    k = 2*k+1;

            }

            k = k - presum;

            ans[pos++] = node[k];

        }

        ans[pos] = '\0';

        printf("S-Tree #%d:
", ++cas);

        puts(ans);

        printf("
");

    }

    return 0;

}