题目的思路就是存一下往左还是往右走，计算出最后的编号，往左右就是num * 2，往右走就是num * 2 + 1，然后查询一下这个位置的值就可以,wrong的一次是没有注意两组之间的空行
A Strange Tree (S-tree) over the variable setXn={x1, x2, . . . , xn}is a binary tree representing aBoolean functionf:{0,1}n→ {0,1}. Each path of the S-tree begins at therootnode and consistsofn+ 1nodes. Each of the S-tree’s nodes has adepth, which is the amount of nodes between itselfand the root (so the root has depth 0). The nodes with depth less thannare callednon-terminalnodes. All non-terminal nodes have two children: theright childand theleft child. Each non-terminalnode is marked with some variablexifrom the variable setXn. All non-terminal nodes with the samedepth are marked with the same variable, and non-terminal nodes with di erent depth are marked withdi erent variables. So, there is a unique variablexi1corresponding to the root, a unique variablexi2corresponding to the nodes with depth 1, and so on. The sequence of the variablesxi1,xi2,. . .,xinis called thevariable ordering. The nodes having depthnare calledterminalnodes. They have nochildren and are marked with either 0 or 1. Note that the variable ordering and the distribution of 0’sand 1’s on terminal nodes are su cient to completely describe an S-tree.
As stated earlier, each S-tree represents a Boolean functionf. If you have an S-tree and values forthe variablesx1,x2,...,xn, then it is quite simple to nd out whatf(x1,x2,...,xn)is: start with theroot. Now repeat the following: if the node you are at is labelled with a variablexi, then depending onwhether the value of the variable is 1 or 0, you go its right or left child, respectively. Once you reach aterminal node, its label gives the value of the function.
Figure 1: S-trees for the functionx1∧(x2∨x3)
On the picture, two S-trees representing the same Boolean function,f(x1, x2, x3) =x1∧(x2∨x3),are shown. For the left tree, the variable ordering isx1, x2, x3, and for the right tree it isx3, x1, x2.
The values of the variablesx1,x2,...,xn, are given as aVariable Values Assignment(VVA)(x1=b1,x2=b2,...,xn=bn)
withb1,b2,...,bn∈ {0,1}. For instance, (x1= 1,x2= 1,x3= 0) would be a valid VVA forn= 3,resulting for the sample function above in the valuef (1,1,0) = 1∧(1∨0) = 1. The correspondingpaths are shown bold in the picture.
Your task is to write a program which takes an S-tree and some VVAs and computesf(x1, x2, . . . , xn)as described above.
Input
The input le contains the description of several S-trees with associated VVAs which you have toprocess. Each description begins with a line containing a single integern,1≤n≤7, the depth of theS-tree. This is followed by a line describing the variable ordering of the S-tree. The format of that lineisxi1xi2. . . xin. (There will be exactlyndi erent space-separated strings). So, forn= 3and thevariable orderingx3,x1,x2, this line would look as follows:
x3 x1 x2
In the next line the distribution of0’s and1’s over the terminal nodes is given. There will be exactly2ncharacters (each of which can be ‘0’ or ‘1’), followed by the new-line character. The characters aregiven in the order in which they appear in the S-tree, the rst character corresponds to the leftmostterminal node of the S-tree, the last one to its rightmost terminal node.
The next line contains a single integerm, the number of VVAs, followed bymlines describingthem. Each of themlines contains exactlyncharacters (each of which can be ‘0’ or ‘1’), followedby a new-line character. Regardless of the variable ordering of the S-tree, the rst character alwaysdescribes the value ofx1, the second character describes the value ofx2, 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 withn= 0. This test case should not be processed.
Output
For each S-tree, output the line ‘S-Tree #j:’, wherej is the number of the S-tree. Then print a linethat contains the value off(x1,x2,...,xn)for each of the givenmVVAs, wherefis the functionde ned 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
//NCC_31060 Majestic
#include <iostream>
#include <deque>
#include <string>
#include <vector>
#include <queue>
#include <cstdio>
#include <stack>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cstdio>
#include <cmath>
#include <cstdlib>
using namespace std;
#define input(a) scanf("%d",&a);
#define print_one_blank printf(" ");
int n;
string value;
int order[8];
void Init() {
for (int i = 0; i < n; ++i) {
string s;
cin >> s;
order[i] = s[1] - '0';
}
cin >> value;
}
void Cal_Value() {
string move;
cin >> move;
int pos = 1;
int head = 1;
for (int i = 0; i < n; ++i) {
if (move[order[i] - 1] == '0')pos *= 2;
else pos = pos * 2 + 1;
head *= 2;
}
pos = pos - head;
printf("%c", value[pos]);
}
int main() {
#ifdef LOCAL
freopen("IN.txt", "r", stdin);
#endif
int cases = 1;
while (scanf("%d", &n) != EOF) {
if(n == 0)break;
Init();
int m;
scanf("%d", &m);
printf("S-Tree #%d:\n",cases++);
for (int i = 0; i < m; ++i) {
Cal_Value();
}
printf("\n\n");
}
return 0;
}