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這題很有意思哇。給出一個這樣的東西：
A---+
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+---:\
: >o---:\
+---:/ : )---?
| C--o:/
B---+
其中：
:\ :\ -:\ -o:\ A-o:\
: ) : > : )- : )o- : )o-?
:/ :/ -:/ --:/ B--:/
AND gate OR gate Gate with inputs An inverted top input Two logic input
and an inverted output and the output
然后叫你分析它的輸出。 其實哥比較喜歡這種有創(chuàng)意的題，做這種題代碼也寫得比較有意思。
 #include <stdio.h>
#include <string.h>
  struct node  {
 char type, n[3];
struct node *child[2];
 };
struct node nodes[128], *root;
int nodes_cnt;
char map[128][128], input[32];
int W[128], H;
int in_range(int dx, int dy)
{
return !(dx < 0 || dx >= W[dy] || dy < 0 || dy >= H);
}
void follow_path(int x, int y, struct node **p_node);
int find_node(int x, int y, struct node **p_node)
{
struct node *t;
  if (map[y][x] >= 'A' && map[y][x] <= 'Z') {
t = &nodes[nodes_cnt++];
t->type = map[y][x];
*p_node = t;
return 1;
 }
if (map[y][x] == 'o')
x--;
if (map[y][x] == ')' || map[y][x] == '>') {
t = &nodes[nodes_cnt++];
memset(t, 0, sizeof(*t));
t->type = map[y][x];
*p_node = t;
if (in_range(x - 3, y - 1) && map[y - 1][x - 3] == 'o')
t->n[0] = 1;
if (in_range(x - 3, y + 1) && map[y + 1][x - 3] == 'o')
t->n[1] = 1;
if (in_range(x + 1, y) && map[y][x + 1] == 'o')
t->n[2] = 1;
follow_path(x - 3, y - 1, &t->child[0]);
follow_path(x - 3, y + 1, &t->child[1]);
return 1;
}
return 0;
}
void follow_path(int x, int y, struct node **p_node)
{
int i, dx, dy;
const struct {
char ch;
int dx, dy;
} dir[4] = {
{'-', -1, 0},
{'-', 1, 0},
{'|', 0, 1},
{'|', 0, -1}
};
while (1) {
for (i = 0; i < 4; i++) {
dx = x + dir[i].dx;
dy = y + dir[i].dy;
if (!in_range(dx, dy))
continue;
if (map[dy][dx] == dir[i].ch || map[dy][dx] == '+')
break;
if (find_node(dx, dy, p_node))
return ;
}
if (i == 4)
break;
map[y][x] = '.';
x = dx;
y = dy;
}
*p_node = NULL;
}
int calc(struct node *t)
{
if (!t)
return 0;
if (t->type >= 'A' && t->type <= 'Z')
return input[t->type - 'A'] - '0';
if (t->type == ')')
return ((calc(t->child[0]) ^ t->n[0]) &
(calc(t->child[1]) ^ t->n[1])) ^
t->n[2];
if (t->type == '>')
return ((calc(t->child[0]) ^ t->n[0]) |
(calc(t->child[1]) ^ t->n[1])) ^
t->n[2];
return 0;
}
void find_root()
{
int x, y;
root = NULL;
for (y = 0; y < H; y++)
for (x = 0; x < W[y]; x++) {
if (map[y][x] == '?') {
follow_path(x, y, &root);
return ;
}
}
}
int main()
{
freopen("e:\\test\\in.txt", "r", stdin);
while (1) {
nodes_cnt = 0;
H = 0;
while (1) {
if (!fgets(map[H], sizeof(map[H]), stdin))
return 0;
if (map[H][0] == '*')
break;
W[H] = strlen(map[H]);
H++;
}
find_root();
while (1) {
fgets(input, sizeof(input), stdin);
if (input[0] == '*')
break;
printf("%d\n", calc(root));
}
printf("\n");
}
return 0;
}
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