模拟 HDOJ 4552 Running Rabbits

 

题目传送门

 1 /*
 2     模拟：看懂题意，主要是碰壁后的转向，笔误2次
 3 */
 4 #include <cstdio>
 5 #include <algorithm>
 6 #include <cstring>
 7 #include <vector>
 8 using namespace std;
 9 
10 const int MAXN = 1e3 + 10;
11 const int INF = 0x3f3f3f3f;
12 struct Rabbit
13 {
14     char c;
15     int d, s, t;
16     int x, y;
17 }r[3];
18 
19 int main(void)        //HDOJ 4552 Running Rabbits
20 {
21 //    freopen ("K.in", "r", stdin);
22 
23     int n;
24     while (scanf ("%d", &n) == 1)
25     {
26         if (n == 0)    break;
27         getchar ();
28         for (int i=1; i<=2; ++i)
29         {
30             scanf ("%c %d %d", &r[i].c, &r[i].s, &r[i].t);
31             if (r[i].c == 'E')    r[i].d = 0;
32             else if (r[i].c == 'N')    r[i].d = 1;
33             else if (r[i].c == 'W')    r[i].d = 2;
34             else    r[i].d = 3;
35             getchar ();
36         }
37         int k;    scanf ("%d", &k);
38         r[1].x = r[1].y = 1;    r[2].x = r[2].y = n;
39 
40         for (int i=1; i<=k; ++i)
41         {
42             for (int j=1; j<=2; ++j)
43             {
44                 if (r[j].d == 0)
45                 {
46                     if (r[j].y + r[j].s > n)
47                     {
48                         r[j].y = 2 * n - (r[j].y + r[j].s);
49                         r[j].d = 2;
50                     }
51                     else    r[j].y += r[j].s;
52                 }
53                 else if (r[j].d == 1)
54                 {
55                     if (r[j].x - r[j].s < 1)
56                     {
57                         r[j].x = 2 + r[j].s - r[j].x;
58                         r[j].d = 3;
59                     }
60                     else    r[j].x -= r[j].s;
61                 }
62                 else if (r[j].d == 2)
63                 {
64                     if (r[j].y - r[j].s < 1)
65                     {
66                         r[j].y = 2 + r[j].s - r[j].y;
67                         r[j].d = 0;
68                     }
69                     else    r[j].y -= r[j].s;
70                 }
71                 else if (r[j].d == 3)
72                 {
73                     if (r[j].x + r[j].s > n)
74                     {
75                         r[j].x = 2 * n - (r[j].x + r[j].s);
76                         r[j].d = 1;
77                     }
78                     else    r[j].x += r[j].s;
79                 }
80             }
81 
82             if (r[1].x == r[2].x && r[1].y == r[2].y)    swap (r[1].d, r[2].d);
83             else
84             {
85                 if (i % r[1].t == 0)    r[1].d = (r[1].d + 1) % 4;
86                 if (i % r[2].t == 0)    r[2].d = (r[2].d + 1) % 4;
87             }
88         }
89 
90         for (int i=1; i<=2; ++i)    printf ("%d %d\n", r[i].x, r[i].y);
91     }
92 
93     return 0;
94 }

 

posted @ 2015-07-17 21:39  Running_Time  阅读(182)  评论(0)  编辑  收藏  举报