题目传送门
1 /*
2 简单模拟；考虑边界的情况输出的是不一样的，还有思维好，代码能短很多
3 */
4 #include <cstdio>
5 #include <iostream>
6 #include <algorithm>
7 #include <string>
8 #include <cstring>
9 #include <map>
10 #include <vector>
11 using namespace std;
12
13 const int MAXN = 1e4 + 10;
14 const int INF = 0x3f3f3f3f;
15
16 int main(void) //ACdream 1196 KIDx's Pagination
17 {
18 //freopen ("A.in", "r", stdin);
19
20 int n, cur, d;
21 int a[110];
22 int cas = 0;
23 while (scanf ("%d%d%d", &n, &cur, &d) == 3)
24 {
25 printf ("Case #%d: ", ++cas);
26 if (cur == 1) printf ("[<<]");
27 else printf ("(<<)");
28
29 if (cur - d > 1) printf ("[...]");
30 for (int i=max(cur-d, 1); i<cur; ++i) printf ("(%d)", i);
31 printf ("[%d]", cur);
32 for (int i=cur+1; i<=min(cur+d, n); ++i) printf ("(%d)", i);
33 if (cur + d < n) printf ("[...]");
34
35 if (cur == n) printf ("[>>]");
36 else printf ("(>>)");
37 puts ("");
38 }
39
40 return 0;
41 }
42
43 /*
44 Case #1: (<<)[...](3)(4)[5](6)(7)[...](>>)
45 Case #2: [<<][1](2)(3)[...](>>)
46 */