模拟 HDOJ 5099 Comparison of Android versions

 1 /*
 2     题意：比较型号的大小
 3     模拟：坑点在长度可能为5，此时设为'A'
 4 */
 5 #include <cstdio>
 6 #include <algorithm>
 7 #include <iostream>
 8 #include <cstring>
 9 #include <cmath>
10 #include <string>
11 #include <vector>
12 #include <queue>
13 #include <map>
14 #include <set>
15 #include <ctime>
16 #include <cstdlib>
17 using namespace std;
18 
19 const int MAXN = 1e4 + 10;
20 const int INF = 0x3f3f3f3f;
21 char s1[10], s2[10];
22 
23 char check(void)
24 {
25     for (int i=2; i<=4; ++i)
26     {
27         if (s1[i] < s2[i])    return '<';
28         else if (s1[i] > s2[i])    return '>';
29     }
30 
31     if (s1[1] == s2[1])
32     {
33         if (s1[5] == '')    s1[5] = 'A';
34         if (s2[5] == '')    s2[5] = 'A';
35         if (s1[5] < s2[5])    return '<';
36         else if (s1[5] > s2[5])    return '>';
37     }
38 
39     return '=';
40 }
41 
42 int main(void)        //HDOJ 5099    Comparison of Android versions
43 {
44     //freopen ("J.in", "r", stdin);
45 
46     int t, cas = 0;    scanf ("%d", &t);
47     while (t--)
48     {
49         scanf ("%s%s", s1, s2);
50         char ch1 = (s1[0] == s2[0]) ? '=' : (s1[0] < s2[0]) ? '<' : '>';
51         char ch2 = check ();
52         printf ("Case %d: %c %c
", ++cas, ch1, ch2);
53     }
54 
55     return 0;
56 }
57 
58 /*
59 Case 1: > >
60 Case 2: = <
61 */
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