本文共 1192 字，大约阅读时间需要 3 分钟。
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 6 #include 7 #include 8 #include 9 #include 10 #include 11 #include 12 #include 13 #include 14 #include 15 #include 16 #include 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] == '\0') s1[5] = 'A';34 if (s2[5] == '\0') 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 versions43 {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\n", ++cas, ch1, ch2);53 }54 55 return 0;56 }57 58 /*59 Case 1: > >60 Case 2: = <61 */
转载于:https://www.cnblogs.com/Running-Time/p/4514900.html