HDU-3788

Posted 2020-12-24 1625--h

Problem Description

对给定的字符串(只包含‘z‘,‘o‘,‘j‘三种字符),判断他是否能AC。

是否AC的规则如下：
1. zoj能AC；
2. 若字符串形式为xzojx，则也能AC，其中x可以是N个‘o‘ 或者为空；
3. 若azbjc 能AC，则azbojac也能AC，其中a,b,c为N个‘o‘或者为空；

 

Input

输入包含多组测试用例，每行有一个只包含‘z‘,‘o‘,‘j‘三种字符的字符串，字符串长度小于等于1000；

 

Output

对于给定的字符串，如果能AC则请输出字符串“Accepted”，否则请输出“Wrong Answer”。

 

Sample Input

zoj ozojo ozoojoo oozoojoooo zooj ozojo oooozojo zojoooo

 

Sample Output

Accepted Accepted Accepted Accepted Accepted Accepted Wrong Answer Wrong Answer

 

Source

浙大计算机研究生复试上机考试-2010年

 

 

 1 #include <iostream>
 2 #include <string>
 3 using namespace std;
 4 int main()
 5 {
 6     int end, count1, count2, count3, flag;
 7     string a;
 8     while (cin >> a)
 9     {
10         flag = 1;
11         count1 = count2 = count3 = 0;
12         end = 0;
13         for (int i = 0; i < a.length(); i++)
14         {
15             if (flag == 1)
16             {
17                 if (a.at(i) == ‘o‘)
18                 {
19                     count1++;
20                     continue;
21                 }
22                 else if (a.at(i) == ‘z‘)
23                 {
24                     flag = 2;
25                     continue;
26                 }
27                 else if (a.at(i) == ‘j‘)
28                 {
29                     end = 1;
30                     break;
31                 }
32             }
33             if (flag == 2)
34             {
35                 if (a.at(i) == ‘o‘)
36                 {
37                     count2++;
38                     continue;
39                 }
40                 else if (a.at(i) == ‘j‘)
41                 {
42                     flag = 3;
43                     continue;
44                 }
45                 else if (a.at(i) == ‘z‘)
46                 {
47                     end = 1;
48                     break;
49                 }
50             }
51             if (flag == 3)
52             {
53                 if (a.at(i) == ‘o‘)
54                 {
55                     count3++;
56                     continue;
57                 }
58                 else
59                 {
60                     end = 1;
61                     break;
62                 }
63             }
64         }
65         if (end)
66         {
67             printf("Wrong Answer
");
68         }
69         else
70         {
71             if (flag ==3&&count1*count2 == count3&&count2)
72                 printf("Accepted
");
73             else
74                 printf("Wrong Answer
");
75         }
76     }
77     return 0;
78 }