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HDU-3788

时间:2018-07-21 22:44:10      阅读:206      评论:0      收藏:0      [点我收藏+]

标签:inpu   color   else   chmod   cep   div   include   判断   main   

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
 
 
 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\n");
68         }
69         else
70         {
71             if (flag ==3&&count1*count2 == count3&&count2)
72                 printf("Accepted\n");
73             else
74                 printf("Wrong Answer\n");
75         }
76     }
77     return 0;
78 }

 

 

HDU-3788

标签:inpu   color   else   chmod   cep   div   include   判断   main   

原文地址:https://www.cnblogs.com/1625--H/p/9348126.html

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