http://lightoj.com/volume_showproblem.php?problem=1051

1051 - Good or Bad

A string is called bad if it has 3 vowels in a row, or 5 consonants in a row, or both. A string is called good if it is not bad. You are given a string s, consisting of uppercase letters (‘A‘-‘Z‘) and question marks (‘?‘). Return "BAD" if the string is definitely bad (that means you cannot substitute letters for question marks so that the string becomes good), "GOOD" if the string is definitely good, and "MIXED" if it can be either bad or good.

The letters ‘A‘, ‘E‘, ‘I‘, ‘O‘, ‘U‘ are vowels, and all others are consonants.

Input

Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case begins with a non-empty string with length no more than 50.

Output

For each case of input you have to print the case number and the result according to the description.

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <cstdio>
 6 #include <vector>
 7 #include <ctime>
 8 #include <queue>
 9 #include <list>
10 #include <set>
11 #include <map>
12 using namespace std;
13 #define INF 0x3f3f3f3f
14 typedef long long LL;
15
16 int dp[55][10][10]; //dp[i][j][k]，表示以第i位为结尾，元音个数为j或辅音个数为k的情况是否存在
17 char s[55];
18 int main()
19 {
20     int t;
21     scanf("%d", &t);
22     for(int ca = 1; ca <= t; ca++)
23     {
24         scanf("%s", s+1);
25         int len = strlen(s+1);
26         memset(dp, 0, sizeof(dp));
27         dp[0][0][0] = 1;
28         int Bad = 0, Good = 0;
29         for(int i = 1; i <= len; i++)
30         {
31             for(int j = 0; j < 3; j++)
32             {
33                 if(dp[i-1][j][0] == 0)
34                     continue;
35                 if(s[i] == ‘?‘)
36                     dp[i][j+1][0] = dp[i][0][1] = 1;
37                 else if(s[i]==‘A‘||s[i]==‘E‘||s[i]==‘I‘||s[i]==‘O‘||s[i]==‘U‘)
38                     dp[i][j+1][0] = 1;
39                 else
40                     dp[i][0][1] = 1;
41             }
42             if(dp[i][3][0])
43                 Bad = 1;
44             for(int j = 0; j < 5; j++)
45             {
46                 if(dp[i-1][0][j] == 0)
47                     continue;
48                 if(s[i] == ‘?‘)
49                     dp[i][0][j+1] = dp[i][1][0] = 1;
50                 else if(s[i]==‘A‘||s[i]==‘E‘||s[i]==‘I‘||s[i]==‘O‘||s[i]==‘U‘)
51                     dp[i][1][0] = 1;
52                 else
53                     dp[i][0][j+1] = 1;
54             }
55             if(dp[i][0][5])
56                 Bad = 1;
57         }
58         for(int i = 1; i < 3; i++)
59             if(dp[len][i][0])
60                 Good = 1;
61         for(int i = 1; i < 5; i++)
62             if(dp[len][0][i])
63                 Good = 1;
64         printf("Case %d: ", ca);
65         if(Bad && Good)
66             puts("MIXED");
67         else if(Bad)
68             puts("BAD");
69         else
70             puts("GOOD");
71     }
72     return 0;
73 }