HDU 4256 The Famous Clock

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The Famous Clock

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1399    Accepted Submission(s): 940


Problem Description
Mr. B, Mr. G and Mr. M are now in Warsaw, Poland, for the 2012’s ACM-ICPC World Finals Contest. They’ve decided to take a 5 hours training every day before the contest. Also, they plan to start training at 10:00 each day since the World Final Contest will do so. The scenery in Warsaw is so attractive that Mr. B would always like to take a walk outside for a while after breakfast. However, Mr. B have to go back before training starts, otherwise his teammates will be annoyed. Here is a problem: Mr. B does not have a watch. In order to know the exact time, he has bought a new watch in Warsaw, but all the numbers on that watch are represented in Roman Numerals. Mr. B cannot understand such kind of numbers. Can you translate for him?
 

 

Input
Each test case contains a single line indicating a Roman Numerals that to be translated. All the numbers can be found on clocks. That is, each number in the input represents an integer between 1 and 12. Roman Numerals are expressed by strings consisting of uppercase ‘I’, ‘V’ and ‘X’. See the sample input for further information.
 

 

Output
For each test case, display a single line containing a decimal number corresponding to the given Roman Numerals.
 

 

Sample Input
I
II
III
IV
V
VI
VII
VIII
IX
X
XI
XII
Sample Output
Case 1: 1
Case 2: 2
Case 3: 3
Case 4: 4
Case 5: 5
Case 6: 6
Case 7: 7
Case 8: 8
Case 9: 9
Case 10: 10
Case 11: 11
Case 12: 12
Source
分析:这种Case的模式不太会写,然后我直接枚举出来,WA了,我也不知道为啥错了,然后看别人写,去统计,感觉自己好傻,根本没必要那样!
下面给出AC代码:
 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int main()
 4 {
 5     char s[10];
 6     int k=0;
 7     int ans;
 8     while(gets(s))
 9     {
10         ans=0;
11         int len=strlen(s);
12         for(int i=0;i<len;i++)
13         {
14             if(s[i]==I)ans++;
15             if(s[i]==V)ans=5-ans;
16             if(s[i]==X)ans=10-ans;
17         }
18         printf("Case %d: %d\n",++k,ans);
19     }
20     return 0;
21 }

 

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