HDU 4256 翻译罗马数字

参考自：https://www.cnblogs.com/ECJTUACM-873284962/p/6414173.html

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

 

 解法：

#include <bits/stdc++.h>
using namespace std;
int main(){
    char s[10];
    int ans,k=0;
    while(gets(s)){
        ans=0;
        int len=strlen(s);
        for(int i=0;i<len;i++){
            if(s[i]=='I')ans++;
            if(s[i]=='V')ans=5-ans;
            if(s[i]=='X')ans=10-ans;
            }
        printf("Case %d: %d\n",++k,ans);
        }
    return 0;
}