NYOJ 484 The Famous Clock

傷城~ 2022-08-25 01:40 95阅读 0赞

## The Famous Clock ##

时间限制： 1000 ms | 内存限制： 65535 KB

难度： 1

描述

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?

输入

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.

输出

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

样例输入

I
    II
    III
    IV
    V
    VI
    VII
    VIII
    IX
    X
    XI
    XII

样例输出

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

思路：字符串的比较。

#include <stdio.h>
    #include <string.h>
    
    int main()
    {
    	int i,total = 1;
    	char str[5];
    	const char *s[] = 
    	{
    		"I",
    		"II",
    		"III",
    		"IV",
    		"V",
    		"VI",
    		"VII",
    		"VIII",
    		"IX",
    		"X",
    		"XI",
    		"XII",
    	};
    	while (scanf("%s",str) != EOF)
    	{
    		for (i=0; i<12; i++)
    		{
    			if (strcmp(str,s[i]) == 0)
    			{
    				printf("Case %d: %d\n",total,i+1);
    				total++;
    				break;
    			}
    		}	
    	}
    
    	return 0;
    }