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  • The Famous Clock

     

    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
    我写的代码有点小长,但是很容易就理解了。
    #include<stdio.h>
    int main()
    {
    	char a[5];
    	int i=0;
    	while(scanf("%s",&a)!=EOF)
    	{
    		i+=1;
    		if(a[0]=='I'&&a[1]=='')
    			printf("Case %d: 1
    ",i);
    		else if(a[0]=='I'&&a[1]=='I'&&a[2]=='')
    			printf("Case %d: 2
    ",i);
    		else if(a[0]=='I'&&a[1]=='I'&&a[2]=='I'&&a[3]=='')
    			printf("Case %d: 3
    ",i);
    		else if(a[0]=='I'&&a[1]=='V'&&a[2]=='')
    			printf("Case %d: 4
    ",i);
    		else if(a[0]=='V'&&a[1]=='')
    			printf("Case %d: 5
    ",i);
    		else if(a[0]=='V'&&a[1]=='I'&&a[2]=='')
    			printf("Case %d: 6
    ",i);
    		else if(a[0]=='V'&&a[1]=='I'&&a[2]=='I'&&a[3]=='')
    			printf("Case %d: 7
    ",i);
    		else if(a[0]=='V'&&a[1]=='I'&&a[2]=='I'&&a[3]=='I'&&a[4]=='')
    			printf("Case %d: 8
    ",i);
    		else if(a[0]=='I'&&a[1]=='X'&&a[2]=='')
    			printf("Case %d: 9
    ",i);
    		else if(a[0]=='X'&&a[1]=='')
    			printf("Case %d: 10
    ",i);
    		else if(a[0]=='X'&&a[1]=='I'&&a[2]=='')
    			printf("Case %d: 11
    ",i);
    		else if(a[0]=='X'&&a[1]=='I'&&a[2]=='I'&&a[3]=='')
    			printf("Case %d: 12
    ",i);
    	}	
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/wangyouxuan/p/3218032.html
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