The Famous Clock

最新推荐文章于 2025-06-16 14:28:30 发布

转载最新推荐文章于 2025-06-16 14:28:30 发布 · 46 阅读

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原文链接：http://www.cnblogs.com/wangyouxuan/p/3218032.html

本文介绍了一个简单的C语言程序，该程序能够将罗马数字转换为阿拉伯数字，特别适用于钟表上的数字转换。通过一系列if-else条件判断，程序能够准确地解析输入的罗马数字，并输出对应的阿拉伯数字。

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]=='\0')
			printf("Case %d: 1\n",i);
		else if(a[0]=='I'&&a[1]=='I'&&a[2]=='\0')
			printf("Case %d: 2\n",i);
		else if(a[0]=='I'&&a[1]=='I'&&a[2]=='I'&&a[3]=='\0')
			printf("Case %d: 3\n",i);
		else if(a[0]=='I'&&a[1]=='V'&&a[2]=='\0')
			printf("Case %d: 4\n",i);
		else if(a[0]=='V'&&a[1]=='\0')
			printf("Case %d: 5\n",i);
		else if(a[0]=='V'&&a[1]=='I'&&a[2]=='\0')
			printf("Case %d: 6\n",i);
		else if(a[0]=='V'&&a[1]=='I'&&a[2]=='I'&&a[3]=='\0')
			printf("Case %d: 7\n",i);
		else if(a[0]=='V'&&a[1]=='I'&&a[2]=='I'&&a[3]=='I'&&a[4]=='\0')
			printf("Case %d: 8\n",i);
		else if(a[0]=='I'&&a[1]=='X'&&a[2]=='\0')
			printf("Case %d: 9\n",i);
		else if(a[0]=='X'&&a[1]=='\0')
			printf("Case %d: 10\n",i);
		else if(a[0]=='X'&&a[1]=='I'&&a[2]=='\0')
			printf("Case %d: 11\n",i);
		else if(a[0]=='X'&&a[1]=='I'&&a[2]=='I'&&a[3]=='\0')
			printf("Case %d: 12\n",i);
	}	
	return 0;
}

转载于:https://www.cnblogs.com/wangyouxuan/p/3218032.html