The Garlic Challenge question 14th: Roman numerals converted to integers

question 14th: convert roman numerals into integers

Memory limit 10000 K time limit 1000 ms

Given a Roman numeral S, (I<=S<=MMMCMXCIX) (i.e. 1 to 3999), converts Roman numerals to integers.

such as the Roman numeral i,ii,iii,iv,v respectively represent the numbers 1, 2, 3, 4, 5.

Format:

The first line enters a Roman digit, and then the corresponding integer is output.

Tips:

First of all, to understand the Roman numeral notation, there are 7 basic characters: I,v,x,l,c,d,m, respectively, representing 1,5,10,50,100,500,1000.

In the form of numbers, there are the following rules:

1, the same number of ligatures, the number represented is equal to the number of these numbers, such as: ⅲ= 3;

2, the small number in the large number of the right, the number represented equal to the number of these numbers, such as: ⅷ= 8;ⅻ= 12;

3, small number, (limited to Ⅰ, X and C) on the left of large numbers, the number represented is equal to the number of reduced number of large numbers, such as: ⅳ= 4;ⅸ= 9;

4, normal use, the number of ligatures should not repeat more than three times.

Sample input

Cxxiii

Sample output

123

for reference only:

#include <stdio.h>
#include <string.h>

void Main ()
{
    char s[100];
	int i,len,count=0;
    scanf ("%s", &s);
	len= strlen (s);

	
	for (i=0;i<len;i++)
	{
		switch (s[i])
		{case
		' M ':
			count+=1000;
			break;
		Case ' D ':
			count+=500;
			break;
		Case ' C ':
			if (s[i+1]== ' D ' | | s[i+1]== ' M ')
				count-=100;
			else
				count+=100;
			break;
		Case ' L ':
			count+=50;
			break;
		Case ' X ':
			if (s[i+1]== ' L ' | | s[i+1]== ' C ')
				count-=10;
			else
				count+=10;
			break;
		Case ' V ':
			count+=5;
			break;
		Case ' I ':
			if (s[i+1]== ' V ') | | s[i+1]== ' X ')
				count--;
			else
				count++;
			break;
		Default:
			printf ("error!\n");
		}

	printf ("%d\n", count);
}