ZOJ-2939

罗马数字转换，也没啥难度，直接按规则写就行了

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

static int map[26];

static void print(int num)
{
	int i, d = num / 1000;
	for (i = 0; i < d; i++)
		putchar('M');
	d = (num / 100) % 10;
	if (d == 9)
		printf("%s", "CM");
	else if (d == 4)
		printf("%s", "CD");
	else
	{
		if (d >= 5)
			putchar('D');
		for (i = 0; i < d % 5; i++)
			putchar('C');
	}
	d = (num / 10) % 10;
	if (d == 9)
		printf("%s", "XC");
	else if (d == 4)
		printf("%s", "XL");
	else
	{
		if (d >= 5)
			putchar('L');
		for (i = 0; i < d % 5; i++)
			putchar('X');
	}
	d = num % 10;
	if (d == 9)
		printf("%s", "IX");
	else if (d == 4)
		printf("%s", "IV");
	else
	{
		if (d >= 5)
			putchar('V');
		for (i = 0; i < d % 5; i++)
			putchar('I');
	}
}

static int number(char *s)
{
	int i, res = 0;
	for (i = 0; s[i] != '\0'; i++)
		if ((s[i] == 'I' || s[i] == 'X' || s[i] == 'C')
				&& map[s[i] - 'A'] < map[s[i + 1] - 'A'])
		{
			res += map[s[i + 1] - 'A'] - map[s[i] - 'A'];
			i++;
		}
		else
			res += map[s[i] - 'A'];
	return res;
}

int main()
{
	map['I' - 'A'] = 1;
	map['V' - 'A'] = 5;
	map['X' - 'A'] = 10;
	map['L' - 'A'] = 50;
	map['C' - 'A'] = 100;
	map['D' - 'A'] = 500;
	map['M' - 'A'] = 1000;

	int n, i, count = 0;
	char s[20];
	while (scanf("%d", &n), n)
	{
		getchar();
		int sum = 0;
		for (i = 0; i < n; i++)
		{
			scanf("%s", s);
			sum += number(s);
		}
		printf("Case ");
		print(++count);
		printf(": ");
		print(sum);
		putchar('\n');
	}
	return 0;
}