按照题意来说是多个猜错的相同字符只算一次，猜已经猜过的算错；

但是下面的代码ac了，其逻辑是猜已猜对的不算错，多个猜错的算多次错误。

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <ctype.h>

char info[4][20] = { "You win.",
					 "You chickened out.",
					 "You lose.",
					 "You lose." };

int main()
{

	int n;
	int flag[26];
	char s[105], t[105];
	while (scanf("%d", &n) && n != -1)
	{
		scanf("%s %s", s, t);
		memset(flag, 0, sizeof(flag));
		int ch_count = 0;
		int wrong_count = 0;
		for (int i = 0; i < strlen(s); i++)
		{
			int index = s[i] - 'a';
			if (!(flag[index]++))
				ch_count++;
		}
		for (int i = 0; i < strlen(t); i++)
		{
			int index = t[i] - 'a';
			if (flag[index] == 0)
			{
				wrong_count++;
				if (wrong_count >= 7)
					break;
			}
			else
			{
				if (flag[index] > 0)
				{
					flag[index] *= -1;
					ch_count--;
				}
				if (!ch_count)
					break;
			}
		}
		printf("Round %d\n%s\n", n, info[(wrong_count >= 7) * 2 + (ch_count != 0)]);
	}

	return 0;
}