NB的解法,已经按照大人的要求实现出来了,一遍过.高明啊

/*
先把light,heavy设为全集，每次得到light'和heavy'进行以下操作
light=light交light',light=light-heavy
heavy=heavy交heavy，heavy=heavy-light
最后在light和heavy里面找只剩下一个元素的那个集合，然后把这个元素打印出来就ok
even的就是。。
light=light-even
right=right-even
*/

#include <iostream>
#include <string.h>
using namespace std;

int main()
{
    int i, j, k, l;
    int n;
    char light[91], heavy[91], tleft[10], tright[10], temp[10];
    cin>>n;
    for (i = 0; i < n; i++)
    {
        for (j = 65; j < 77; j++)
        {
            light[j] = 1;
            heavy[j] = 1;
        }
        for (j = 0; j < 3; j++)
        {
            cin>>tleft>>tright>>temp;
            if (strcmp(temp, "even") == 0)
            {
                for (k = 0; k < strlen(tleft); k++)
                {
                    light[tleft[k]] = 0;
                    light[tright[k]] = 0;
                }
                for (k = 0; k < strlen(tright); k++)
                {
                    heavy[tleft[k]] = 0;
                    heavy[tright[k]] = 0;
                }
            }
            if (strcmp(temp, "up") == 0)
            {
                for (k = 0; k < strlen(tleft); k++)
                    heavy[tleft[k]]++;
                for (k = 65; k < 77; k++)
                    if (heavy[k] == 2)
                        heavy[k] = 1;
                    else
                        heavy[k] = 0;
                for (k = 0; k < strlen(tright); k++)
                    heavy[tright[k]] = 0;
                for (k = 0; k < strlen(tright); k++)
                    light[tright[k]]++;
                for (k = 65; k < 77; k++)
                    if (light[k] == 2)
                        light[k] = 1;
                    else
                        light[k] = 0;
                for (k = 0; k < strlen(tleft); k++)
                    light[tleft[k]] = 0;
            }
            if (strcmp(temp, "down") == 0)
            {
                for (k = 0; k < strlen(tright); k++)
                    heavy[tright[k]]++;
                for (k = 65; k < 77; k++)
                    if (heavy[k] == 2)
                        heavy[k] = 1;
                    else
                        heavy[k] = 0;
                for (k = 0; k < strlen(tleft); k++)
                    heavy[tleft[k]] = 0;
                for (k = 0; k < strlen(tleft); k++)
                    light[tleft[k]]++;
                for (k = 65; k < 77; k++)
                    if (light[k] == 2)
                        light[k] = 1;
                    else
                        light[k] = 0;
                for (k = 0; k < strlen(tright); k++)
                    light[tright[k]] = 0;
            }            
        }
        for (k = 65; k < 77; k++)
            if (light[k] == 1)
                cout<<k<<" is the counterfeit coin and it is light."<<endl;
        for (k = 65; k < 77; k++)
            if (heavy[k] == 1)
                cout<<k<<" is the counterfeit coin and it is heavy."<<endl;
    }
	return 0;
}