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//poj2692假币问题#include 
   
    #include
    
     #include
     
      #include
      
       using namespace std;int n;int status[12]; //每个硬币的状态char lef[3][7], righ[3][7], res[3][7];bool balance(){    for(int i = 0; i < 3; i++){        int rightW = 0, leftW = 0;        for(int j = 0; righ[j] != 0 && j < 6; j++)            rightW += status[righ[i][j]-'A'];        for(int j = 0; left[j] != 0 && j < 6; j++)            leftW += status[lef[i][j]-'A'];        if(res[i][0] == 'e' && rightW != leftW)            return false;        if(res[i][0] == 'u' && rightW >= leftW)            return false;        if(res[i][0] == 'd' && rightW <= leftW)            return false;    }    return true;}int main(){//0表示未遍历，1表示为真，-1表示可能为假且轻，1表示可能为假且重    scanf("%d",&n);    char ans[n+2][43];    for(int index = 0; index < n; index++){        memset(status, 0, sizeof(status));        memset(lef, 0, sizeof(lef));        memset(righ, 0, sizeof(righ));        for(int i = 0; i < 3; i++)            scanf("%s %s %s", lef[i], righ[i], res[i]);        for(int i = 0; i < 12; i++){            //分别假设每个硬币为轻，重，真，依次枚举，如果满足条件则跳出            status[i] = -1;            if(balance()){                strcpy(ans[index], "0 is the counterfeit coin and it is light. ");                ans[index][0] = 'A'+i;                break;            }            status[i] = 1;            if(balance()){                strcpy(ans[index], "0 is the counterfeit coin and it is heavy. ");                ans[index][0] = 'A'+i;                break;            }            status[i] = 0;        }    }    for(int i = 0; i < n; i++)        printf("%s\n",ans[i]);    return 0;}
      
     
    
   

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