枚举－称硬币POJ1013

 #include <iostream>
 #include<string.h>
 using namespace std;
 char Lleft[][];
 char Lright[][];
 char result[][];
 bool wight(char c,bool b);

 int main(){
     int n;
     cin>>n;
     for(int i=;i<=n;i++) {

         for (int j = ; j < ; ++j) {
             cin>>Lleft[i]>>Lright[i]>>result[i];

         }

         char c;
         for (c = 'A'; c <= 'L'; c++) {
             if (wight(c, )) {
                 cout << c << "is  ," << "heavy";
                 break;//第一次忘记加break了，感觉哪里不对劲但是就是忘记了
             }
             if (wight(c, )) {

                 cout << c << "is," << "light";
                 break;
             }

         }
     }
 }
 bool wight(char c,bool b){

     for (int i = ; i < ; ++i) {

         //这步把轻的中的左边赋值给了一个指针，这样两个问题可以化成一个
         //本来写的是弄成两个分开挨个想，唉其实还是错误率高啊

         char *Left,*Right;
         if(b){
             Left=Lleft[i];
             Right=Lright[i];
         }
         else{
             Right=Lleft[i];
             Left=Lright[i];

         }
         switch(result[i][]){
             case'u':
                 if(strchr(Right,c)==NULL){
                     return false;
                 }
                 break;
             case 'e':
                 if(strchr(Right,c)||strchr(Left,c)){
                     return false;
                 }
                 break;
             case 'd':
                 if(strchr(Left,c)==NULL){
                     return false;
                 }
                 break;
         }

     }
     return true;
 }