计算24点很简单的小程序，这一段用于解决不改变顺序的24点计算。不过最后总是有一些冗余的括号，这里为了通过SB的POJ3983，只写了一个去除最外层括号的函数，因为对这类问题的完善不太热心，就不关心去括号的事了

编译环境：G++4.6 under Fedora

/** This piece is used to solve the POJ3983.It's another 24-point game. The code piece is oriented from the Beauty of Programming, yet with adjustment to solve 3983 and to make it more like C++ style.Optimization still needed. Bug Log: Declared LIMEN as int, and equal() never returned true. **/ #include 
     
       #include 
      
        #include 
       
         #include 
        
          #include 
         
           using namespace std; //Judge whether two double value is equal bool equal(double,double); //Try an operator and deal with num array and expressions inline bool tryExp(double *num,string *exp,         const double &a,const double &b,         const string &expA,const string &expB,         int i,int n,char opt); //Recursive body of the solution,return true and stop if solution found bool findExp(double *,string*,int); //Test functions void printExp(string *exp,int n) {
        for(int i=0;i
          
           0); if(n==1) { if(equal(num[0],24)) return true; else return false; } for(int i=0;i
           
            i;j--) { num[j]=num[j-1]; exp[j]=exp[j-1]; } num[i]=a; num[i+1]=b; exp[i]=expA; exp[i+1]=expB; } return false; } #include 
            
              void removeOutsideBrackets(string &exp){ char s[50]; strcpy(s,exp.c_str()); int len=strlen(s); if(s[0]=='(' && s[len-1]==')') strncpy(s,s+1,len-1); s[len-2]=0; exp=s; } int main() { double num[4]; string exp[4]; //Input the initial numbers. As we need expression form, //here input string and transform that to number later //will hit two birds with one stone. for(int i=0;i<4;i++) { cin>>exp[i]; num[i]=atoi(exp[i].c_str()); } if(findExp(num,exp,4)) { removeOutsideBrackets(exp[0]); cout<
             
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