求200行的C++程序源代码

答案：一个计算24点的c++程序
#include <iostream>
#include <sys/timeb.h>
#include <vector>
#include <string>
#include <stdio.h>
using std::vector;
using std::string;
using namespace std;
class Node
{
public:
    enum OpType
    {
        UNARY, PLUS, SUB, SUB2, MUL, DIV, DIV2
    };
    double& value(){ return value_; }
    double value() const { return value_; } 
    bool Add() { op_ = PLUS; value_ = left_->value_ + right_->value_; return true; }
    bool Sub() { op_ = SUB; value_ = left_->value_ - right_->value_; return true; }
    bool Sub2()  { op_ = SUB2; value_ = right_->value_ - left_->value_; return true; }
    bool Mul() { op_ = MUL; value_ = left_->value_ * right_->value_; return true; }
    bool Div() { if(right_->value_==0)return false; op_ = DIV; value_ = left_->value_ / right_->value_; return true; }
    bool Div2() { if(left_->value_==0)return false; op_ = DIV2; value_ = right_->value_ / left_->value_;return true; }
    Node( double d = 0 )
        : left_(0), right_(0), op_(UNARY), value_(d)
    { }
    Node( Node* l, Node *r )
        : left_(l), right_(r), op_(UNARY), value_(0)
    { }
    string getExpr()
    {
        if( op_ == UNARY )
        {
            char buf[32];
            return "";
        }
        string str;
        switch( op_ )
        {
        case PLUS:
        case SUB:
        case MUL:
        case DIV:
            str += "(";
            str += left_->getExpr();
            str += " +--*//"[op_];
            str += right_->getExpr();
            str += ")";
            break;
        case SUB2:
        case DIV2:
        default:
            str += "("; 
            str += right_->getExpr();
            str += " +--*//"[op_];
            str += left_->getExpr();
            str += ")";
        }
        return str;
    }
    double value_;
    OpType op_;
    Node *left_;
    Node *right_;
};
template<unsigned int L>
bool query( double target, Node** cand, string& expr )
{
    for( unsigned int k=0; k<L-1; ++k )
    {
        Node* newcand[L-1];
        for( unsigned int j=k+1; j<L; ++j )
        {
            for( unsigned int i=k+1; i<j; ++i )
            {
                if( cand[i]->value() == cand[j]->value() )
                    continue;
            }
            Node node( cand[k], cand[j] );
            Node **pNode = newcand;
            for( unsigned int i=0; i<L; ++i )
            {
                if( i!=k && i!=j )
                    *pNode++ = cand[i];
            }
            newcand[L-2] = &node;
            if( node.Add() && query<L-1>( target, newcand, expr ) )
                return true;
            if( node.Sub() && query<L-1>( target, newcand, expr ) )
                return true;
            if( node.Sub2() && query<L-1>( target, newcand, expr ) )
                return true;
            if( node.Mul() && query<L-1>( target, newcand, expr ) )
                return true;
            if( node.Div() && query<L-1>( target, newcand, expr ) )
                return true;
            if( node.Div2() && query<L-1>( target, newcand, expr ) )
                return true;
        }
    }
    return false;
}
template<>
bool query<1>( double target, Node** cand, string& expr )
{
    double diff =  target - cand[0]->value();
    if( diff < 0.000001 && diff > -0.000001 )
    {
        //expr = cand[0]->getExpr();
        return true;
    }
    return false;
}
int main1( int argc, char** argv)
{
 const int LEN = 4;
 if( argc != LEN+1 )
 {
  exit( -1 );
 }
 double target = 24;
 Node nodes[LEN];
 Node *cands[LEN];
 for( int k=0; k<LEN; ++k )
 {
  int i = atoi( argv[k+1] );
  if( i == 0 )

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