/*錯(cuò)誤文件，利用遞歸對(duì)算術(shù)的“全運(yùn)算”，對(duì)計(jì)算符號(hào)的控制和中間變量的控制比較麻煩；
主要是對(duì)對(duì)遞歸的不熟練，更多心得在另一個(gè)頭文件中*/


/*#include<iostream>
#include<vector>
#include<string>
using namespace std;
const int Ans = 24;
/*遞歸的控制思考
	對(duì)于要在回溯過程中利用的返回變量，應(yīng)該在程序中設(shè)置一些中間變量如a，b，c，利用它們可以很
	好地返回到遞歸的上一層。
	而對(duì)于要在全過程中利用的或者輸出的變量，則考慮將其設(shè)置為形式參數(shù)，并在程序中自己改變其結(jié)果
	如下面程序中的Exp，Number，這兩個(gè)變量分別都需要在全過程中利用到
*/
/*bool Calu_24(vector<double>Number,string Exp)
{
	if(Number.size() == 1)
	{
		if(Number[0] == Ans)
		{
			cout<<"Success...."<<endl;
		 	cout<<Exp<<endl;
			return true;
		}
		else
		{
			return false;
		}
	}
	else
	{	
		double a,b,c;
		char ch_a,ch_b;
		a = Number[Number.size()-1];
		ch_a = a+48;
		Number.pop_back();
		b = Number[Number.size()-1];
		ch_b = b+48;
		Number.pop_back();
		c = a*b;
		Number.push_back(c);
		if(Number.size() == 3)
		{
			Exp = ch_a;
		}
		Exp = '('+Exp+'*'+ch_b+')';
		if(Calu_24(Number,Exp)) return true;		
		
		c = a+b;
		Number.pop_back();
		Number.push_back(c);
		Exp = Exp.erase(0,1);
		Exp = Exp.erase(Exp.size()-3,3);
		if(Number.size() == 3)
		{
			Exp = ch_a;
		}
		Exp = '('+Exp+'+'+ch_b+')';
		if(Calu_24(Number,Exp)) return true;
		
		if(Number.size() == 3)
		{
			Exp = ch_a;
		}
		if(a>b)
		{
			c = a-b;
		}
		else
		{
			c = b-a;
		}
		Exp = Exp.erase(0,1);
		Exp = Exp.erase(Exp.size()-3,3);
		Exp = '('+Exp+'-'+ch_b+')';
		Number.pop_back();
		Number.push_back(c);
		if(Calu_24(Number,Exp)) return true;
	
		c = a/b;
		Number.pop_back();
		Number.push_back(c);
		Exp = Exp.erase(0,1);
		Exp = Exp.erase(Exp.size()-3,3);
		if(Number.size() == 3)
		{
			Exp = ch_a;
		}
		Exp = '('+Exp+'/'+ch_b+')';
		if(Calu_24(Number,Exp)) return true;
		else
			Number.pop_back();
			Exp = " ";
			return false;
	}
}
*/

/*void Insert_Char(char c1,string &str,int i)
{
	if(i == 0)
	{
		str = c1 + str;
	}
	else
	{
		string Temp_str = str.substr(i);
		str.erase(i,str.size()-i);
		str = str + c1 +Temp_str;
	}
}

void Rotate(vector<int>Number,string Exp)
{
	int count = Number.size();
	
	vector<string>temp;
	vector<string>store;

	store.push_back("0");
	
	string T_Str;
	char ch_i;

	for(int i = 1;i<count;i++)
	{
		for(int j = 0;j<store.size();j++)
		{
			T_Str = store[j];
			for(int t = 0;t<=store[j].size();t++)
			{
				ch_i = i+48;
				Insert_Char(ch_i,T_Str,t);
				temp.push_back(T_Str);
				T_Str = store[j];
			}
		}
		while(!store.empty())
		{
			store.pop_back();
		}
  		store = temp;
		while(!temp.empty())
		{
			temp.pop_back();
		}
	}
//以上得出下標(biāo)排列
	
	vector<int>Sort_Number;
	int T_int;
	bool test = false;
	for(i = 0;i<store.size();i++)
	{
		for(int j = 0;j<Number.size();j++)
		{
			T_int = store[i][j];
			T_int = T_int-48;
			Sort_Number.push_back(Number[T_int]);
		}
		if(Calu_24(Sort_Number,Exp))
		{
			test = true;
			break;
		}
	}
	if(!test)
		cout<<"Fail To Compute..."<<endl;
	else
		cout<<"Success!!"<<endl;

}
*/	
/*
#include <iostream> 
#include <string> 
#include <cmath> 

using namespace std; 

const double PRECISION = 1E-6; 
const int COUNT_OF_NUMBER  = 4; 
const int NUMBER_TO_CAL = 24; 

double number[COUNT_OF_NUMBER]; 
string expression[COUNT_OF_NUMBER]; 

bool Search(int n) 
{ 
    if (n == 1) { 
        if ( fabs(number[0] - NUMBER_TO_CAL) < PRECISION ) { 
            cout << expression[0] << endl; 
            return true; 
        } else { 
            return false; 
        } 
    } 

    for (int i = 0; i < n; i++) { 
        for (int j = i + 1; j < n; j++) { 
            double a, b; 
            string expa, expb; 

            a = number[i]; 
            b = number[j]; 
            number[j] = number[n - 1]; 

            expa = expression[i]; 
            expb = expression[j]; 
            expression[j] = expression[n - 1]; 

            expression[i] = '(' + expa + '+' + expb + ')'; 
            number[i] = a + b; 
            if ( Search(n - 1) ) return true; 
            
            expression[i] = '(' + expa + '-' + expb + ')'; 
            number[i] = a - b; 
            if ( Search(n - 1) ) return true; 
            
            expression[i] = '(' + expb + '-' + expa + ')'; 
            number[i] = b - a; 
            if ( Search(n - 1) ) return true; 
                        

            expression[i] = '(' + expa + '*' + expb + ')'; 
            number[i] = a * b; 
            if ( Search(n - 1) ) return true; 

            if (b != 0) { 
                expression[i] = '(' + expa + '/' + expb + ')'; 
                number[i] = a / b; 
                if ( Search(n - 1) ) return true; 
            } 
            if (a != 0) { 
                expression[i] = '(' + expb + '/' + expa + ')'; 
                number[i] = b / a; 
                if ( Search(n - 1) ) return true; 
            } 

            number[i] = a; 
            number[j] = b; 
            expression[i] = expa; 
            expression[j] = expb; 
        } 
    } 
    return false; 
} 

void main() 
{ 
    for (int i = 0; i < COUNT_OF_NUMBER; i++) { 
        char buffer[20]; 
        int  x; 
        cin >> x; 
        number[i] = x; 
        itoa(x, buffer, 10); 
        expression[i] = buffer; 
    } 

    if ( Search(COUNT_OF_NUMBER) ) { 
        cout << "Success." << endl; 
    } else { 
        cout << "Fail." << endl; 
    }        
}
*/