//CIS 610-251: Data Structures and Algorithms.
//Assignment #1 Sept 9,1999
//Object: Ternary to Binary Vice-versa Conversion
//Platform: Visual C++ 5.0

#include<iostream.h>
#include<math.h>


class String
{
private:
	char * value;
public:
	String() ;
	String(char*);
	int Length();
	int BaseNTo10(int);
	String  Base10ToN(int,int);
	char * getValue();
};


String::String()
{
	value=new char[33];
	int i=0;
	for(;i<32;i++) value[i]=48;
	value[32]=NULL;
}


String::String(char* Str)
{
	value = new char[33];
	value[32]=NULL;
	String temp;temp.value=Str;
	int LengthofStr=temp.Length();

	if (LengthofStr > 32) 
	{
		temp.value[32]=NULL;value=temp.value;
	}
	else if (LengthofStr < 32) 
	{
		int i=LengthofStr;
	
		for(;i>-1;i--) value[32-i]=temp.value[LengthofStr-i];
		for(i=0;i<32-LengthofStr;i++) value[i]='0';
	}
	else 
	{
		value=temp.value;
	}

}


int String::Length(void)
{ 
	int len=0; 
	for(;*(value+len);len++); 
	return len;
}


//Given Base N convert to decimal
int String::BaseNTo10(int baseN)
{
	int total=0,i=31;
	for(;i>-1;i--) total+=(this->value[i]-48)* pow(baseN,31-i);
	return total;
}


//Given Base 10 convert to base N.
String String::Base10ToN(int dec,int base)
{
	String result; 
	int i=31;
	while (i>-1 && dec >=base)
	{
		result.value[i]=dec%base+48;
		dec=int(dec/base);i--;
		
	}
	if (dec) result.value[i--]=dec+48;
	return result;

}

char * String::getValue()
{
	return value;
}

main()
{
	
	//Binary Input String
	char str1[33], str2[33];
	cout << "Enter Binary String:";
	cin >> str1;
	cout << "Enter Ternary String:";
	cin >> str2;

	String bs(str1);
	cout << "Binary String =" << bs.getValue(); 
	cout << "\nTernary Value =" << (bs.Base10ToN(bs.BaseNTo10(2),3)).getValue();

	//ternary Input String
	String ts(str2);
	cout << "\n\nTernary String =" << ts.getValue();
	cout << "\nBinary Value =" << (ts.Base10ToN(ts.BaseNTo10(3),2)).getValue() <<"\n";
	
	return 0;
	
}