高精度数的加减乘法

/*  处理2进制的高精度加,减和乘法的C++程序.Prositive HighPricistion—Operation of baniry system  */
#include <iostream.h>
#include <mem.h>
const int MAXSIZE = 20;     //max length of the number
const int K = 2;             //baniry system 在这里可以修改K进制的高精度
class hp{
	int len;               //length of number
	int s[MAXSIZE];      //store high precistion number
public:
	hp();
	hp hp::operator = (hp C);
};

hp::hp()
{
	len = 0;
	memset(s, 0, MAXSIZE*sizeof(int));
}
istream &operator >> (istream &in, hp &HP)
{
	char s[MAXSIZE]; 
	int i;  
	cout << "Input Number = ";
	cin >> s;
	HP.len = strlen(s);
	for (i = 0; i < HP.len; i++)
		HP.s[i] = s[HP.len-i-1] - '0'; //change string to high precisition
	return in;
}

ostream &operator << (ostream &out, hp &HP)
{
	int i;
	for (i = HP.len-1; i >= 0; i--)
		cout << HP.s[i];
	return out;
}

hp operator +(hp A, hp B)
{
	int i, len;
	hp C;

	if (A.len > B.len)
		len = A.len;
	else
		len = B.len;           //get the bigger length of A,B
	for (i = 0; i < len; i++){
		C.s[i] += A.s[i] + B.s[i];
		if (C.s[i] >= K){
			C.s[i] -= K;
			++C.s[i+1];        //add 1 to a higher position
		}
	}
	if (C.s[len] > 0)
		C.len = len+1;
	else
		C.len = len;
	return C;
}

hp operator - (hp A, hp B)  //different of the two HighPrecision Numbers
{
	int len, i;
	hp C;

	if (A.len > B.len)
		len = A.len;
	else
		len = B.len;C.len = 4;
	for (i = 0; i < len; i++){
		C.s[i] += A.s[i] - B.s[i];
		if (C.s[i] < 0){
			C.s[i] += K;
			--C.s[i+1];        //subtract 1 to higher position
		}
	}
	while (C.s[len-1] == 0 && len > 1)
	--len;
	C.len = len;
	return C;
}

hp operator * (const hp &A, const hp &B)
{
	int len, i, j;
	hp C;
	for (i = 0; i < A.len; i++)
		for (j = 0; j < B.len; j++){
			len = i+j;                  
			C.s[len] += A.s[i] * B.s[j];
			C.s[len+1] += C.s[len] / K;
			C.s[len] %= K;
		}
		len = A.len + B.len + 1;   
/* 
the product of a number with i digits and a number with j digits
  can only have at most i+j+1 digits
*/
	while (len > 1 && C.s[len-1] == 0)
		--len;
	C.len = len;
	return C;
}

hp hp::operator = (hp C)
{
	int i;
	len = C.len;
	for (i = 0; i < MAXSIZE; i++)
		s[i] = C.s[i];
	return *this;
}

int main()
{
	hp A, B, C;
	cin >> A >> B;
	C = A+B;
	cout << A << ‘+’ << B << “ = “ << C << endl;
	C = A-B;
	cout << A << ‘-’ << B << “ = “ << C << endl;
	C = A*B;
	cout << A << '*' << B << " = " << C << endl;    
	return 0;  
}