Learning records - high precision addition and multiplication

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

const int maxn = 1000;
int A[maxn], B[maxn], C[maxn];

int main()
{
	string a, b;
	cin >> a >> b;
	int len = max(a.size(), b.size());// Number of digits of the result 

	// Reverse save 
	for (int i = a.length() - 1; i >= 0; --i)
		A[a.length()-i] = a[i] - '0';
	for (int i = b.length() - 1; i >= 0; --i)
		B[b.length()-i] = b[i] - '0';

	// Add up 
	for (int i = 1; i <= len; ++i)
	{
		C[i] += A[i] + B[i];
		// Processing carry 
		C[i + 1] = C[i] / 10;
		C[i] %= 10;
	}

	if (C[len+1])// If it is carried, the length is added 1
		len++;
	for (int i = len ; i >= 1; --i)// Reverse output 
		cout << C[i];

	return 0;
}

#include <iostream>
using namespace std;

const int maxn =10000;
int a[maxn], b[maxn], c[maxn];

int main()
{
	string x, y;
	cin >> x >> y;
	int lx = x.length();
	int ly = y.length();

	// Reverse save 
	for (int i = lx - 1; i >= 0; --i)
		a[lx - i] = x[i] - '0';
	for (int i = ly - 1; i >= 0; --i)
		b[ly - i] = y[i] - '0';

	// Computation contribution --a[i]*b[j] The result of is in page i+j-1 On a 
	for (int i = 1; i <= lx; ++i)
		for (int j = 1; j <= ly; ++j)
			c[i + j - 1] += a[i] * b[j];

	int len = lx + ly;// The number of digits of the product shall not exceed the sum of two digits 

	// Processing carry 
	for (int i = 1; i <= len; ++i)
	{
		c[i + 1] += c[i] / 10;
		c[i] %= 10;
	}

	// Remove the lead 0
	while (!c[len])
		len--;

	// Output 
	for (int i = max(1,len); i >= 1; --i)
		cout << c[i];

	return 0;
}