B1017 A除以B （20 分）高精度与低精度的除法规范解法

#include
#include
#include
#include
using namespace std;
struct bign
{
	int d[1010];
	int len;
	bign()
	{
		memset(d, 0, sizeof(d));
		len = 0;
	}
};
bign change(char str[])
{
	bign a;
	a.len = strlen(str);
	for (int i = 0; i < a.len; i++)
	{
		a.d[i] = str[a.len - i - 1] - '0';
	}
	return a;
}
bign divide(bign a, int b, int& r)
{
	bign c;
	c.len = a.len;
	for (int i = a.len - 1; i >= 0; i--)
	{
		r = r * 10 + a.d[i];
		if (r < b) c.d[i] = 0;
		else
		{
			c.d[i] = r / b;
			r = r % b;
		}
	}
	while (c.len - 1 >= 1 && c.d[c.len - 1] == 0)
	{
		c.len--;
	}
	return c;
}
int main()
{
	char str[1010];
	int b, r = 0;
	scanf("%s%d", str, &b);
	bign a = change(str);
	bign result = divide(a, b, r);
	for (int i = result.len - 1; i >= 0; i--)
	{
		printf("%d", result.d[i]);
	}
	printf(" %d", r);
	return 0;
}