1017 A除以B (20 分)

//本题关键就在于怎么存不超过1000位的数字
//模仿除法的整个过程
#include
#include
using namespace std;


int main()
{
	string str;
	int a;
	cin >> str >> a;
	int q, r = 0;//商、余数
	string str1;
	
	for (unsigned int i = 0; i < str.length(); i++)
	{
		q = ((str[i] - '0') + r * 10) / a;
		r = ((str[i] - '0') + r * 10) % a;
		string s = to_string(q);
		str1.insert(i, s);
	}
	
	//把前面的零去了
	if (str.length() == 1)
	{
		cout << str1 << " " << r << endl;
	}
	else
	{
		if (str1[0] != '0')
		{
			cout << str1[0];
		}
		for (unsigned int i = 1; i < str1.length(); i++)
		{
			cout << str1[i] ;
		}
		cout << " " << r << endl;
	}
	
	//system("pause");
	return 0;
}

上面自己写的代码，求出每一位的商之后用了一个字符串装入，最后在输出  空间复杂度不是很好

改进后的代码：

int main()
{
	string str;
	int a;
	cin >> str >> a;
	int q, r = 0;//商、余数
	string str1;

	if (str.length() == 1)
	{
		q = ((str[0] - '0') + r * 10) / a;
		r = ((str[0] - '0') + r * 10) % a;
		cout << q << " " << r << endl;
	}
	else
	{
		q = ((str[0] - '0') + r * 10) / a;
		r = ((str[0] - '0') + r * 10) % a;
		if (q != 0)
		{
			cout << q;
		}
		for (unsigned int i = 1; i < str.length(); i++)
		{
			q = ((str[i] - '0') + r * 10) / a;
			r = ((str[i] - '0') + r * 10) % a;
			cout << q;
		}
		cout << " " << r << endl;
	}

	//system("pause");
	return 0;
}