两个参数都是高精度的大整数加减乘除（vector）

话不多说，这次写的是两个参数均是vector的高精度加减乘除，直接上代码：

vector<int> PlusH(vector<int>a, vector<int>b)//此参数的vector的a如果真实值为123那么vector里变成321
{
	vector<int>c;
	c.clear();
	for (int i = 0, t = 0; i < a.size() || i < b.size() || t; i++)
	{
		if (i < a.size()) t += a[i];
		if (i < b.size())t += b[i];
		c.push_back(t % 10);
		t = t / 10;
	}
	return c;//还是倒着的
}//时间复杂度为O（n）

vector<int> Plus(vector<int>a, int b)
{
	vector<int>c;
	c.clear();
	a[0] += b;
	for (int i = 0, t = 0; i < a.size() || t; i++)
	{
		t += a[i];
		c.push_back(t % 10);
		t = t/ 10;
	}
	return c;
}

vector<int>  Mul(vector<int>a,int b)
{
	vector<int> c;
	if (b == 0)
	{
		c.push_back(0);
		return c;
	}
	c.clear();
	for (int i = 0, t = 0; i < a.size()||t; i++)
	{
		if (i < a.size()) t += b * a[i];
		c.push_back(t % 10);
		t = t / 10;
	}
	return c;
}//时间复杂度为O(n)

vector<int>  mulH(vector<int>a, vector<int>b)
{
	vector<int>c;
	c.clear();
	vector<int>d;
	for (int i = 0; i < a.size()||d.size(); i++)
	{
		d = PlusH(d, Mul(b, a[i]));
		c.push_back(d[0]);
		d.erase(d.begin(), d.begin() + 1);
	}
	return c;
}//时间复杂度O(n^2)

vector<int> reduceH(vector<int>a, vector<int>b)
{
	vector<int>c;
	c.clear();
	if (a.size() > b.size())//最后的值大于零
	{
		for (int i = 0,t=0; i < a.size(); i++)
		{
			a[i] += t;
			int d;
			if (a[i] < b[i])
			{
				d = abs(a[i] - b[i]);
				t--;
			}
			else
				d = a[i] - b[i];
			c.push_back(d);
		}
		if (c[c.size() - 1] == 0)//去除末尾0
			c.erase(c.end() - 1, c.end());
	}
	else if (a.size() < b.size())//最后的值小于零
	{
		for (int i = 0, t = 0; i < b.size(); i++)
		{
			b[i] += t;
			int d;
			if (b[i] < a[i])
			{
				d = abs(a[i] - b[i]);
				t--;
			}
			else
				d = b[i] - a[i];
			c.push_back(d);
		}
		if (c[c.size() - 1] == 0)
			c.erase(c.end() - 1, c.end());
		c[c.size() - 1] = -c[c.size() - 1];
	}
	else //两个位数相同通过比较得出答案
	{
		int i;
		for (i = a.size() - 1; i >= 0; i--)
		{
			if (a[i] != b[i])//找到从最高位开始的第一个不同的数
				break;
		}
		if (a[i] > b[i])//a的值大于b的值
		{
			for (int j = 0,t=0; j <= i; j++)
			{
				a[j] += t;
				if (a[j] < b[j]) t--;
				c.push_back(abs(a[j] - b[j]));
			}
			if (c[c.size() - 1] == 0)
				c.erase(c.end() - 1, c.end());
		}
		else//a的值小于b的值
		{
			for (int j = 0, t = 0; j <= i; j++)
			{
				b[j] += t;
				if (b[j] < a[j]) t--;
				c.push_back(abs(b[j] - a[j]));
			}
			if (c[c.size() - 1] == 0)
				c.erase(c.end() - 1, c.end());
			c[c.size() - 1] = -c[c.size() - 1];
		}
	}
	return c;
}

vector<int>  MaxH(vector<int>a, vector<int>b)
{
	if (a.size() > b.size())  return a;
	else if (a.size() < b.size()) return b;
	else
	{
		for (int i = a.size() - 1; i >= 0; i--)
		{
			if (a[i] > b[i])
				return a;
			else if (a[i] < b[i])
				return b;
		}
		return a;
	}
}