【模板】高精度

ACM模板

#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
struct bign
{
	int d[200010], len;
	bign() { memset(d, 0, sizeof d); len = 1; }
	bign(int num) { *this = num; }
	bign(char* num) { *this = num; }
	bign operator=(const char* num)
	{
		len = strlen(num);
		for (int i = 0; i < len; i++) d[i] = num[len - i - 1] - '0';
		return *this;
	}
	bign operator=(int num)
	{
		char c[1010];
		sprintf(c, "%d", num);
		*this = c;
		return *this;
	}
	void clear()
	{
		while (len > 1 && !d[len - 1]) len--;
	}
	bign operator+(const bign& b)
	{
		bign c; c.len = 0;
		for (int i = 0, t = 0; t || i < len || i < b.len; i++)
		{
			if (i < len) t += d[i];
			if (i < b.len) t += b.d[i];
			c.d[c.len++] = t % 10;
			t /= 10;
		}
		return c;
	}
	bign operator-(const bign& b)
	{
		bign c; c.len = 0;
		for (int i = 0, t = 0; i < len; i++)
		{
			t += d[i];
			if (i < b.len) t -= b.d[i];
			c.d[c.len++] = (t + 10) % 10;
			if (t >= 0) t = 0;
			else t = -1;
		}
		c.clear();
		return c;
	}
	bign operator*(const bign& b)
	{
		bign c; c.len = len + b.len;
		for (int i = 0; i < len; i++) for (int j = 0; j < b.len; j++) c.d[i + j] += d[i] * b.d[j];
		for (int i = 0; i < c.len - 1; i++) c.d[i + 1] += c.d[i] / 10, c.d[i] %= 10;
		c.clear();
		return c;
	}
	bool operator < (const bign& b)
	{
		if (len != b.len) return len < b.len;
		for (int i = len - 1; i >= 0; i--)
			if (d[i] != b.d[i]) return d[i] < b.d[i];
		return false;
	}
	bool operator <= (const bign& b)
	{
		if (len != b.len) return len < b.len;
		for (int i = len - 1; i >= 0; i--)
			if (d[i] != b.d[i]) return d[i] < b.d[i];
		return true;
	}
	bool operator > (const bign& b)
	{
		if (len != b.len) return len > b.len;
		for (int i = len - 1; i >= 0; i--)
			if (d[i] != b.d[i]) return d[i] > b.d[i];
		return false;
	}
	bool operator >= (const bign& b)
	{
		if (len != b.len) return len > b.len;
		for (int i = len - 1; i >= 0; i--)
			if (d[i] != b.d[i]) return d[i] > b.d[i];
		return true;
	}
	bign operator+=(const bign& b)
	{
		*this = *this + b;
		return *this;
	}
	bign operator-=(const bign& b)
	{
		*this = *this - b;
		return *this;
	}
	bign operator*=(const bign& b)
	{
		*this = *this * b;
		return *this;
	}
	void print()
	{
		for (int i = len - 1; i >= 0; i--) std::cout << d[i];
		cout << '\n';
	}
	string str()
	{
		string res = "";
		for (int i = 0; i < len; i++) res = (char)(d[i] + '0') + res;
		return res;
	}
};
istream& operator >>(istream& in, bign& x)
{
	string s;
	in >> s;
	x = s.c_str();
	return in;
}
ostream& operator <<(ostream& out, bign& x)
{
	out << x.str();
	return out;
}
int main()
{
    bign a,b,c;
    cin>>a>>b;
    c=a+b;
    cout<<c<<'\n';
}

标签：return,高精度,int,len,bign,operator,const,模板
来源： https://blog.csdn.net/Fighting_Peter/article/details/113826132