大整数/高精度（暂不支持除法）

• 前言
• 使用方法
• 实现代码
• 部分习题

前言

有些题的数据范围很大，以至于开了long long还是会爆…
为此我写了大整数，大致思想是开long long数组存储每九位数字（多了在处理乘法的时候会爆），然后用一个布尔值来判断正负。
目前只支持赋值，加法，减法，乘法，比较。
感觉除法太难，暂时没想出来怎么写，可能以后用到了会补上吧（orz）。

使用方法

声明：

bignum a,b,c;

初始化：

a.clear();
b.clear();

赋值：

a = 123; //赋值int类型
a = "123"; //赋值字符串类型(char *)
a = b; //赋值bignum类型

运算（支持 + - *）：

c = a + b; //bignum + bignum
c = a + 1; //bignum + int
c += a; //bignum += bignum
c += 1; //bignum += int
c = -a; //bignum的负数

比较：

if (a < c)
{
	//...
}
if (a >= c)
{
	//...
}

实现代码

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
const int maxn = 100;
const long long max_digit = 1e9;
struct big_num
{
	long long digits[maxn];
	bool minus = 0;
	void clear(void)
	{
		this->minus = 0;
		memset(this->digits, 0, sizeof(this->digits));
	}
	big_num operator=(const big_num b)
	{
		for (int i = 0; i < maxn; i++)
			this->digits[i] = b.digits[i];
		this->minus = b.minus;
		return *this;
	}
	big_num operator=(const int b)
	{
		this->clear();
		this->minus = b < 0;
		if (b < 0)
			this->digits[0] = -b;
		else
			this->digits[0] = b;
		return *this;
	}
	big_num operator=(const char* b)
	{
		int len = strlen(b), st = 0;
		this->clear();
		if (b[0] == '-')
		{
			st = 1;
			this->minus = 1;
		}
		for (int i = st; i < len; i++)
		{
			this->digits[(len - i - 1 + st) / 9] *= 10;
			this->digits[(len - i - 1 + st) / 9] += b[i] - '0';
		}
		return *this;
	}
	bool operator<(const big_num b)
	{
		if (this->minus != b.minus) return this->minus > b.minus;
		for (int i = 0; i < maxn; i++)
			if (this->digits[i] != b.digits[i])
				return (this->digits[i] < b.digits[i]) ^ minus;
		return false;
	}
	bool operator>(const big_num b)
	{
		if (this->minus != b.minus) return this->minus < b.minus;
		for (int i = 0; i < maxn; i++)
			if (this->digits[i] != b.digits[i])
			{
				cout << i;
				return (this->digits[i] > b.digits[i]) ^ this->minus;
			}
		return false;
	}
	bool operator==(const big_num b)
	{
		if (this->minus != b.minus) return false;
		for (int i = 0; i < maxn; i++)
			if (this->digits[i] != b.digits[i])
				return false;
		return true;
	}
	bool operator<=(const big_num b)
	{
		return *this < b || *this == b;
	}
	bool operator>=(const big_num b)
	{
		return *this > b || *this == b;
	}
	const big_num operator-() const
	{
		big_num temp = *this;
		temp.minus ^= 1;
		return temp;
	}
	big_num operator+(const big_num b)const
	{
		big_num temp = *this;
		if (temp.minus == b.minus)
		{
			for (int i = 0; i < maxn; i++)
			{
				temp.digits[i] += b.digits[i];
			}
			for (int i = 1; i < maxn; i++)
			{
				temp.digits[i] += temp.digits[i - 1] / max_digit;
				temp.digits[i - 1] %= max_digit;
			}
			return temp;
		}
		else
		{
			return temp - (-b);
		}
	}
	big_num operator+(const int b)const
	{
		big_num temp = *this;
		big_num b2;
		b2 = b;
		return temp + b2;
	}
	big_num operator-(const big_num b) const
	{
		big_num temp = *this;
		if (temp.minus == b.minus)
		{
			if (temp >= b && temp.minus == 0 || temp <= b && temp.minus == 1)
			{
				for (int i = 0; i < maxn; i++)
				{
					temp.digits[i] -= b.digits[i];
				}
				for (int i = 1; i < maxn; i++)
				{
					if (temp.digits[i - 1] < 0)
					{
						temp.digits[i - 1] += max_digit;
						temp.digits[i]--;
					}
				}
				return temp;
			}
			else
			{
				return -(b - temp);
			}
		}
		else
		{
			return temp + (-b);
		}
	}
	big_num operator-(const int b) const
	{
		big_num temp = *this;
		big_num b2;
		b2 = abs(b);
		b2.minus = b < 0;
		return temp - b2;
	}
	big_num operator*(const big_num b)const
	{
		big_num temp;
		temp.clear();
		temp.minus = this->minus ^ b.minus;
		for (int i = maxn - 1; i >= 0; i--)
		{
			for (int i2 = maxn - i - 1; i2 >= 0; i2--)
			{
				temp.digits[i + i2] += this->digits[i] * b.digits[i2];
				if (temp.digits[i + i2] >= max_digit && i + i2 + 1 < maxn)
				{
					temp.digits[i + i2 + 1] += temp.digits[i + i2] / max_digit;
				}
				temp.digits[i + i2] %= max_digit;
			}
		}
		for (int i = 1; i < maxn; i++)
		{
			temp.digits[i] += temp.digits[i - 1] / max_digit;
			temp.digits[i - 1] %= max_digit;
		}
		return temp;
	}
	big_num operator*(const int b)const
	{
		big_num temp = *this;
		temp.minus ^= (b < 0);
		for (int i = 0; i < maxn; i++)
		{
			temp.digits[i] *= b;
		}
		for (int i = 1; i < maxn; i++)
		{
			temp.digits[i] += temp.digits[i - 1] / max_digit;
			temp.digits[i - 1] %= max_digit;
		}
		return temp;
	}
	big_num operator+=(const big_num b)
	{
		*this = *this + b;
		return *this;
	}
	big_num operator-=(const big_num b)
	{
		*this = *this - b;
		return *this;
	}
	big_num operator*=(const big_num b)
	{
		*this = *this * b;
		return *this;
	}
	big_num operator+=(const int b)
	{
		*this = *this + b;
		return *this;
	}
	big_num operator-=(const int b)
	{
		*this = *this - b;
		return *this;
	}
	big_num operator*=(const int b)
	{
		*this = *this * b;
		return *this;
	}
	void show(void)
	{
		bool is_begun = 0;
		for (int i = maxn - 1; i >= 0; i--)
		{

			if (is_begun)
				printf("%0.9lld", this->digits[i]);
			if (this->digits[i] != 0 && !is_begun)
			{
				if (this->minus == 1)
					putchar('-');
				is_begun = 1;
				printf("%lld", this->digits[i]);
			}
		}
		if (is_begun == 0) putchar('0');
		putchar('\n');
	}
};

部分习题

洛谷1005 - 矩阵取数游戏
洛谷1797 - 克鲁斯的加减法