大整数的四则运算

int 表示的数据范围有限，当涉及大整数的运算时候，我们就只能借助数组来存储，借助数组就依靠手工加减乘除，就小学学的那种来运算，然后一个个存入。
先来个结构体bign，含有一个数组和数组长度。我们用字符串读入的数字可以存入bign。头尾颠倒存入，因为模拟手工加减乘这三个都是从低位开始的，颠倒后方便操作

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;
}

加法

bign add(bign a, bign b)
{
	bign c;
	int carry = 0;
	for (int i = 0; i < a.len || i < b.len; i++)
	{
		int temp = a.d[i] + b.d[i] + carry;
		c.d[c.len++] = temp % 10;
		carry = temp / 10;
	}
	if (carry != 0)
	{
		c.d[c.len++] = carry;
	}
	return c;
}

减法

记得去掉前面的多余的0。

bign sub(bign a, bign b)
{
	bign c;
	for (int i = 0; i < a.len || i < b.len; i++)
	{
		if (a.d[i] < b.d[i])
		{
			a.d[i + 1]--;
			a.d[i] += 10;
		}
		c.d[c.len++] = a.d[i] - b.d[i];
	}
	while (c.len - 1 >= 1 && c.d[c.len - 1] == 0)
		c.len--;  //删除除个位上的0
	return c;
}

乘法（高精度与低精度）

乘法与手工乘法有些不同，他是把被乘数当成了一个整体，所得个位数做该位结果，高位部分作为新的进位。
然后乘法进位可能不止一位了。用while
前面bign 后面int

bign multi(bign a, int b)
{
	bign c;
	int carry = 0;
	for (int i = 0; i < a.len; i++)
	{
		int temp = a.d[i] * b + carry;
		c.d[c.len++] = temp % 10;
		carry = temp / 10;
	}
	while (carry != 0)
	{
		c.d[c.len++] = carry % 10;
		carry /= 10;
	}
	return c;
}

除法（高精度与低精度）

是余数和除数相除，注意从尾部开始计算，也要除去多余的0
前面bign 后面int

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;
}

全部代码奉上

//找质数，找能除的质数
#include<iostream>
#include<cstring>
#include<algorithm>
#include<string>
#include<time.h>
#include<math.h>
using namespace std;
typedef long long ll;
struct bign {
	int d[1000];
	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;
}
int compare(bign a, bign b)
{
	if (a.len > b.len) return 1;
	else if (a.len < b.len) return -1;
	else
	{
		for (int i = a.len - 1; i >= 0; i--)
		{
			if (a.d[i] > b.d[i]) return 1;
			else if (a.d[i] < b.d[i]) return -1;
		}
		return 0;
	}
}
bign add(bign a, bign b)
{
	bign c;
	int carry = 0;
	for (int i = 0; i < a.len || i < b.len; i++)
	{
		int temp = a.d[i] + b.d[i] + carry;
		c.d[c.len++] = temp % 10;
		carry = temp / 10;
	}
	if (carry != 0)
	{
		c.d[c.len++] = carry;
	}
	return c;
}
bign sub(bign a, bign b)
{
	bign c;
	for (int i = 0; i < a.len || i < b.len; i++)
	{
		if (a.d[i] < b.d[i])
		{
			a.d[i + 1]--;
			a.d[i] += 10;
		}
		c.d[c.len++] = a.d[i] - b.d[i];
	}
	while (c.len - 1 >= 1 && c.d[c.len - 1] == 0)
		c.len--;  //删除除个位上的0
	return c;
}
//
bign multi(bign a, int b)
{
	bign c;
	int carry = 0;
	for (int i = 0; i < a.len; i++)
	{
		int temp = a.d[i] * b + carry;
		c.d[c.len++] = temp % 10;
		carry = temp / 10;
	}
	while (carry != 0)
	{
		c.d[c.len++] = carry % 10;
		carry /= 10;
	}
	return c;
}
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;
}
void print(bign a)
{
	for (int i = a.len - 1; i >= 0; i--)
	{
		cout << a.d[i];
	}
}
int main()
{
	//char str1[1010], str2[1010];
	//int b, r = 0;
	//cin >> str1 >> b;
	//bign a = change(str1);
	//print(divide(a, b, r));
	//cout << " " << r << endl;
	char str3[1010], str4[1010];
	cin >> str3 >> str4;
	int q3;
	bign q1 = change(str3);
	bign q2 = change(str4);
	cin >> q3;
	print(add(q1, q2));
	cout << endl;
	print(sub(q1, q2));
	cout << endl;
	print(multi(q1, q3));
	cout << endl;
}