2019.1.27小白之路：大整数的四则运算

#include <stdio.h>
#include <string.h>

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;
}

bign add(bign a, bign b) {  //高精度加法
	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) {  //高精度减法
	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--;
	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--)
		printf("%d", a.d[i]);
}

int main() {
	char str1[1000], str2[1000];
	scanf("%s%s", str1, str2);
	bign a = change(str1);
	bign b = change(str2);
	print(sub(a, b));
	return 0;
}