大整数高精度运算C++实现

定义

//定义和初始化
struct bign
{
	int d[1000];
	int len;
	bign() {
		memset(d, 0, sizeof(d));
		len = 0;
	}
};
//字符串转成结构体
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

//大数加法
bign add(bign a, bign b) {
	bign c;
	int carry = 0;
	for (int i = 0; i < a.len || 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

//大数减法
bign sub(bign a, bign b) {
	bign c;
	for (int i = 0; i < a.len || b.len; i++) {
		if (a.d[i] < b.d[i]) {//如果不够减
			a.d[i + 1]--;
			a.d[i] += 10;
		}
		c.d[c.len++] = a.d[i] - b.d[i];
	}
	//去除高位的0
	while (c.len - 1 >= 1 && c.d[c.len - 1] == 0) {
		c.len--;
	}
}

乘法

//乘法,高精度数a除以int型b
bign multi(bign a, int b) {
	bign c;
	int carry = 0;//进位
	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;//乘法的进位可能不止一位
		carry /= 10;
	}
	return c;
}

除法 

//除法
//除数a,被除数b,余数r
bign divide(bign a, int b, int& r) {
	bign c;
	c.len = a.len;//被除数的每一位和商的每一位是对应的，因此先令长度相等
	for (int i = a.len - 1; i >= 0; i--) {
		r = r*10 + a.d[i];
		if (r < b) c.d[i] = 0;//不够除，该位为0
		else {
			c.d[i] = r/b;//商
			r = r%b;//获得新的余数
		}
	}
	while (c.len - 1 >= 1 && c.d[len - 1] == 0) {
		c.len--;//去掉先导零
	}
	return c;
}

逆向输出

//逆向输出，因为是逆向存储的
void print(bign a) {
	for (int i = a.len-1; i >= 0; i--) {
		printf("%d", a.d[i]);
	}
}