蓝桥杯算法提高 高精度除高精度

题目链接

问题描述
　　给定a, b，求a/b。
输入格式
　　输入两行，分别包含一个整数。
输出格式
　　输出一行，为a/b的商。
样例输入
62349
64
样例输出
974
数据规模和约定
　　1<=a<=10^10000, 1<=b<=10^10000。

代码1：（采用ceil() 向上取整函数 和floor() 向下取整函数）

#include
using namespace std;
int main()
{
     
	double a, b;
	cin >> a >> b;
	double c = a / b; 
	printf("%.f", floor(c));
	return 0;
}

代码2：（运算符重载）

#include
using namespace std;
const int maxn = 1000;
struct bign{
     
    int d[maxn], len;
	void clean() {
      while(len > 1 && !d[len-1]) len--; }
    bign() 			{
      memset(d, 0, sizeof(d)); len = 1; }
    bign(int num) 	{
      *this = num; } 
	bign(char* num) {
      *this = num; }
    bign operator = (const char* num){
     
        memset(d, 0, sizeof(d)); len = strlen(num);
        for(int i = 0; i < len; i++) d[i] = num[len-1-i] - '0';
        clean();
		return *this;
    }
    bign operator = (int num){
     
        char s[20]; sprintf(s, "%d", num);
        *this = s;
		return *this;
    }
    bign operator + (const bign& b){
     
        bign c = *this; int i;
        for (i = 0; i < b.len; i++){
     
        	c.d[i] += b.d[i];
        	if (c.d[i] > 9) c.d[i]%=10, c.d[i+1]++;
		}
		while (c.d[i] > 9) c.d[i++]%=10, c.d[i]++;
		c.len = max(len, b.len);
		if (c.d[i] && c.len <= i) c.len = i+1;
        return c;
    }
    bign operator - (const bign& b){
     
        bign c = *this; int i;
        for (i = 0; i < b.len; i++){
     
        	c.d[i] -= b.d[i];
        	if (c.d[i] < 0) c.d[i]+=10, c.d[i+1]--;
		}
		while (c.d[i] < 0) c.d[i++]+=10, c.d[i]--;
		c.clean();
		return c;
    }
    bign operator * (const bign& b)const{
     
        int i, j; bign c; c.len = len + b.len; 
        for(j = 0; j < b.len; j++) for(i = 0; i < len; i++) 
			c.d[i+j] += d[i] * b.d[j];
        for(i = 0; i < c.len-1; i++)
            c.d[i+1] += c.d[i]/10, c.d[i] %= 10;
        c.clean();
		return c;
    }
    bign operator / (const bign& b){
     
    	int i, j;
		bign c = *this, a = 0;
    	for (i = len - 1; i >= 0; i--)
    	{
     
    		a = a*10 + d[i];
    		for (j = 0; j < 10; j++) if (a < b*(j+1)) break;
    		c.d[i] = j;
    		a = a - b*j;
    	}
    	c.clean();
    	return c;
    }
    bign operator % (const bign& b){
     
    	int i, j;
		bign a = 0;
    	for (i = len - 1; i >= 0; i--)
    	{
     
    		a = a*10 + d[i];
    		for (j = 0; j < 10; j++) if (a < b*(j+1)) break;
    		a = a - b*j;
    	}
    	return a;
    }
	bign operator += (const bign& b){
     
        *this = *this + b;
        return *this;
    }
    bool operator <(const bign& b) const{
     
        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) const{
     return b < *this;}
    bool operator<=(const bign& b) const{
     return !(b < *this);}
    bool operator>=(const bign& b) const{
     return !(*this < b);}
    bool operator!=(const bign& b) const{
     return b < *this || *this < b;}
    bool operator==(const bign& b) const{
     return !(b < *this) && !(b > *this);}
    string str() const{
     
        char s[maxn]={
     };
        for(int i = 0; i < len; i++) s[len-1-i] = d[i]+'0';
        return s;
    }
};
istream& operator >> (istream& in, bign& x)
{
     
    string s;
    in >> s;
    x = s.c_str();
    return in;
}
ostream& operator << (ostream& out, const bign& x)
{
     
    out << x.str();
    return out;
}
int main()
{
     
	bign a, b, c;
	while (cin >> a >> b)
	{
     
		c = a / b;
		cout << c << endl;
	}
	return 0;
}