高精度

/*
1.高精度加法
2.高精度减法 
3.高精度乘法 
4.高精度除以低精度 
5.高精度对低精度的取余 

必要时可以将全局的int替换成long long.除了main函数的返回值int
用到除法和取余的时候可能需要把全局的int替换成long long 
*/
#include <cstdio>
#include <iostream>
#include <cstring>
#include <climits>
using namespace std;

#define maxn 30000

struct bign
{
    int len, s[maxn];

    bign()
    {
        memset(s, 0, sizeof(s));
        len = 1;
    }
    
    bign(int num)
    {
        *this = num;
    }
    
    bign(const char* num) 
    {
        *this = num;
    }
    
    bign operator = (int num) 
    {
        char s[maxn];
        sprintf(s, "%d", num);
        *this = s;
        return *this;
    }
    
    bign operator = (const char* num) 
    {
        len = strlen(num);
        for (int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';
        return *this;
    }
    
    string str() const 
    {
        string res = "";
        for (int i = 0; i < len; i++) res = (char)(s[i] + '0') + res;
        if (res == "") res = "0";
        return res;
    }
    
    /*去除前导0*/ 
    void clean() 
    {
        while(len > 1 && !s[len-1]) len--;
    }
    
    /*高精度的加法*/
    bign operator + (const bign& b) const
    {
        bign c;
        c.len = 0;
        for (int i = 0, g = 0; g || i < max(len, b.len); i++) 
        {
            int x = g;
            if (i < len) x += s[i];
            if (i < b.len) x += b.s[i];
            c.s[c.len++] = x % 10;
            g = x / 10;
        }
        return c;
    }
    
    /*高精度的减法*/ 
    bign operator - (const bign& b) 
    {
        bign c; c.len = 0;
        for (int i = 0, g = 0; i < len; i++) 
        {
            int x = s[i] - g;
            if (i < b.len) x -= b.s[i];
            if (x >= 0) 
                g = 0;
            else 
            {
                g = 1;
                x += 10;
            }
            c.s[c.len++] = x;
        }
        c.clean();
        return c;
    }
    
    /*高精度的乘法*/ 
    bign operator * (const bign& b) 
    {
        bign c; c.len = len + b.len;
        for (int i = 0; i < len; i++)
        for (int j = 0; j < b.len; j++)
        c.s[i+j] += s[i] * b.s[j];
        for (int i = 0; i < c.len-1; i++)
        {
            c.s[i+1] += c.s[i] / 10;
            c.s[i] %= 10;
        }
        c.clean();
        return c;
    } 
    
    /*高精度除以低精度*/    /*用到除法和取余的时候可能需要把全局的int替换成long long*/
    bign operator / (long long b) const
    {
        assert(b > 0);
        bign c;c.len = len;
        for (long long i = len-1, g = 0; i >= 0; --i)
        {
            long long x = 10*g+s[i];    //这里可能会超过int 故用long long 
            
            c.s[i] = x/b;                //这里可能会超过int
            
            g = x-c.s[i]*b;                //这里可能会超过int 
        }
        c.clean();
        return c;
    }
    
    /*高精度对低精度取余*/    /*用到除法和取余的时候可能需要把全局的int替换成long long*/
    bign operator % (long long b)
    {
        assert(b > 0);
        bign d = b;
        bign c = *this-*this/b*d;
        return c;
    }
    
    bool operator < (const bign& b) const
    {
        if (len != b.len) return len < b.len;
        for (int i = len-1; i >= 0; i--)
        if (s[i] != b.s[i]) return s[i] < b.s[i];
        return false;
    }
    
    bool operator > (const bign& b) const
    {
        return b < *this;
    }
    
    bool operator <= (const bign& b) 
    {
        return !(b > *this);
    }
    
    bool operator >= (const bign& b) 
    {
        return !(b < *this);
    }
    
    bool operator == (const bign& b) 
    {
        return !(b < *this) && !(*this < b);
    }
    
    bool operator != (const bign& b) 
    {
        return (b < *this) || (*this < b);
    }
    
    bign operator += (const bign& b) 
    {
        *this = *this + b;
        return *this;
    }
};

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;
}