高精度重载运算符

Posted 2020-12-30 sun915

高精度重载运算符模板

struct bign{
        //存在性定义
    int len, s[2010];
        //初始化
    bign () {memset(s, 0, sizeof(s)), len = 1;} 
    bign (int num){ *this = num;} 
    bign (char *num) {*this = num;} 
    bign (const char *num){ *this = num;}
        //重载 =
    bign operator = (int num) {
        char s[2010];
        sprintf(s, "%d", num);
        *this = s;
        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;
    }
    void clear(){ while(len > 1 && !s[len - 1]) len --;}
    bign operator = (const char *num) {
        len = strlen(num);
        for (int i = 0; i < len; i++)
            s[i] = num[len - i - 1] - ‘0‘;
            return *this;
    }
        //四则运算的重载
    bign operator + (const bign &b) const {
        bign c = *this;
        int i = 0;
        for (int i = 0; i < b.len; i++) {
            c.s[i] += b.s[i];
            if (c.s[i] > 9) c.s[i] %= 10, c.s[i + 1] ++;
        }
        while (c.s[i] > 9) c.s[i++] %= 10, c.s[i]++;
        c.len = max(len, b.len);
        if (c.s[i] && c.len <= i) c.len = i + 1;
        return c;
    }
    bign operator - (const bign &b) const {
        bign c = *this;
        int i = 0;
        for (int i = 0; i < b.len; ++i) {
            c.s[i] -= b.s[i];
            if (c.s[i] < 0) c.s[i] += 10, c.s[i + 1] --;
        }
        while (c.s[i] < 0) c.s[i++] += 10, c.s[i] --;
        c.clear();
        return c;
    }
    bign operator * (const bign &b) const {
        int i, j;
        bign c;
        c.len = len + b.len;
        for (int j = 0; j < b.len; j++)
            for (int i = 0; i < len; i++)
                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.clear();
        return c;
    }
    bign operator / (const bign &b) const {
        bign c = *this, a = 0;
        int i, j;
        for (i = len - 1; i >= 0; i --) {
            a = a * 10 + s[i];
            for (j = 0; j < 10; j++) 
                if (a < b * (j + 1)) break;
            c.s[i] = j;
            a = a - b * j;
        }
        c.clear();
        return c;
    }
    bign operator % (const bign &b) const {
        int i, j;
        bign c = 0;
        for (int i = len - 1; i >= 0; i--) {
            c = c * 10 + s[i];
            for (int i = len - 1; i >= 0; --i){
                if (c < b * (j + 1)) break;
                c = c - b * j;
            }
        }
        return c;
    }
    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 (s[i] != b.s[i]) return s[i] < b.s[i];
        return false;
    }
        //比较符重载
    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) || (*this < b);}
};
        //流输入输出重载
istream& operator >> (istream &in, bign &x) {
    string s;
    in >> s;
    x = s.c_str();
    return in;
}

ostream& operator << (ostream &out, bign &x){
    out << x.str();
    return out;
}