Bigint Library by Jan vai

Thanks to Jane Alam Jan vai for this, It will be a great help... :)

/*
    Author       :    Jan
    Problem Name :    Big int for contest
    Algorithm    :
    Complexity   :
*/

#include <cstdio>
#include <string>
#include <algorithm>
using namespace std;

struct Bigint {
    string a;
    int sign;

    Bigint() {}
    Bigint( string b ) { (*this) = b; }
    int size() { return a.size(); }
    Bigint inverseSign() { sign *= -1; return (*this); }
    Bigint normalize( int newSign ) {
        sign = newSign;
        for( int i = a.size() - 1; i > 0 && a[i] == '0'; i-- ) a.erase(a.begin() + i);
        if( a.size() == 1 && a[0] == '0' ) sign = 1;
        return (*this);
    }
    void operator = ( string b ) {
        a = b[0] == '-' ? b.substr(1) : b;
        reverse( a.begin(), a.end() );
        this->normalize( b[0] == '-' ? -1 : 1 );
    }
    bool operator < ( const Bigint &b ) const {
        if( a.size() != b.a.size() ) return a.size() < b.a.size();
        for( int i = a.size() - 1; i >= 0; i-- ) if( a[i] != b.a[i] ) return a[i] < b.a[i];
        return false;
    }
    Bigint operator + ( Bigint b ) {
        if( sign != b.sign ) return (*this) - b.inverseSign();
        Bigint c;
        for( int i = 0, carry = 0; i < (int)a.size() || i < (int)b.size() || carry; i++ ) {
            carry += (i < (int)a.size() ? a[i] - 48 : 0) + (i < (int)b.a.size() ? b.a[i] - 48 : 0);
            c.a += (carry % 10 + 48);
            carry /= 10;
        }
        return c.normalize(sign);
    }
    Bigint operator - ( Bigint b ) {
        if( sign != b.sign ) return (*this) + b.inverseSign();
        if( (*this) < b ) return (b - (*this)).inverseSign();
        Bigint c;
        for( int i = 0, borrow = 0; i < (int)a.size(); i++ ) {
            borrow = a[i] - borrow - (i < b.size() ? b.a[i] : 48);
            c.a += borrow >= 0 ? borrow + 48 : borrow + 58;
            borrow = borrow >= 0 ? 0 : 1;
        }
        return c.normalize(sign);
    }
    Bigint operator * ( Bigint b ) {
        Bigint c("0");
        for( int i = 0, k = a[i]; i < (int)a.size(); i++, k = a[i] ) {
            while(k-- - 48) c = c + b;
            b.a.insert(b.a.begin(), '0');
        }
        return c.normalize(sign * b.sign);
    }
    Bigint operator / ( Bigint b ) {
        if( b.size() == 1 && b.a[0] == '0' ) b.a[0] /= ( b.a[0] - 48 ) ;
        Bigint c("0"), d;
        for( int j = 0; j < (int)a.size(); j++ ) d.a += "0";
        int dSign = sign * b.sign; b.sign = 1;
        for( int i = a.size() - 1; i >= 0; i-- ) {
            c.a.insert( c.a.begin(), '0');
            c = c + a.substr( i, 1 );
            while( !( c < b ) ) c = c - b, d.a[i]++;
        }
        return d.normalize(dSign);
    }
    Bigint operator % ( Bigint b ) {
        if( b.size() == 1 && b.a[0] == '0' ) b.a[0] /= ( b.a[0] - 48 ) ;
        Bigint c("0");
        int cSign = sign * b.sign; b.sign = 1;
        for( int i = a.size() - 1; i >= 0; i-- ) {
            c.a.insert( c.a.begin(), '0');
            c = c + a.substr( i, 1 );
            while( !( c < b ) ) c = c - b;
        }
        return c.normalize(cSign);
    }
    void print() {
        if( sign == -1 ) putchar('-');
        for( int i = a.size() - 1; i >= 0; i-- ) putchar(a[i]);
    }
};

int main() {
    Bigint a, b, c;
    a = "511";
    b = "10";

    c = a + b;
    c.print();
    putchar('\n');

    c = a - b;
    c.print();
    putchar('\n');

    c = a * b;
    c.print();
    putchar('\n');

    c = a / b;
    c.print();
    putchar('\n');

    c = a % b;
    c.print();
    putchar('\n');

    return 0;
}

Hope this helps my friends as well. Actually many of us are afraid of big integers, and also the above code may look hard, but this is actually what we do when performing these operations in real life... Use with caution, not a very first algorithm. Do not attempt to use *, /, % if numbers are 10⁴ or more digits long.