Handling large numbers in C++?

What is the best way to handle large numeric inputs in C++ (for example 10^100)?

For algorithms I usually switch over to ruby and I sometimes use strings.

Any other good methods?

Solution 1:

It sounds like you're looking for a way to enter Arbitrary Precision numbers. here are two libraries you could use: GMP and MAPM

Solution 2:

Check out The Large Integer Case Study in C++.pdf by Owen Astrachan. I found this file extremely useful with detail introduction and code implementation. It doesn't use any 3rd-party library. I have used this to handle huge numbers (as long as you have enough memory to store vector<char>) with no problems.

Idea: It implements an arbitrary precision integer class by storing big int in a vector<char>.

vector<char> myDigits; // stores all digits of number

Then all operations related to the big int, including <<, >>, +, -, *, ==, <, !=, >, etc., can be done based on operations on this char array.

Taste of the code: Here is the header file, you can find its cpp with codes in the pdf file.

#include <iostream>
#include <string> // for strings
#include <vector> // for sequence of digits
using namespace std;

class BigInt
{
public:
    BigInt(); // default constructor, value = 0
    BigInt(int); // assign an integer value
    BigInt(const string &); // assign a string
    // may need these in alternative implementation
    // BigInt(const BigInt &); // copy constructor
    // ~BigInt(); // destructor
    // const BigInt & operator = (const BigInt &);
    // assignment operator
    // operators: arithmetic, relational
    const BigInt & operator += (const BigInt &);
    const BigInt & operator -= (const BigInt &);
    const BigInt & operator *= (const BigInt &);
    const BigInt & operator *= (int num);
    string ToString() const; // convert to string
    int ToInt() const; // convert to int
    double ToDouble() const; // convert to double
    // facilitate operators ==, <, << without friends
    bool Equal(const BigInt & rhs) const;
    bool LessThan(const BigInt & rhs) const;
    void Print(ostream & os) const;
private:
    // other helper functions
    bool IsNegative() const; // return true iff number is negative
    bool IsPositive() const; // return true iff number is positive
    int NumDigits() const; // return # digits in number
    int GetDigit(int k) const;
    void AddSigDigit(int value);
    void ChangeDigit(int k, int value);
    void Normalize();
    // private state/instance variables
    enum Sign{positive,negative};
    Sign mySign; // is number positive or negative
    vector<char> myDigits; // stores all digits of number
    int myNumDigits; // stores # of digits of number
};

// free functions
ostream & operator <<(ostream &, const BigInt &);
istream & operator >>(istream &, BigInt &);
BigInt operator +(const BigInt & lhs, const BigInt & rhs);
BigInt operator -(const BigInt & lhs, const BigInt & rhs);
BigInt operator *(const BigInt & lhs, const BigInt & rhs);
BigInt operator *(const BigInt & lhs, int num);
BigInt operator *(int num, const BigInt & rhs);
bool operator == (const BigInt & lhs, const BigInt & rhs);
bool operator < (const BigInt & lhs, const BigInt & rhs);
bool operator != (const BigInt & lhs, const BigInt & rhs);
bool operator > (const BigInt & lhs, const BigInt & rhs);
bool operator >= (const BigInt & lhs, const BigInt & rhs);
bool operator <= (const BigInt & lhs, const BigInt & rhs);

Solution 3:

Are you looking for how to perform operations on the large inputs you receive? There is a big integer C++ library (similar to Java) that allows you to perform arithmetic operations...

Solution 4:

If you wish to make your own code for the purpose try using strings to store big numbers... you can then create basic ops like + - / * on them... for example -

#include <iostream>

using namespace std;

string add (string &s1, string &s2){
    int carry=0,sum,i;

    string  min=s1,
    max=s2,
    result = "";

    if (s1.length()>s2.length()){
        max = s1;
        min = s2;
    } else {
        max = s2;
        min = s1;
    }

    for (i = min.length()-1; i>=0; i--){
        sum = min[i] + max[i + max.length() - min.length()] + carry - 2*'0';

        carry = sum/10;
        sum %=10;

        result = (char)(sum + '0') + result;
    }

    i = max.length() - min.length()-1;

    while (i>=0){
        sum = max[i] + carry - '0';
        carry = sum/10;
        sum%=10;

        result = (char)(sum + '0') + result;
        i--;
    }

    if (carry!=0){
        result = (char)(carry + '0') + result;
    }       

    return result;
}

int main (){
    string a,b;

    cin >> a >> b;

    cout << add (a,b)<<endl;

    return 0;
}

Solution 5:

assuming you are talking about inputting numbers, double precision would get you up to 1.7976931348623157 x 10^308