CareerCup

Better way to store long integer data type data is array. So that we can use this number for any arithmetic operation. eg) if the number is 123456789, keep the data in the form of (1,2,3,4,5,6,7,8,9)

We need max 4 bits to represent digits from 1 to 9.
Use a structure with bit-field.

struct b
{
        unsigned int num:4;
};

Use an array of such structure where each slot will be handling each digit.

This will conserve memory.

vector<int>, first element is the least significant part, following elements are added dynamically. It is relatively easy to implement most arithmetic operations

#include<iostream>
using namespace std;

struct DT{
    unsigned int d:4;
};

int main()
{

    int size,digit;
    DT *dt;

    cout<<"Enter number of digits to store : ";
    cin>>size;
    dt = new DT[size];

    for(int i = 0; i<size; i++)
    {
        switch(i+1)
        {
            case 1 :
                cout<<"Enter "<<i+1<<"st digit from left : ";
                break;
            case 2 :
                cout<<"Enter "<<i+1<<"nd digit from left : ";
                break;
            case 3 :
                cout<<"Enter "<<i+1<<"rd digit from left : ";
                break;
            default :
                  cout<<"Enter "<<i+1<<"th digit from left : ";
        }

        while(true)
        {
            cin>>digit;
            if(digit>=0 && digit<10)
                break;
            cout<<"Renter this digit please  :";
        }
        dt[size-1-i].d=digit;
    }

    cout<<"Your number is : ";
    for(int i = 0; i<size; i++)
        cout<<dt[size-1-i].d;

    return 0;
}

Use Strings to represent long data.

use a 3-D array

int crazyLongData [nth_number_of_looooooong_data][(max_digits)/(sizeof(int)*8)][max_digits]

Anonymous on July 15, 2012 is = softy !

ArrayList in little endian format

I think the best answer would be 'file' . Open a file and write to it the entire input and we can access any digit through the file pointer

struct data{
char ch[100]; // this will store 200 Digits
struct data *next;
};

if digit in 100 Multiple
overhead of memory= (200+ 2*4)/ 200= 1.04%

what about trie?

Steps:
1. Count number of digit in the number(i.e. length of the number), Let it be N.
2. Create 10 Bit vectors(i.e. for each num "0, 1, 2, 3, 4,5 6, 7, 8, 9") of lenth N.
    BOOLEAN VECTORS[10][N] = {0};
3. For each number(i.e. "0, 1, 2, 3, 4,5 6, 7, 8, 9") in Large number, create boolean vector.
    for(int j=0; j<N; j++){
        m = MSB(LARGE_NUMBER);
		remove m from LARGE_NUMBER;
        VECTORS[m][i] = 1;
    }
4. For each number(i.e. "0, 1, 2, 3, 4,5 6, 7, 8, 9") compress the corresponding boolean vector.
    for(i = 0; i<10; i++)
        compress(VECTORS[i])
5. Now store the compressed vectors of the 10 digits of the number in HASH Table.

NOTE:
Data Structure Use Here
1. Bit Vector Representation for each number.
2. Hash Table to store all number.

I like the bitfield approach, but what about converting the number to hexadecimal so that no bit patterns will be wasted. In decimal representation you have to waste 6 bit patterns per 4 bits.

I would say a balance binary tree structure like an AVL tree where everything happens in logarithmic time.

using linked list, we can store the data till available memory.