CODEWITHSUNDEEP
c++c++11numeric
TosinAl
TosinAl

Reputation: 166

Large digit numbers

The program is simple. The user inputs n and n amount of numbers and i try to add zeros in between adjacent digits. For example if the user enters 9(n) digits as 1 2 3 4 5 6 7 8 9 (spaced out), the program outputs 10203040506070809. The program works well for up to n=8 digits but i get funny answers from n=9 digits upwards. The range of n should be 3<=n<=15 . My program is as follows:

int main()
{
    cout << "\nEnter n and n values: \n";
    int n;
    cin >> n;
    vector<long long>nums;
    int en = n;
    while (en > 0)
    {
        long long x;
        cin >> x;
        nums.push_back(x);
        --en;
    }

    int r = 2 * n - 2;
    long long new_val = 0;
    int j = 0;
    for (int i = 0; i < n; ++i)
    {
        new_val = new_val + nums[i] * (pow(10, r - j));
        j += 2;
    }

    cout << new_val << endl;
}

I don't know how to solve the issue of funny answers from n=9 to n=15.

Upvotes: 0

Views: 98

Answers (2)

Marvel_Maro
Marvel_Maro

Reputation: 28

Allow me to suggest a different approach to this problem. Since containers will give you trouble holding those large numbers, then don't with them as numbers, try string instead:

After you get the numbers from the user and store them in the nums vector, do the following:

string output = "";
char temp;
for (int i = 0; i < nums.size(); i++){
   temp = nums [i] + '0';
   output += temp;
   if (i != (nums.size()-1))
      output += "0";
}
cout << output;

Using this method, the user can enter any number as large as they want. If you want the number to be from 9 to 15, you can simply add validation at the beginning without having to deal with complicated containers.

Upvotes: 0

Remy Lebeau
Remy Lebeau

Reputation: 595827

The main problem is long long is only 64-bits in size, and thus can only hold up to 19 digits. Its maximum value is 9,223,372,036,854,775,807.

It is possible to make your code work correctly up to n=10 if you simply remove pow() (which operates on floating-point types, not integer types). For the 2nd and subsequent loop iterations, you can simply multiply new_val by 100 before adding nums[i]:

#include <iostream>
#include <vector>

using namespace std;

int main()
{
    cout << "\nEnter n and n values: \n";
    int n;
    cin >> n;
    vector<long long> nums;
    int en = n;
    while (en > 0)
    {
        long long x;
        cin >> x;
        nums.push_back(x);
        --en;
    }

    long long new_val = 0;
    if (n > 0)
    {
        new_val = nums[0];
        for (int i = 1; i < n; ++i)
        {
            new_val *= 100;
            new_val += nums[i];
        }
    }

    cout << new_val << endl;
}

Live Demo

However, 1020304050607080901 is 19 digits, so n>=11 will overflow past the max value of long long.

Live Demo

For such high values, you need to use a BigNumber library (as most compilers do not yet have a native 128-bit numeric type). Or, just use std::string instead of long long:

#include <iostream>
#include <vector>
#include <string>
#include <sstream>

using namespace std;

int main()
{
    cout << "\nEnter n and n values: \n";
    int n;
    cin >> n;
    vector<int> nums;
    int en = n;
    while (en > 0)
    {
        int x;
        cin >> x;
        nums.push_back(x);
        --en;
    }

    ostringstream new_val;
    if (n > 0)
    {
        new_val << nums[0];
        for (int i = 1; i < n; ++i)
            new_val << '0' << nums[i];
    }

    cout << new_val.str() << endl;
}

Live Demo

Upvotes: 2

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