JohnSmithEr - 4 months ago 19

C++ Question

I have problem solving this problem.

The task is simple at first line I enter how many examples I have.

On second line I need to enter how many numbers im going to read.

and then we enter all the numbers separate by space.

The task itselfs do , sorting the string array wtih numbers from smalles to the biggest one. After that if we have even numbers entered we print the middle number -1, if they are uneven we just print the middle number.

So far if I use the exact same code with long long it works perfectly , but it is limited only to 19 digit number and I want to expand the program so it can use bigger numbers.

Using that way the sort func , when I try to sort 16 elements from 160 to 10 , they all messed it start from 110 then in the midle is 160 and so one , which makes absolutly non sense, using 5 numbers or 8 works perfectly w/o any problem , using more numbers fails.

`#include <iostream>`

#include <algorithm>

#include <string>

using namespace std;

int main() {

int examples;

cin >> examples;

for (size_t i = 0; i < examples; i++)

{

long long unsigned int n;

cin >> n;

string * numbers = new string[n];

for (size_t i = 0; i < n; i++)

{

cin >> numbers[i];

}

sort(numbers, numbers + n);

if (n % 2 == 0) {

cout << numbers[n / 2 - 1];

}

else

cout << numbers[n / 2];

}

system("pause");

return 0;

}

Answer

First, if you allocate memory with operator new, you must release it with operator delete[].

Second, when you sort strings instead of values, they are sorted just like strings would do, and here is where your problem lies. You see, 100 is alphabetically less than 2 or 20, that's why it would appear earlier.

Here's the output your program gives. Check this rule out, and you'll see that i'm right.

```
10 100 110 120 130 140 150 160 20 30 40 50 60 70 80 90
```

Third, using operator new is discouraged for pretty much anything. You have STL, and you seem to be using it extensively - why not vector?

Fourth, you don't check if anything we write into numbers[i] is actually a number. Think on that.

Fifth, for N being long enough(more than 2^sizeof(size_t)) your problem will NEVER stop due to integer overflow.

Sixth, you don't check for n == 0, and you will ultimately get memory access violation if you enter it.

A fast-right-off-the-bat fix for your problem:

```
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
using namespace std;
int main() {
int examples;
cin >> examples;
for (size_t i = 0; i < examples; i++)
{
size_t n;
cin >> n;
if (n <= 0)
break;
vector<string> numbers(n);
for (size_t i = 0; i < n; i++)
cin >> numbers[i];
//here we add a predicate for string checking,
//which evaluates the length of string
//before using the usual operator<.
sort(numbers.begin(), numbers.end(), [](const string& s1, const string& s2){
if (s1.length() < s2.length())
return true;
if (s2.length() < s1.length())
return false;
else
return (s1 < s2);
});
if (n % 2 == 0) {
cout << numbers[n / 2 - 1];
}
else
cout << numbers[n / 2];
}
system("pause");
return 0;
}
```

Still, it has a number of problems:

Checking if numbers[i] is actually a number

I'm not sure that predicate I wrote doesn't have bugs - I'm just trying to give you the idea of how it should work.