Memset, Big size array global declaration confusion

#1

#include
#include
using namespace std;
#define ll long

ll a[1000005],pre[1000005];

int main()
{

int t;
cin>>t;
while (t--)
{
    int n;
    cin>>n;
    memset(pre,0,sizeof(pre)); //sets the memory block from pre till end of pre as 0

    pre[0]=1;
    int sum=0;
    for (int i = 0; i < n; i++)
    {
        cin>>a[i];
        sum+=a[i];
        sum%=n;
        sum= (sum+n)%n;
        pre[sum]++; //storing values of sum in frequency array
    }
    ll ans=0;
    for (int i = 0; i < n; i++)
    {
        ll m=pre[i];
        ans+=(m)*(m-1)/2;
    }
    cout<<ans<<endl;

}

return 0;

}

  1. In this program i tried using pre[1000005] = {0} which should do the same thing that is initializing whole array with 0 but test cases failed. What is the difference between memset and {0} ?
  2. If we declare pre and a array in the main function the program terminates immediately why is that and why is global declaration working ?
    Please elaborate
#2

For your second question you can follow this https://discuss.codechef.com/t/maximum-size-of-an-array/3633.
And for the first point are you declaring your array inside the main? Because that will give error anyway.

#3
  1. Ok so arrays declared globally can have greater array size because they are stored in heap segment of the memory ??
  2. No i tried with declaring globally also but it did not work with pre[100000] = {0}.
#4
  1. Yes.
  2. You had to reinitialise the array for each testcase, were you doing that?
#5

Yes i was .
This is what i was doing

#include
using namespace std;
#define ll long long int
ll a[1000005], pre[1000005];

int main()
{

int t;
cin >> t;
while (t--)
{
    int n;
    cin >> n;
    pre[1000005] = {0};
    pre[0] = 1;
    //  read input
    ll sum = 0;
    for (int i = 0; i < n; i++)
    {
        cin >> a[i];
        sum += a[i];
        sum %= n;
        sum = (sum + n) % n;
        pre[sum]++; //We are directly building the frequency bucket array
    }
    ll ans = 0;
    for (int i = 0; i < n; i++)
    {
        ll m = pre[i];
        ans += m * (m - 1) / 2;
    }
    cout << ans << endl;
}

return 0;

}

#6

How is this reinitialisation? It’s like accessing an element and setting it 0.

#7

It is reinitialisation . We are not just accessing an element , we are initializing whole array to 0 with pre[100005] = {0};

Look at this example and try to run it you will understand:

#include
#include <bits/stdc++.h>

using namespace std;
int arr[15];

int main()

{

int t;

cin >> t;

while (t--)

{

    int arr[15] = {0};

    for (int i = 0; i < 15; i++)

    {

        cout << arr[i] << " ";

    }

    cout << endl;

}
return 0;

}
at each iteration array is getting reinitialised.

#8

you are declaring the array here.
int arr[15] = {0), works.

this arr[15] = {0} doesn’t do anything.

Also in your case, you are not using global array, instead you are creating a new array for every t.

#9

Screenshot (125)
Screenshot (125)1920×1080 416 KB

Read the highlighted text in the screenshot and also have a look at this:

And see the following program we are not declaring a new array in this .
#include

#include <bits/stdc++.h>

using namespace std;

int arr[15];

int main()

{

int t;

cin >> t;

while (t--)

{

     arr[15] = {0};

    for (int i = 0; i < 15; i++)

    {

        cout << arr[i] << " ";

    }

    cout << endl;

}

return 0;

}

#10

Can’t you see that DATA TYPE that they are giving before the array?
it’s always int arr[] = {0} or char arr[] = {}. This is called initialisation.
As far as your code goes. I made the change, set an element equal to 1, and you can clearly see the output, it stays 1.

include <bits/stdc++.h>

using namespace std;

int arr[15] = {};

int main()

{

int t;

cin >> t;

arr[1] = 1;

while (t–)

{

 arr[15] = {0};

for (int i = 0; i < 15; i++)

{

    cout << arr[i] << " ";

}

cout << endl;

}

return 0;

}

This is because we are not REINITIALISING it.

#11

yes i got it.

#12

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