Average of a stream of numbers

PROBLEM :

Given a stream of numbers, print average or mean of the stream at every point.

Input:

The first line of input contains an integer T denoting the number of test cases.
The first line of each test case is N,N is the size of array.
The second line of each test case contains N input C[i].

Output:

Print the average of the stream at every point (in integer).

Constraints:

1 = T = 20
1 = N = 50
1 = C[i] = 100

Example:

Input
2
5
10 20 30 40 50
2
12 2

Output
10 15 20 25 30
12 7

--------------------------------------------------------------------------------
SIMPLE c++ IMPLEMENTATION :
--------------------------------------------------------------------------------
SIMPLE SOLUTION O(n^2)

#include<iostream>
using namespace std;
int main()
 {
int t,no,a[100],i,sum,j ;
cin>>t ;
while(t--)
{
   cin>>no ;
   for(i=0;i<no;i++)
   cin>>a[i] ;
 
   for(i=0;i<no;i++)
   {
       sum=0 ;
       for(j=0;j<=i;j++)
       {
           sum=sum+a[j] ;
       }
       cout<<sum/(i+1)<<" " ;
   }
   cout<<endl ;
}
return 0;
}

--------------------------------------------------------------------------------
BETTER SOLUTION O(n)
--------------------------------------------------------------------------------

#include<iostream>
using namespace std;
#include<math.h>
void print_avarage(int ,float* ,int ) ;
int main()
 {
int t,no,a[50],i ;
float avg ;
cin>>t ;
while(t--)
{
   cin>>no ;
   for(i=0;i<no;i++)
       cin>>a[i] ;
     
   avg=0 ;
   for(i=0;i<no;i++)
   {
       print_avarage(a[i],&avg,i) ;
   }
   cout<<endl ;
}
return 0;
}

void print_avarage(int ele,float *avg,int no)
{
    float r=(ele+(*avg)*no)/(no+1) ;
    cout<<floor(r)<<" " ;
    (*avg)=r ;
}

---------------------------------------------------------------------------------