Number by 3#919

Number by 3 (Editorial)

In this problem we have to find number and sum of real numbers that, positive and it's index is divisible by $3$ .

For this problem we'll use array. In array indexing is starting from $0$ , so we have to create an array with $1$ more extra element, like $a [n + 1]$ a[n+1]. After creating, we'll check that if the number's index is divisible by $3$ and the number is positive. If it is, we'll add this to sum and increase the value of number by $1$ . After all of these, we'll print the number and the sum with $2$ decimal points.

Implementation

double a[n + 1], number = 0, sum = 0;

Creating an array called $a$ and setting it's number of elements to $n + 1$ n+1, because indexing is starting from $0$ and in problem it wants from us to start it from $1$ . Setting the value of $n u mb er$ to $0$ . Also setting the value of $s u m$ to $0$ too.

for (int i = 1; i <= n; i++)

Checking all the elements of our array.

cin >> a[i];

Input the element of array.

if (i % 3 == 0 && a[i] > 0)
{
    number++;
    sum += a[i];
}

If the index of $a [i]$ , which one is $i$ , is divisible by $3$ and $a [i]$ is positive, that means we can increase the value of $n u mb er$ by $1$ and also add $a [i]$ to current value of $s u m$ .

cout << number << ' ';

Printing the $n u mb er$ .

cout.precision(2);
cout << fixed << sum;

Printing the sum with $2$ decimal points.