How can I calculate the moving average of the array? - c

Question

In Question, When an array X of size n and a degree k are input,
Write a program that calculates the k-th moving average of array X.
The kth-order moving average of the array X consisting of primitive data values is the average of the last k elements up to the i-th point of X.
That is, A [i] = (X [i-k + 1] + X [i-k + 2] + ... + X [i]) / k.
If the number of the preceding element (including itself) is smaller than k,
Calculate as an average.
For example, if array X is as follows and k is 3, X = 1 3 2 10 6 8
The third moving average is as follows.
A = 1 2 2 5 6 8 A [1] = (1 + 3) / 2, A [2] = (1 + 3 + 2) / 3
However, the program must have the execution time of O (n), not O (nk).
Round off the decimal point in the average calculation and obtain it as an integer.
For exact rounding, do not use% .f, but round it using the int property.
int main()
{
int i, n1, k;
int *array1;
scanf("%d", &n1);
array1 = (int *)malloc(sizeof(int)*n1);
scanf("%d", &k);
for (i = 0; i < n1; i++)
{
scanf("%d", &array1[i]);
}
double tmp = 0;
for (int i = 0; i < n1; i++)
{
tmp += array1[i];
if (i >= k)
{
tmp -= array1[i - k];
}
if (i >= k - 1)
{
double average = tmp / k;
printf("%2lld ", llrint(average));
}
return 0;
}
The program does not work because the problem is not understood.
I would like to know how to solve it.
add) Thank you for answer but the output required by the problem is as follows.
Input : 9 4 (n = 9, k = 3)
2 7 4 5 6 8 2 8 13
Output : 2 5 4 5 6 6 5 6 8

Answer 1

After Modifying your code
int main()
{
int i, n1, k;
int *array1, *array2;
scanf("%d", &n1);
array1 = (int *)malloc(sizeof(int)*n1);
scanf("%d", &k);
for (i = 0; i < n1; i++)
{
scanf("%d", &array1[i]);
}
double tmp = 0;
for (int i = 0; i < n1; i++)
{
tmp += array1[i];
// now tmp contains exactly k + 1 elements sum
// so subtract elements outside of k sized window(leftmost element)
if(i >= k) {
tmp -= array1[i - k];
}
if(i >= k - 1) {
double average = tmp / k;
printf("%lf\n", average);
}
}
return 0;
}