Q1 double sumskip10 double array int n n size of the array A

Q1.
double sum_skip10 (double array[], int n)
//n: size of the array. Assume n is divisible by 10
{
double sum=0;
for (int i=0; i<n; i=i+10)
sum = sum + array[ i ];
return sum;

}

Solution

sum=0 - 1 unit - 1 time - 1 unit

i=0 - 1 unit - 1 time - 1unit

i<n - 1 unit - n/10 times - n/10 units

i=i+10- 1 unit - n/10 times - n/10 units

sum=sum + array[i]-1unit -n/10 times - n/10 units

Total time = 3n/10 + 2 = O(n)