Program doesn't work with 10 values in array - c

Question

Basically, the max size of the array is 10 and the user is allowed to enter up to 10 values. If the user enters -1 or 0, before entering the ten values, then the loop stops and goes to the next loop. My problem is is that it works perfectly UNTIL I enter 10 values. The result will divide by 9 instead of 10 and print that there are 9 values in the array.
#include <stdio.h>
int main(void)
{
float numbers[10];
int i;
int n;
int count =0;
for(i = 0; i<10; i++)
{
scanf("%f", &numbers[i]);
if(numbers[i] == -1)
break;
if(numbers[i] == 0)
break;
count++;
}
n = count-1;
float sum = 0;
float average;
for(i = 0; i<=n; i++)
{
sum = sum + numbers[i];
}
average = sum/count;
printf("The average price of the %d products is %.2f.\n", count, average);
return 0;
} //this is the fixed solution.

Answer 1

You should not write n=count-1; write n=count; instead. And move count++ to the end of loop.
#include <stdio.h>
int main(void)
{
float numbers[10];
int i;
int n;
int count =0;
for(i = 0; i<10; i++)
{
scanf("%f", &numbers[i]);
if(numbers[i] == -1)
break;
if(numbers[i] == 0)
break;
count++;
}
n = count;
float sum = 0;
float average;
for(i = 0; i<n; i++)
{
sum = sum + numbers[i];
}
average = sum/a;
printf("The average price of the %d products is %.2f.\n", n, average);
return 0;
}

Answer 2

I'm pretty sure it is because of this line
n = count-1
Can you explain why you are subtracting 1? In the case where you input 10 numbers, count will equal 10. After subtracting 1 you will then only iterate thru the first 9 indexes in the array.
If you need to subtract 1 (to account for the user inputting a 0 or -1), then change the condition in the last for loop to be <= instead.
for(i = 0; i<=n; i++)

Answer 3

There are two things has to be considered in this program.
1 . n = count-1;
for(i = 0; i<n; i++)
The count variable contains the number of element in the array, while n has been assigned to count - 1 to access from the 0th position , but while iterating the loop condition `i<n` make the loop to run n-1 time (i.e 9 times in this case).
So that the sum calculation failed to calculate the last array element.
average = sum/a;
The variable average and sum are float , whereas n is an int, so the type conversion has to be made while calculating the average.
average = sum/(float)a;
NOTE : a should be replaced by count , which hold the exact count of the element in array.
The complete corrected code is,
#include <stdio.h>
int main(void)
{
float numbers[10];
int i;
int n;
int count =0;
for(i = 0; i<10; i++)
{
scanf("%f", &numbers[i]);
count++;
if(numbers[i] == -1)
break;
if(numbers[i] == 0)
break;
}
n = count-1;
float sum = 0;
float average;
for(i = 0; i<=n; i++)
{
sum = sum + numbers[i];
}
average = sum/(float)count;
printf("The average price of the %d products is %.2f.\n", count, average);
return 0;
}