数组中每三个连续数字的平均值

Question

How can I calculate the average for each three consecutive numbers in an array and then to print each result? (in C)

I tried to do this.

#include <stdio.h>

int main() {
  int n,v[100],i;
  float average;

  int sum=0;
  scanf("%d",&n);

  for (i = 0; i < n; ++i) {
    scanf("%d",&v[i]);
  }
      
  for(i=0;i<n;i+=3) {
    sum=v[i]+v[i+1]+v[i+3];
    average=sum/3;
    printf("%.2f ", average);
  }

  return 0;
}

Answer 1

sum/3 is an int division with an int quotient.

To also have a fractional part, use floating point division.

// average=sum/3;
average = sum / 3.0f;

---

for(i=0;i<n;i+=3) { sum=v[i]+v[i+1]+v[i+3]; risks accessing data out of v[] bounds. v[i]+v[i+1]+v[i+3] are not consecutive. I'd expect v[i]+v[i+1]+v[i+2]

for (i = 0; i + 2 < n; i += 3) {
  sum = v[i] + v[i+1] + v[i+2];

---

No good reason to use float . Use double unless float is needed for space/ maybe speed, etc.

Also avoid overflow and precision loss.

for(int i = 0; i+2 < n; i += 3) {
  double sum = 0.0 + v[i] + v[i+1] + v[i+2]; // Addition here uses FP math.
  double average = sum/3.0;
  printf("%.2f ", average);
}

---

Better code writes a '\n' at the end and does not end with a dangling " " .

const char *separator = "";
for(int i = 0; i+2 < n; i += 3) {
  ...
  printf("%s%.2f ", separator, average);
  separator = " ";
}
printf("\n");