I think I've just found a solution to your problem. In what follows I will list the whole code with comments to make it easier to follow.

#include <stdio.h>
 
int main()
{
  // input no. of students
  int n, i;
  printf("no. of students = "); scanf("%d", &n);
 
  // allocate an array for exactly "n" marks
  int *marks = new int[n];
 
  // input marks
  printf("marks: ");
  for (i = 0; i < n; i++)
    scanf("%d", &marks[i]);
 
  // compute frequencies
  int freq[101] = {0};
  for (i = 0; i < n; i++)
    freq[marks[i]]++;
 
  // display non-zero frequencies
  printf("\nMark Frequency\n\n");
  for (i = 0; i < 101; i++)
    if (freq[i])
      printf("%d %d\n", i, freq[i]);
 
  // display all intervals between identical marks
  printf("\nIntervals\n\n");
  int pos[101] = {0};
  for (i = 0; i < n; i++)
  {
    if (pos[marks[i]])
      printf("%d %d\n", marks[i], i - pos[marks[i]]);
    pos[marks[i]] = i + 1;
  }
  printf("\n");
 
  // free memory used by the marks[] array
  delete[] marks;
 
  return 0;
}

I've tried inputing your example and here is what has been displayed (notice that the frequencies are displayed considering the ascending order of the marks).

no. of students = 10
marks: 46 37 28 87 76 83 76 38 76 37
 
Mark Frequency
 
28 1
37 2
38 1
46 1
76 3
83 1
87 1
 
Intervals
 
76 1
76 1
37 7

The algorithm I've used is very quick even for a large number of marks. It goes two times through the elements of the marks array - one time to compute the frequencies and the other to find the intervals. Thus the algorithm has linear complexity. As you may have noticed I've used two additional arrays "freq" and "pos" of 101 elements, which help at minimising the total running time.

Please let me know if you understand how the algorithm works or not. I am ready to give you any further information concerning the algorithm or the C implementation.