Code that splits an array and counts repetitions - c

Question

The following code nearly works but for some reason the first value in the array printed is also the last value and also the first value of the next set of data that is printed. I know that might sound confusing.
Here is the code
#include <stdio.h>
#define MAX 288
int nThArrayPrint(double prev, int count, int i, double *array) {
int j = MAX*i;
for (i = j; i <(j+MAX); i++) {
if (array[i] == prev) {
count++;
} else {
printf("%.4lf= %d\t", prev, count);
prev = array[i];
count = 1;
}
}
return count;
}
int main()
{
FILE *fp;
fp = fopen("C:\\Users\\niall\\Desktop\\question2.txt", "r");
double array[MAX * 5]; // Initializes array to all 0.0s
int i, x, j, count;
double prev = 0;
if (fp != NULL) {
for (i = 0; i < MAX * 5; i++) { //Load values from file into an array
fscanf(fp, "%lf", &array[i]);
}
fclose(fp);
prev = array[0]; // initialize
for (i = 0; i < 5; i++){
count = 1;
printf("\tNumber of times each value is repeated for the %dth set of values\n\n", i+1);
count = nThArrayPrint(prev, count, i, array);
prev = array[MAX*i];
printf("%.4lf= %d\n\n", prev, count);
}
} else {
printf("There was a probem opening the file.");
}
}

Answer 1

modify the below segment of the code
prev = array[0]; // initialize
for (i = 0; i < 5; i++){
count = 1;
printf("\tNumber of times each value is repeated for the %dth set of values\n\n", i+1);
count = nThArrayPrint(prev, count, i, array);
prev = array[MAX*i];
printf("%.4lf= %d\n\n", prev, count);
}
into
count = 1;
for (i = 0; i < 5; i++){
printf("\tNumber of times each value is repeated for the %dth set of values\n\n", i+1);
for (j = i*MAX; j < (i*MAX)+MAX; j++) {
if (array[j] == array[j+1]) {
count++;
} else {
printf("%.4lf= %d\n", array[j], count);
count = 1;
}
}
if(count > 1){
printf("%.4lf= %d\n", array[j], count-1);
count = 1;
}
}
Tested with below input with MAX as 9
12 12 12 12 12 1 1 1 1 4 5 5 6 7 8 9 3 3 3 2 2 2 1 1 1 1 1 1 6 62 2.2 2.2 2.2 3.2 3.2 3.2 3.2 22 22 22 22 22 1 22 22 22 22
output:
Number of times each value is repeated for the 1th set of values
12.0000= 5
1.0000= 4
Number of times each value is repeated for the 2th set of values
4.0000= 1
5.0000= 2
6.0000= 1
7.0000= 1
8.0000= 1
9.0000= 1
3.0000= 2
Number of times each value is repeated for the 3th set of values
3.0000= 1
2.0000= 3
1.0000= 5
Number of times each value is repeated for the 4th set of values
1.0000= 1
6.0000= 1
62.0000= 1
2.2000= 3
3.2000= 3
Number of times each value is repeated for the 5th set of values
3.2000= 1
22.0000= 5
1.0000= 1
22.0000= 2