The behind idea of this problem is . Since the list is unsorted, we can only use linear search to find the maximum and minimum. After that, the logic is simple.
#include <stdio.h>
#include <stdlib.h>
#include <errno.h>
long read_long()
{
long a;
char buf[1024]; // use 1KiB just to be sure
int success; // flag for successful conversion
do
{
if (!fgets(buf, 1024, stdin))
{
// reading input failed:
return 1;
}
// have some input, convert it to integer:
char *endptr;
errno = 0; // reset error number
a = strtol(buf, &endptr, 10);
if (errno == ERANGE)
{
printf("Sorry, this number is too small or too large.\n");
success = 0;
}
else if (endptr == buf)
{
// no character was read
success = 0;
}
else if (*endptr && *endptr != '\n')
{
// *endptr is neither end of string nor newline,
// so we didn't convert the *whole* input
success = 0;
}
else
{
success = 1;
}
} while (!success); // repeat until we got a valid number
return a;
}
long find_min(long *list, long n)
{
long min = list[0];
for (long i = 1; i < n; i += 1)
{
if (list[i] < min)
{
min = list[i];
}
}
return min;
}
long find_max(long *list, long n)
{
long max = list[0];
for (long i = 0; i < n; i += 1)
{
if (list[i] > max)
{
max = list[i];
}
}
return max;
}
int main()
{
long n;
scanf("%ld\n", &n);
long *list = calloc((size_t)n, sizeof(long));
if (list == NULL)
{
return 1;
}
for (long i = 0; i < n; i += 1)
{
list[i] = read_long();
}
long min = find_min(list, n);
long max = find_max(list, n) / 2;
if (min <= max)
{
printf("%ld\n", (long)0);
}
else
{
printf("%ld\n", min - max);
}
}
