I did as well at first. But it's easier if you break it into two parts:

1. Traverse all possible substrings within string
2. Check if any two substrings of equal length are anagrams

It's much more digestable that way.

C Solution:

int check_anagram(char a[], char b[])
{
   int first[26] = {0}, second[26] = {0}, c = 0;

   while (a[c] != '\0') {
      first[a[c]-'a']++;
      c++;
   }
   c = 0;
   while (b[c] != '\0') {
      second[b[c]-'a']++;
      c++;
   }

   for (c = 0; c < 26; c++) {
      if (first[c] != second[c])
         return 0;
   }

   return 1;
}

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        char s[100];
        char sub1[100] = {0};
        char sub2[100] = {0};
        scanf("%s", s);

        int count = 0;
        for (int len = 1; len < strlen(s); len++) {
            memset(sub1, 0, len);
            for (int i = 0; i < strlen(s) - len; i++) {
                strncpy(sub1, &s[i], len);
                memset(sub2, 0, len);
                for (int j = i + 1; j < strlen(s) - len + 1; j++) {
                    strncpy(sub2, &s[j], len);
                    if (check_anagram(sub1, sub2) == 1) {
                        count++;
                    }  
                }
            }     
        }

        printf("%d\n", count);


}

return 0;

}