**Solution Notes:** This problem has a nice recursive solution.
We first write a recursive function to compute the length of S(k)
(given by twice the length of S(k-1) plus the length of the middle
section, k+3). Afterwards, we can figure out the nth chracter in S(k)
by checking if n lies in the left copy of S(k-1) (in which case we
can proceed by recursion), in the middle section, or in the right
copy of S(k-1) (in which case we can again proceed by recursion).

```
#include <stdio.h>
/* Compute the length of S(k) */
int len(int k)
{
int x;
if (k==-1) return 0;
x = len(k-1);
return x + k+3 + x;
}
/* Return nth character in S(k) */
char solve(int n, int k)
{
if (n > len(k)) return solve(n,k+1);
if (n <= len(k-1)) return solve(n,k-1);
n = n - len(k-1); /* Discount S(k-1) from beginning of string */
if (n <= k+3) /* n in middle section? */
return (n==1) ? 'm' : 'o';
n = n - (k+3);
return solve(n,k-1);
}
int main(void)
{
int n;
freopen ("moo.in", "r", stdin);
freopen ("moo.out", "w", stdout);
scanf ("%d", &n);
printf ("%c\n", solve(n,0));
return 0;
}
```