int count8(int n) {
if (n == 0)
return 0;
if ((n / 10) % 10 == 8 && n % 10 == 8)
return 2 + count8(n / 10);
if (n % 10 == 8)
return 1 + count8(n / 10);
return count8(n / 10);
}
// Given a non - negative int n, compute recursively( no loops ) the count of the occurrences of 8 as a digit,
// except that an 8 with another 8 immediately to its left counts double, so 8818 yields 4. ( no loops )
int count8( int n )
{
if( n <= 0 )
return 0;
if( n % 100 == 88 )
return 2 + count8( n / 10 );
if( n % 10 == 8 )
return 1 + count8( n / 10 );
return count8( n / 10 );
}
int main() {
int n;
printf("Vnesi n:");
scanf("%d", &n);
printf("count8(%d) = %d\n", n, count8(n));
return 0;
}
