Problem:
Score : 200 points

Problem Statement

Given is an integer N. Find the number of positive integers less than or equal to N that have an odd number of digits (in base ten without leading zeros).

Constraints

1 \leq N \leq 10^5

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of positive integers less than or equal to N that have an odd number of digits.

Sample Input 1

11

Sample Output 1

9

Among the positive integers less than or equal to 11, nine integers have an odd number of digits: 1, 2, \ldots, 9.

Sample Input 2

136

Sample Output 2

46

In addition to 1, 2, \ldots, 9, another 37 integers also have an odd number of digits: 100, 101, \ldots, 136.

Sample Input 3

100000

Sample Output 3

90909

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

// Ver19.03
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#define INF (1 << 29) * 2
#define LLINF 4545454545454545454
#define MOD 1000000007
#define ll long long
#define ull unsigned long long
#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))
int upll(const void *a, const void *b) { return *(ll *)a < *(ll *)b ? -1 : *(ll *)a > *(ll *)b ? 1 : 0; }
int downll(const void *a, const void *b) { return *(ll *)a < *(ll *)b ? 1 : *(ll *)a > *(ll *)b ? -1 : 0; }
void sortup(ll *a, int n) { qsort(a, n, sizeof(ll), upll); }
void sortdown(ll *a, int n) { qsort(a, n, sizeof(ll), downll); }

int main()
{
  int n, ans = 0;
  char a[10];
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
  {
    sprintf(a, "%d", i);
    if (strlen(a) % 2 == 1)
      ans++;
  }
  printf("%d\n", ans);
}