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)

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <string.h>
#include <math.h>
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) > (b) ? (b) : (a))
#define abs(x) ((x) > 0 ? (x) : -(x))
#define rep(i, n) for(int i = 0; i < (n); i++)
#define INF 1000000000000 //10^12
#define MOD 1000000007 //10^9 + 7
#define endl printf("\n")
typedef long long ll;

int sort(const void *a, const void *b){
  int A = *(int *)a, B = *(int *)b;
  if(A > B) return  1;
  if(A < B) return -1;
  return 0;
}

int
main(int argc, char *argv[])
{
  int n;
  scanf("%d", &n);
  int cnt = 0;
  int m = n;
  while(m > 0){
    m /= 10;
    cnt++;
  }
  int d[100]; d[1] = 9; d[2] = 90;
  for(int i = 3; i < 7; i++){
    d[i] = d[i - 1] * 10;
  }
  int ans = 0;

  for(int i = 0; 2 * i + 1 < cnt; i++){
    ans += d[2 * i + 1];
  }

  int k = 0; int s = 1;
  for(int i = 0; i < cnt - 1; i++){
    k += 9 * s;
    s *= 10;
  }
  if(cnt % 2 == 1) ans += n - k;
  printf("%d\n", ans);

  return 0;
}