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>
int min(int a,int b){
  return a<b ? a:b;
}
int max(int a,int b){
  return a>b ? a:b;
}

int main() {
  int N;
  scanf("%d",&N);
  if (N<10) {
    printf("%d",N);
  }
  else if (N<100) {
    printf("9");
  }
  else if (N<1000) {
    printf("%d",9+N-99);
  }
  else if (N<10000) {
    printf("909");
  }
  else if (N<100000) {
    printf("%d\n",909+N-9999);
  }
  else
  printf("90909\n");
  return 0;
}