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: Wrong Answer)

#include<stdio.h>
int main(){
  int n, i, j, a, b, c;
  i = 0;
  a = 0;
  b = 0;
  c = 0;
  scanf("%d", &n);
  if(i<10){
    for(j = 0;j = n;j++){
      i = i + 1;
    }
    i = a;
  }
  if(i>1000){
    while(i<100){
      i=i-100;
    }
    i = b;
  }
   if(i>100000){
    while(i<10000){
      i=i-10000;
    }
    i = c;
      while(i<100){
      i=i-100;
    }
    i = b;
  }
  printf("%d",a + b + c);
  
  return 0;
}