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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<math.h>
#include<string.h>

#define FOR(n) for(int i=0;i<n;i++)
#define PRN(n) printf("%d\n",n)
#define PRS(s) printf("%s\n",s)

int kt(int a){
  int b=a;
  int c=0;
  while(b){
    b/=10;
    c++;
  }
  return c;
}
int f(int a){
  int c=0;
  FOR(a-2){
    c+=9*(int)pow(10,i);
    i++;
  }
  return c;
}
int main(void){
  int n;
  scanf("%d",&n);
  int k=kt(n);
  int ans;
  if(k%2){
    ans=f(k);
    ans+=n%(int)pow(10,k);
  }else{
    ans=f(k+1);
  }
  PRN(ans);
  return 0;
}