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>
#include <math.h>
int main (void){
  /* 変数の宣言 */
  int number,num2;
  int digit = 0;
  int sum=0;
  int q;
 
  /* 数値の入力 */
  scanf("%d", &number);
  num2=number;
 
  /* 桁数を計算 */
  while(number!=0){
    number = number/10;
    ++digit;
  }
  for (int i=3;i<digit;i=i+2){
      sum=sum+(pow(10,i-1)+1)*9-8;
      q=(pow(10,i-1)+1);
      printf("%d\n",sum);
  }
    sum=sum+9;
    if(digit%2==1){sum=num2-pow(10,digit-1)+1+sum;}
    printf("%d",sum);
}