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)

/*
 * C言語のサンプルプログラム - Webkaru
 * - 入力した数値の桁数を表示 -
 */
#include <stdio.h>
 
int main(void)
{
  /* 変数の宣言 */
  int number,souko;
  int digit = 0;
 
  /* 数値の入力 */
  scanf("%d", &number);
  souko=number;
  /* 桁数を計算 */
  while(souko!=0){
    souko = souko / 10;
    ++digit;
  }
  if(digit==1){
    printf("%d",number);
  }else if(digit==2){
        printf("9");

  }else if(digit==3){
        printf("%d",9+number-99);

  }else if(digit==4){
        printf("%d",9+899);

  }else if (digit==5)
  {
        printf("%d",9+899+number);

  }else {
    printf("90909");
}
  
 
  /* 数値の桁数を表示 */
 
  return 0;
}