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: Accepted)

#include <stdio.h>
int main(void){
    int i,j;            //  ループ文に使用
    int x,y,z;
    int n;
    int count,count2;   //  カウント用に使用
    char s[256];        //  文字列
    
    count = 0;
    count2 = 0;
    
    //  入力処理
    
    scanf("%d",&n);
    
    //  計算処理
    
    if (n < 10){
        count = n;
    }else if (n < 100){
        count = 9;
    }else if (n < 1000){
        for (i = 100 ; i <= n; i++){
            count++;
        }
        count = count + 9;
    }else if (n < 10000){
        count = 9 + 900;
    }else if (n < 100000){
        for (i = 10000 ; i <= n; i++){
            count++;
        }
        count = count + 9 + 900;
    }else if (n == 100000){
        count = 9 + 900 + 90000;
    }
    //  出力処理
    
    printf("%d",count);
    
    
}