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: Accepted)

#include<stdio.h>

int main(void){
    int N;
    scanf("%d",&N);
    if(N>=1&&N<=9){
        printf("%d\n",N);
    }
    else if(N>=10&&N<=99){
        printf("9\n");
    }
    else if(N>=100&&N<=999){
        printf("%d\n",N-90);
    }
    else if(N>=1000&&N<=9999){
        printf("909\n");
    }
    else if(N>=10000&&N<=99999){
        printf("%d\n",N-9090);
    }
    else{
        printf("90909\n");
    }
    return 0;
    
}