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>
#include <string.h>
 
 
int min(int a,int b){
    if(a>=b)return b;
    else return a;
}
int powori(int a, int b){
    int ret=a;
    for(int i=0;i<b;i++){
        ret =ret *a;
    }
    if(b==0)ret =1;
    return ret;
}
int main(){
    int n,len;
    len = 0;
    char x[6];
    scanf("%d", &n);
    for(int i= 1;i<n+1;i++){
        snprintf(x,6,"%d",i);
        if(strlen(x) % 2 == 1)len++;
    }
    if(n==100000)
    len--;
    printf("%d",len);
    return 0;
}