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 <stdlib.h>
#include <math.h>

int main()
{
	int N;
	int digit = 0;

	scanf("%d", &N);

	if(N < 1 || N > 100000) exit(0);

	int tmpN = N;

	/* 桁数を計算 */
  	while(tmpN != 0) {
    	tmpN = tmpN / 10;
    	++digit;
  	}

  	int start;
  	if(digit % 2 == 0) {
  		start = digit - 1;
  	} else {
  		start = digit;
  	}

  	int count = 0;
  	while(1) {
  		if(start == 1) {
  			count += 9;
  		} else if(start == digit) {
  			count += N - pow(10, start - 1) + 1;
  		} else {
  			count += 9 * pow(10, start - 1);
  		}

  		start -= 2;

  		if(start < 0) break;
  	}
 
  	/* 数値の桁数を表示 */
  	printf("%d\n", count);
}