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++14 (GCC 5.4.1) (judge verdict: Accepted)

/**

+++---------+++--------- Don't lose faith ---------+++---------+++
+++---------+++--------- Coder: sea_26 ----------+++---------+++

**/

#pragma GCC optimize("03")
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <bits/stdc++.h>

using namespace __gnu_pbds;
using namespace std;

template <typename T>
    using ordered_set = tree<T, null_type, less<T>,
        rb_tree_tag, tree_order_statistics_node_update>;


using ll = long long;
using ld = long double;

priority_queue<int> pq;

const int M = 1e9 + 7;
const ll MOD = 998244353;
const ld eps = 1e-9;
const ld PI = acos(-1.0);

// const double pi = atan2(0, -1);
// const double PI = 4 * atan(1);
// const double pi = 2*cos(0.0);

const int iinf = 0x3f3f3f3f;
const ll linf = 0x3f3f3f3f3f3f3f3f;

#define mp make_pair
#define eb emplace_back
#define ep emplace
#define fbo find_by_order
#define ook order_of_key
#define f first
#define s second
#define sz size()

#define bp(n)  __builtin_popcount(n)
#define bclz(n) __builtin_clzll(n)
#define bctz(n) __builtin_ctzll(n)
#define bip(n) __builtin_parity(n)

#define vsort(v) sort(v.begin(),v.end())
#define rvsort(v) sort(v.begin(),v.end(),greater<int>())
#define mnv(v) *min_element(v.begin(),v.end())
#define mxv(v) *max_element(v.begin(),v.end())
#define all(v) (v).begin(),(v).end()
#define ita(v) iota(all(v), 0)
#define acm(v, ini) accumulate(all(v),ini)

#define ub upper_bound
#define lb lower_bound
#define cr(n) (floor)cbrt(n)
#define msz(a) memset(a, 0, sizeof(a))
#define msm(a) memset(a, 0xff, sizeof(a))

#define rep(n) for(int i=0; i<n; i++)
#define rrep(n) for(int i=n-1; i>=0; i--)
#define lrep(p,q) for(int j=p; j<=q; j++)
#define ldrep(p,q) for(int i=p; i>=q; i--)


#define yes cout << "YES" << endl
#define no cout << "NO" << endl

#define fs ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);

template< typename T> inline T MAX(T a, T b){ return a ^ ((a ^ b) & -(a < b)); }
template< typename T> inline T MIN(T a, T b){ return b ^ ((a ^ b) & -(a < b)); }
template< typename T> void SWAP(T &a, T &b){ a = ((a) ^= (b)),  b = ((b) ^= (a)),  a = ((a) ^= (b)); }
template< typename T> inline T sqr(T x){ return x * x; }
template< typename T> inline T lcm(T a, T b){ return (a / gcd(a,b)) * b; }

template< typename T> inline T gcd(T a, T b){ while (b) b ^= a ^= b ^= a %= b; return a; }
template< typename T> T exgcd(T a, T b, T& x, T& y){ if(!b){ x = 1; y = 0;return a;} else{T g = exgcd(b,a%b,y,x); y -= (a/b)*x; return g;}}
template< typename T>T gcdextd(T a, T b, T& x, T& y) { if (!a) { x = 0; y = 1; return b; }  T x1,y1,d; d = gcdextd(b % a, a, x1, y1);  x = y1 - (b / a) * x1; y = x1; return d; }

template< typename T> T icbrt(T mul){T r = cbrt(mul) + 1; while(r * r * r > mul) r--; return r; } /// 54229668
template< typename T> void tpns(T n){ rrep(30){ if(n & (1 << i)) pq.push(1 << i); } }

template< typename T> inline T mod(T a, T b){ return a >= b ? a % b : a; }
template< typename T> T binpow(T n, T p){ T res = 1; while(p > 0){ if(p & 1) res *= n; n *= n; p >>= 1; } return res;}
template< typename T> T bigmod(T n, T p, T M){ n %= M; T res = 1; while(p > 0){ if(p & 1)res = ((res%M) * (n % M)) % M; n = ((n%M) * (n%M))%M; p >>= 1; } return res;}
template< typename T> T modInv(T a, T m){return bigmod(a, m-2, m);}


void ReT(){ cout << "\nTime = " << fixed << setprecision(3) << double(clock()) / CLOCKS_PER_SEC  << " Sec"<< endl; exit(0);}


int main(){

    fs;
    int a, n;
    cin >> a;

    int cnt = 0;
    lrep(1,a){
        int x=floor(log10(j) + 1);
        if(x&1) cnt++;
    }

    cout << cnt << endl;
    /// ReT();

return 0;
}