Problem:
Score : 300 points

Problem Statement

Find the maximum possible sum of the digits (in base 10) of a positive integer not greater than N.

Constraints

1\leq N \leq 10^{16}

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the maximum possible sum of the digits (in base 10) of a positive integer not greater than N.

Sample Input 1

100

Sample Output 1

18

For example, the sum of the digits in 99 is 18, which turns out to be the maximum value.

Sample Input 2

9995

Sample Output 2

35

For example, the sum of the digits in 9989 is 35, which turns out to be the maximum value.

Sample Input 3

3141592653589793

Sample Output 3

137

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
//----***やべーやつら***----
using namespace std;
#define int long long

//----***型定義***----
using ll = long long;
using P = pair<int,int>;

//----***Like a Pythonista***----
#define REP(ii,jj,nn) for (ll ii=jj;ii<(nn);ii++)
#define RREP(ii,nn,jj) for (ll ii = nn; jj<ii;ii--)
#define each(i,...) for (auto&& i:__VA_ARGS__)
#define ALL(vec) (vec).begin(),(vec).end()
#define sum(...) accumulate(ALL(__VA_ARGS__),0LL)
#define dsum(...) accumulate(ALL(__VA_ARGS__),0.0)
#define vec(type,name,...) vector<type> name(__VA_ARGS__)
template<class T> inline auto max(const T& a){ return *max_element(ALL(a)); }
template<class T> inline auto min(const T& a){ return *min_element(ALL(a)); }
inline ll gcd(ll a,ll b){if(b == 0) return a;return  gcd(b,a%b);}
inline ll lcm(ll a,ll b){ll g = gcd(a,b);return a / g * b;}

//----***定数***----
#define MOD 1000000007
#define INF 100000000000000000
#define EPS 1e-9
#define PI  acos(-1)
const vector<vector<int>> DXY={{-1,0},{1,0},{0,-1},{0,1}};

//----***パーツ***----
#define fi first
#define se second
#define pb push_back
#define re return
#define br break

//----***入出力***---
void print() { std::cout << "\n"; }
template <class T>void print(const T &x) {std::cout << x<<"\n";}
template <class T, class... Args>void print(const T &x, const Args &... args) {std::cout << x << " ";print(args...);}
#define debug(var)  do{std::cerr << #var << " ↓ "<<"\n";view(var);}while(0);
#define dbg cerr<<"🥺🥺🥺🥺🥺🥺"<<endl;
template<typename T> void view(T e){std::cout << e << std::endl;}
template<typename T> void view(const std::vector<T>& v){for(const auto& e : v){ std::cout << e << " "; } std::cout << std::endl;}
template<typename T> void view(const std::vector<std::vector<T> >& vv){ for(const auto& v : vv){ view(v); } }

//----***初期時読み込み***----
struct initial{initial(){cin.tie(0); ios::sync_with_stdio(0); cout<<fixed<<setprecision(15);};}initial_;


signed main(){
  string S;cin>>S;
  vector<int> num(S.size());
  REP(i,0,S.size()){
    num[i]=S[i]-'0';
  }
  int n=S.size();
  int ans=-1;
  vector<vector<int>> dp(n,vector<int>(2,0));
  dp[0][0]=num[0];dp[0][1]=num[0]-1;
  REP(i,1,n){
    dp[i][0]=dp[i-1][0]+num[i];
    if(num[i]>0){
      dp[i][1]=max(dp[i-1][1]+9,dp[i-1][0]+num[i]-1);
    }
    else{
      dp[i][1]=dp[i-1][1]+9;
    }
  }
  print(max(dp[n-1][0],dp[n-1][1]));

}