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- #include <bits/stdc++.h>
- using namespace std;
- #define cin_2d(vec, n, m) for(int i = 0; i < n; i++) for(int j = 0; j < m && cin >> vec[i][j]; j++);
- #define cout_2d(vec, n, m) for(int i = 0; i < n; i++, cout << "\n") for(int j = 0; j < m && cout << vec[i][j] << " "; j++);
- #define fixed(n) fixed << setprecision(n)
- #define ceil(n, m) (((n) / (m)) + ((n) % (m) ? 1 : 0))
- #define fill(vec, value) memset(vec, value, sizeof(vec));
- #define mul_mod(a, b, m) (((a % m) * (b % m)) % m)
- #define add_mod(a, b, m) (((a % m) + (b % m)) % m)
- #define all(vec) vec.begin(), vec.end()
- #define rall(vec) vec.rbegin(), vec.rend()
- #define sz(x) int(x.size())
- #define debug(x) cout << #x << ": " << (x) << "\n";
- #define fi first
- #define se second
- #define ll long long
- #define ull unsigned long long
- #define Mod 1'000'000'007
- #define OO 2'000'000'000
- #define EPS 1e-9
- #define PI acos(-1)
- template < typename T = int > using Pair = pair < T, T >;
- vector < string > RET = {"NO", "YES"};
- template < typename T = int > istream& operator >> (istream &in, vector < T > &v) {
- for (auto &x : v) in >> x;
- return in;
- }
- template < typename T = int > ostream& operator << (ostream &out, const vector < T > &v) {
- for (const T &x : v) out << x << ' ';
- return out;
- }
- void Solve(){
- ll n;
- cin >> n;
- ll x = -1, y = -1, curr_number = 0, sum_of_digits = 0;
- // best_answer is to get the maximum number contains 9 smaller than n
- // 0 9 99 999 9999 99999 999999 ...
- while(curr_number < n){
- x = curr_number;
- curr_number = curr_number * 10 + 9;
- // if curr_number is 9 new curr_number will be 9 * 10 = 90 + 9 = 99
- }
- // x will save the last valid number
- y = n - x;
- // get summation of digits of x and y
- while(x > 0)
- sum_of_digits += x % 10, x /= 10;
- while(y > 0)
- sum_of_digits += y % 10, y /= 10;
- cout << sum_of_digits << "\n";
- }
- int main(){
- ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
- int t = 1;
- //cin >> t;
- while(t--)
- Solve();
- return 0;
- }
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