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: Wrong Answer)

#include"bits/stdc++.h"

#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define REP(i, n) for (int i = 1; i <= (int)(n); i++)
#define all(v) v.begin(), v.end()
using namespace std;
using ll = long long;
using pi = pair<int, int>;
const ll INF = 1LL << 60;

int main() {
    string s;
    cin >> s;
    int sum = 0;
    ll digit2 = stoll(s);
    digit2 -= 10;
    digit2 /= 10;
    digit2 *= 10;
    digit2 += 9;
    bool replace = false;
    for (auto ss : s)if (ss != '9')replace = true;
    if(replace)for (int i = 1; i < s.size(); i++)s[i] = '0';
    ll digit = stoll(s);
    if(replace)digit--;
    int sum2=0;
    while (digit > 9) {
        sum += digit % 10;
        digit /= 10;
    }
    while (digit2 > 9) {
        sum2 += digit2 % 10;
        digit2 /= 10;
    }
    cout << max(sum + digit,sum2 + digit2) << endl;
}