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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<math.h>


int main(void){
    long long int N;
    long long int n;
    
    int max= 0;
    int keta = 0;
    int count = 0;
    scanf("%lld",&N);
    //printf("%lld\n",N);
    n = N;
    while(n != 0){
        if(n%10==9){
            count++;
        }
        keta ++; 
        max = n;
        n = n/10;
        
    }
    //printf("%d",count);
    /*
    for(long long int i=N;i>=N-20;i--){
        long long int n = i;
        int sum = 0;
        while(n != 0){
            sum += n%10;
            n = n/10;
        }
        if(sum>max){
            max = sum;
        }
    }
    */
    int ans = (keta-1)*9+max-1;
    if(count  == keta){
        ans ++;
    }
    double a = (int)(max+1)*pow(10,keta-1)-1;
    long long int b = (long long int)a;
    //printf("%lld",b);
    if(b==N){
        ans++;
    }
    printf("%d",ans);
    return 0;
}