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: Time Limit Exceeded)

#include<stdio.h>
#include<math.h>
int main(void){
    long int C,Q,R=0,C1,P;
    int j;
    scanf("%ld",&C);
    int X[16],Y[16];
    C1=C;
    for(int i=0;i<16;i++){X[i]=0;Y[i]=0;}
    for(int i=0;i<16;i++){
        X[i]=C%10;
        C=(C-X[i])/10;
        if(C==0){j=i+1;break;}
    }
    Q=X[j-1]-1+9*(j-1);
    for(long int i=Q+1;i<=C1;i++){
        R=0;
        P=i;
        for(int k=0;k<=j-1;k++){
            Y[k]=P%10;
            R+=Y[k];
            P=(P-Y[k])/10;
        }
        if(R>Q)Q=R;
    }
    printf("%ld\n",Q);
    return 0;
}