Codeforces Round #555 (Div. 3)------A. Reachable Numbers(set查找）

探讨一个数学函数f(x)，通过特定规则将x增加1并去除尾部零，直到无法去除为止。挑战在于找出从给定数字n出发，通过多次应用该函数可以达到的不同数字的数量。

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A. Reachable Numbers

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Let's denote a function f(x)f(x) in such a way: we add 11 to xx, then, while there is at least one trailing zero in the resulting number, we remove that zero. For example,

f(599)=6f(599)=6: 599+1=600→60→6599+1=600→60→6;
f(7)=8f(7)=8: 7+1=87+1=8;
f(9)=1f(9)=1: 9+1=10→19+1=10→1;
f(10099)=101f(10099)=101: 10099+1=10100→1010→10110099+1=10100→1010→101.

We say that some number yy is reachable from xx if we can apply function ff to xx some (possibly zero) times so that we get yy as a result. For example, 102102 is reachable from 1009810098 because f(f(f(10098)))=f(f(10099))=f(101)=102f(f(f(10098)))=f(f(10099))=f(101)=102; and any number is reachable from itself.

You are given a number nn; your task is to count how many different numbers are reachable from nn.

Input

The first line contains one integer nn (1≤n≤1091≤n≤109).

Output

Print one integer: the number of different numbers that are reachable from nn.

Examples

input

Copy

output

Copy

input

Copy

output

Copy

Note

The numbers that are reachable from 10981098 are:

1,2,3,4,5,6,7,8,9,11,12,13,14,15,16,17,18,19,1098,10991,2,3,4,5,6,7,8,9,11,12,13,14,15,16,17,18,19,1098,1099.

https://codeforces.com/contest/1157/problem/A

#include<set>
#include<map>
#include<cmath>
#include<vector>
#include<cstdio>
#include<string>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include <fstream>
#include<algorithm>
using namespace std;
int main()
{
    int n;
    cin>>n;
    set<int>s;
    set <int> :: iterator it;;
    s.insert(n);
    n++;
    while(n%10==0)
    {
        if(n%10==0)
            n=n/10;
    }
    while(1)
    {
        if(s.count(n)==0)
        {
            s.insert(n);
              n++;
            while(n%10==0)
            {
                if(n%10==0)
                    n=n/10;
            }
        }
        else if(s.count(n)==1)
        {
            break;
        }
    }
    cout<< s.size()<<endl;
}