Maximum Sum of Digits（CodeForces 1060B）

Description

You are given a positive integer $\mathrm{nn.}$

Let $\mathrm{S(x) be\; sum\; of\; digits\; in\; base\; 10\; representation\; of xx,\; for\; example, S(123)=1+2+3=6, S(0)=0.}$

Your task is to find two integers $\mathrm{a,ba,b,\; such\; that 0\le a,b\le n, a+b=n and S(a)+S(b) is\; the\; largest\; possible\; among\; all\; such\; pairs.}$

Input

The only line of input contains an integer $\mathrm{nn (1\le n\le 1012).}$

Output

Print largest $\mathrm{S(a)+S(b) among\; all\; pairs\; of\; integers a,ba,b,\; such\; that 0\le a,b\le n and a+b=n.}$

Sample Input

Input

35

Output

17

Input

10000000000

Output

91

Hint

In the first example, you can choose, for example, $\mathrm{a=17 and b=18,\; so\; that S(17)+S(18)=1+7+1+8=17.\; It\; can\; be\; shown\; that\; it\; is\; impossible\; to\; get\; a\; larger\; answer.}$

In the second test example, you can choose, for example, $\mathrm{a=5000000001 and b=4999999999,\; with S(5000000001)+S(4999999999)=91.\; It\; can\; be\; shown\; that\; it\; is\; impossible\; to\; get\; a\; larger\; answer.}$

相信不少人wa是这个代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<algorithm>
#define max(a,b)   (a>b?a:b)
#define min(a,b)   (a<b?a:b)
#define swap(a,b)  (a=a+b,b=a-b,a=a-b)
#define maxn 320007
#define N 100000000
#define INF 0x3f3f3f3f
#define mod 1000000009
#define e  2.718281828459045
#define eps 1.0e18
#define PI acos(-1)
#define lowbit(x) (x&(-x))
#define read(x) scanf("%d",&x)
#define put(x) printf("%d
",x)
#define memset(x,y) memset(x,y,sizeof(x))
#define Debug(x) cout<<x<<" "<<endl
#define lson i << 1,l,m
#define rson i << 1 | 1,m + 1,r
#define ll long long
//std::ios::sync_with_stdio(false);
//cin.tie(NULL);
using namespace std;


int main()
{
    ll n,a,b;
    cin>>n;
    ll m=n,k=9;
    while(m>=k)
    {
        k=k*10+9;
    }
    k/=10;
    //cout<<k<<endl;
    a=k;
    b=n-k;
    ll sum=0;
    while(a)
    {
        sum+=a%10;
        a/=10;
    }
    while(b)
    {
        sum+=b%10;
        b/=10;
    }
    cout<<sum<<endl;
    return 0;
}

你们可以尝试一下123这个样例，输出为15,但正确答案为24（99+24）。

$\mathrm{AC代码如下：}$

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<algorithm>
#define max(a,b)   (a>b?a:b)
#define min(a,b)   (a<b?a:b)
#define swap(a,b)  (a=a+b,b=a-b,a=a-b)
#define maxn 320007
#define N 100000000
#define INF 0x3f3f3f3f
#define mod 1000000009
#define e  2.718281828459045
#define eps 1.0e18
#define PI acos(-1)
#define lowbit(x) (x&(-x))
#define read(x) scanf("%d",&x)
#define put(x) printf("%d
",x)
#define memset(x,y) memset(x,y,sizeof(x))
#define Debug(x) cout<<x<<" "<<endl
#define lson i << 1,l,m
#define rson i << 1 | 1,m + 1,r
#define ll long long
//std::ios::sync_with_stdio(false);
//cin.tie(NULL);
using namespace std;


int main()
{
    ll n,a,b;
    cin>>n;
    ll m=n,k=9;
    while(m>=k)
    {
        k=k*10+9;
    }
    k/=10;
    //cout<<k<<endl;
    a=k;
    b=n-k;
    ll sum=0;
    while(a)
    {
        sum+=a%10;
        a/=10;
    }
    while(b)
    {
        sum+=b%10;
        b/=10;
    }
    cout<<sum<<endl;
    return 0;
}