https://ac.nowcoder.com/acm/contest/318/G

题解：本题考验的是简单的贪心思维。通过题目我们可以了解到要使得 f (x)尽可能的大，就是要使得x 中的9尽可能的多。所以对于本题，我们可以考虑构造出一个最大的a，使得这个a满足它的每一位都是9且a<=c，接着我们再令b = c - a，将a 和 b 分别代入 f (x)中，求得最后的答案即可。

C++版本一

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
using namespace std;
typedef long long ll;
typedef __int128 lll;
const int N=10000;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
ll t,n,m,k,q;
int f(ll x){
    int tmp = 0;
    while(x != 0){
    tmp += x % 10;
    x /= 10;
    }
    return tmp;
}
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    while(~scanf("%lld",&n)){
        ll i=9;
        for(;i<=n;i=i*10+9);
        cout << f(i/10)+f(n-i/10) << endl;
    }

    //cout << "Hello world!" << endl;
    return 0;
}

C++版本二

#include<bits/stdc++.h>
#define ll long long
using namespace std;


int main() {
    ll n, a, b;
    while(cin >> n) {
        ll m = n, k = 9;
        while(m >= k) {
            k = k * 10 + 9;
        }
        k /= 10;
        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;
}