Problem:
Score : 200 points

Problem Statement

Takahashi has two positive integers A and B.

It is known that A plus B equals N.
Find the minimum possible value of "the sum of the digits of A" plus "the sum of the digits of B" (in base 10).

Constraints

2 ≤ N ≤ 10^5

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the minimum possible value of "the sum of the digits of A" plus "the sum of the digits of B".

Sample Input 1

15

Sample Output 1

6

When A=2 and B=13, the sums of their digits are 2 and 4, which minimizes the value in question.

Sample Input 2

100000

Sample Output 2

10

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, j, n) for (ll i = j; i < n; i++)
#define all(x) (x).begin(),(x).end()
#define INF (1000000000)
#define MOD (1000000007)
#define MAX (110)
#define pii pair<int, int>

/////////////////////////////////////////////////////////

class XY{
public:
  ll x, y;
  XY() {x = y = 0;}
  XY(ll u, ll v) {x = u; y = v;}
};

template<typename T1, typename T2>
void chmin(T1 &a, T2 b) {if(a > b) a = b;}
template<typename T1, typename T2>
void chmax(T1 &a, T2 b) {if(a < b) a = b;}

template<typename T1, typename T2>
T1 mypow(T1 a, T2 n){
  if(n == 0) return 1;
  if(n == 1) return a;
  if(n % 2) return a * mypow(a, n - 1);
  T1 tmp = mypow(a, n / 2);
  return tmp * tmp;
}

template<typename T>
int BS(vector<T> V, int left, int right, T key){
  int mid = (left + right) / 2;
  if(V[mid] <= key) left = mid;
  else right = mid;
  if(right - mid == 1) return left;
  else return BS(V, left, right, key);
}

bool isPrime(int n){
  if(n == 1) return false;
  rep(i, 2, min((int)(pow(n, 0.5) + 1), n)){
    if(n % i == 0) return false;
  }
  return true;
}

/////////////////////////////////////////////////////////

void Main() {
  ll N; cin >> N;
  ll res = INF;
  rep(i, 1, N){
    ll A = i;
    ll B = N - A;
    int sum = 0;
    while(A > 0){
      sum += A % 10;
      A /= 10;
    }
    while(B > 0){
      sum += B % 10;
      B /= 10;
    }
    chmin(res, sum);
  }
  cout << res << endl;
}

/////////////////////////////////////////////////////////

int main(){
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  cout << std::fixed << std::setprecision(15);

  Main();
}