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 (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

// 標準入力から改行かスペースまで読み取り、数字として返す
int get_num() {
    // 10文字+区切り文字までしか受け付けられない
    char s[11];
    int i, n;
    for (i = 0; i < sizeof(s); i++) {
        s[i] = getchar();
        if (s[i] == ' ' || s[i] == '\n') {
            s[i] = '\0';
            break;
        }
    }
    sscanf(s, "%d", &n);
    return n;
}

int main(void) {
    int A, B, N, i, minimum = INT_MAX, sum;

    N = get_num();
    
    for (i = 1; i < N; i++) {
        A = i;
        B = N - A;
        sum = 0;
        while(A > 0) {
            sum += A % 10;
            A /= 10;
        }
        while(B > 0) {
            sum += B % 10;
            B /= 10;
        }
        if (minimum > sum) {
            minimum = sum;
        }
    }
    
    printf("%d\n", minimum);
    
    return 0;
}