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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int devide(int x){
	int sum;
	if(x>=10000){
		sum=x/10000;	x=x-sum*10000;
		sum+=x/1000;	x=x-(x/1000)*1000;
		sum+=x/100;		x=x-(x/100)*100;
		sum+=x/10;		x=x-(x/10)*10;
		sum+=x;
	}else if(x>=1000){
		sum=x/1000;	x=x-(x/1000)*1000;
		sum+=x/100;		x=x-(x/100)*100;
		sum+=x/10;		x=x-(x/10)*10;
		sum+=x;
	}else if(x>=100){
		sum=x/100;		x=x-(x/100)*100;
		sum+=x/10;		x=x-(x/10)*10;
		sum+=x;
	}else if(x>=10){
		sum=x/10;		x=x-(x/10)*10;
		sum+=x;
	}else{
		sum=x;
	}
	
	return sum;
}

int main(void)
{
	int N;
	scanf("%d",&N);
	int a,min=100;
	for(a=1;a<N/2;a++){
		int b=N-a;
		int suma=devide(a);
		int sumb=devide(b);
		if(sumb+suma<min){
			min=sumb+suma;
		}
	}
	printf("%d\n",min);
	return 0;
}