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 main(void){
    int a,b,sum;
    int min=99999;
    int box=0;

    printf("Input the natural number SUM:");
    scanf("%d",&sum);

    int i,j;
    for(a=1;a<sum;a++){
        b=sum-a;
        if(a>=10000){
            box+=a/10000;
            box+=(a/1000)%10;
            box+=(a/100)%10;
            box+=(a/10)%10;
            box+=a%10;
        }
        else if(a>=1000){
            box+=a/1000;
            box+=(a/100)%10;
            box+=(a/10)%10;
            box+=a%10;
        }
        else if(a>=100){
            box+=a/100;
            box+=(a/10)%10;
            box+=a%10;
        }
        else if(a>=10){
            box+=a/10;
            box+=a%10;
        }
        else{
            box+=a%10;
        }
         if(b>=10000){
            box+=b/10000;
            box+=(b/1000)%10;
            box+=(b/100)%10;
            box+=(b/10)%10;
            box+=b%10;
        }
        else if(b>=1000){
            box+=b/1000;
            box+=(b/100)%10;
            box+=(b/10)%10;
            box+=b%10;
        }
        else if(b>=100){
            box+=b/100;
            box+=(b/10)%10;
            box+=b%10;
        }
        else if(b>=10){
            box+=b/10;
            box+=b%10;
        }
        else{
            box+=b%10;
        }

        if(box<min){
            min=box;
        }
    }
    printf("%d",min);

   
}