Problem:
Score : 100 points

Problem Statement

You are given two positive integers a and b.
Let x be the average of a and b.
Print x rounded up to the nearest integer.

Constraints

a and b are integers.

1 \leq a, b \leq 100

Input

Input is given from Standard Input in the following format:

a b

Output

Print x rounded up to the nearest integer.

Sample Input 1

1 3

Sample Output 1

2

The average of 1 and 3 is 2.0, and it will be rounded up to the nearest integer, 2.

Sample Input 2

7 4

Sample Output 2

6

The average of 7 and 4 is 5.5, and it will be rounded up to the nearest integer, 6.

Sample Input 3

5 5

Sample Output 3

5

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

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

int cmp(const void *a, const void *b) {
    return *(char *)a - *(char *)b;
}

int rcmp(const void *a, const void *b) {
    return *(char *)b - *(char *)a;
}

int main(void) {
    char s[100], t[100];
    int _ign = scanf("%s %s", s, t);
    qsort(s, strlen(s), sizeof(char), cmp);
    qsort(t, strlen(t), sizeof(char), rcmp);
    for (int i = 0; ; i++) {
        if (!s[i] && !t[i]) {
            printf("No");
            break;
        }

        if (!s[i] && t[i]) {
            if (s[i - 1] == t[i - 1]) {
                printf("Yes");
            } else {
                printf("No");
            }
            break;
        }

        if (s[i] && !t[i]) {
            if (s[i - 1] == t[i - 1]) {
                printf("No");
            } else {
                printf("Yes");
            }
            break;
        }

        if (s[i] < t[i]) {
            printf("Yes");
            break;
        }
    }
    return 0;
}