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: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#define ll long long
#define rep(i,l,r) for(ll i=(l);i<(r);i++)
#define max(p,q) ((p)>(q)?(p):(q))
#define min(p,q) ((p)<(q)?(p):(q))
#define chmax(a,b) ((a)=(a)>(b)?(a):(b))
#define chmin(a,b) ((a)=(a)<(b)?(a):(b))
#define abs(p) ((p)>=(0)?(p):(-(p)))
int cmp(const void *a, const void *b);
//your code here!

int main(void){
    ll a, b;
    scanf("%lld %lld", &a, &b);
    if(a%2==b%2) printf("%lld\n", (a+b)/2);
    else printf("%lld\n", (a+b)/2+1);
    return 0;
}

int cmp(const void *a, const void *b){
    ll A=*(ll *)a, B=*(ll *)b;
    if(A==B)return 0;
    else return A>B ? 1:-1;//昇順ソート
}