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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define REP(i, n) for(ll i = 0; i < (ll)n; i++)
#define FOR(i, a, b) for(ll i = (a); i < (ll)b; i++)
#define ALL(obj) (obj).begin(), (obj).end()
#define INF 1000000000000000
using namespace std;
typedef long long ll;
typedef double db;
typedef string str;
typedef pair<ll, ll> p;
constexpr int MOD = 1000000007;
template <class T> inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}
template <class T> inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}

void print(const std::vector<int> &v) {
    std::for_each(v.begin(), v.end(), [](int x) { std::cout << x << " "; });
    std::cout << std::endl;
}

int main() {
    int a, b;
    cin >> a >> b;
    if((a + b) % 2 == 0) {
        cout << (a + b) / 2 << endl;
    } else {
        cout << (a + b) / 2 + 1 << endl;
    }
    return 0;
}