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)

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<sstream>
#include<vector>
#include<algorithm>
#include<string.h>
#include<set>
#include<fstream>
#include<map>
#include<string>
#include <stdio.h>
#include<bitset>
#include<queue>
#include<iomanip>
#include<cmath>
#include<stack>
#include<climits>

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<ll>vl;
typedef vector<int>vi;
typedef vector<bool>vb;
typedef vector<double>vd;
typedef vector<char>vc;
typedef vector<string>vs;
#define MP make_pair
#define all(v) v.begin(),v.end()
const ll mod = 1000000007;
const ll OO = (ll)1e18;
const int dx[] = { 0, 1, -1, 0 };
const int dy[] = { 1, 0, 0, -1 };
ll gcd(ll a, ll b){ if (b == 0){ return a; }return gcd(b, a % b); }
ll fast_power(ll base, ll power){
	if (power == 1) return base;
	if (power % 2 == 0) return fast_power((base*base), power / 2);
	else return(base*fast_power((base*base), power / 2));
}
//#pragma warning (disable : 4996)
void Qalbaz()
{
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif
	
	cout << fixed << setprecision(2);
}

int main(){
	//Qalbaz();
ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);	
	int a, b, ans = 0, x;
	cin >> a >> b;
	for (int i = min(a, b); i <= max(a, b); i++)
		ans += i;
	x = abs(a - b) + 1;
	double y = (double)ans / x;
	cout << (int)(y + .5);
	return 0;
}