Problem:
Score : 200 points

Problem Statement

We have a board with H horizontal rows and W vertical columns of squares.
There is a bishop at the top-left square on this board.
How many squares can this bishop reach by zero or more movements?

Here the bishop can only move diagonally.
More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds:

r_1 + c_1 = r_2 + c_2

r_1 - c_1 = r_2 - c_2

For example, in the following figure, the bishop can move to any of the red squares in one move:

Constraints

1 \leq H, W \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

H \ W

Output

Print the number of squares the bishop can reach.

Sample Input 1

4 5

Sample Output 1

10

The bishop can reach the cyan squares in the following figure:

Sample Input 2

7 3

Sample Output 2

11

The bishop can reach the cyan squares in the following figure:

Sample Input 3

1000000000 1000000000

Sample Output 3

500000000000000000

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

typedef unsigned long long int ull;
typedef long double ldb;
typedef long long int ll;

#define MOD1        1000000007
#define MOD2        998244353
#define PI          3.1415926535897932384626433832795
#define FILL(a, value) memset(a, value, sizeof(a))
#define BOOST       ios_base::sync_with_stdio(false); cin.tie(NULL)
#define what_is(x)  cerr << #x << " is " << x << endl;
#define ASSERT(left,operator,right) { if(!((left) operator (right))){ std::cerr << "ASSERT FAILED: " << #left << #operator << #right << " @ " << __FILE__ << " (" << __LINE__ << "). " << #left << "=" << (left) << "; " << #right << "=" << (right) << std::endl; } }

#define rep01(k,n)      for (ll i=k; i<n; i++)
#define rep02(k,n)      for (ll j=k; j<n; j++)
#define rep03(k,n)      for (ll k=k; k<n; k++)
#define rep11(k,n)      for (ll i=k; i<=n; i++)
#define rep12(k,n)      for (ll j=k; j<=n; j++)
#define rep13(k,n)      for (ll k=k; k<=n; k++)
#define repIT(c,it)     for(typeof(c.begin()) it = c.begin(); it != c.end(); ++it)
#define all(c)      c.begin(), c.end()

using namespace std;

const double pi = acos(-1.0);
const double epsilon = 1e-9;        // Add epsilon when casting double to int to avoid rounding pitfall issues, eg, 0.94*8178 in int
const int maxn = 1e6+10;


void solve()
{
    ll h, w;
    cin >> h >> w;

    if(h == 1 || w == 1)
        cout << 1;
    else
    {
        ll ans = ceil(w*0.5)*(ceil(h*0.5)) + floor(w*0.5)*floor(h*0.5);
        cout << ans;
    }
    
}

int main()
{
    

    int t = 1;
    //cin >> t;
    while(t--)
        solve();

    cerr << "Time elapsed: " << clock() / (double)CLOCKS_PER_SEC << endl;
    return 0;
}