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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <algorithm>
#include <climits>
#include <cmath>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <unordered_map>
#include <vector>

const int mod = 1e9 + 7;
const int kmax = 510000;
const int last_days[] = { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };

long long fact[kmax], fact_inv[kmax], inv[kmax];

void init_comb()
{
    fact[0] = fact[1] = 1;
    fact_inv[0] = fact_inv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < kmax; i++) {
        fact[i] = fact[i - 1] * i % mod;
        inv[i] = mod - inv[mod % i] * (mod / i) % mod;
        fact_inv[i] = fact_inv[i - 1] * inv[i] % mod;
    }
}

long long comb(int n, int r)
{
    if (n < r) {
        return 0;
    }

    if (n < 0 || r < 0) {
        return 0;
    }

    return fact[n] * (fact_inv[r] * fact_inv[n - r] % mod) % mod;
}

template <typename T, T N>
class UnionFind {
    T parent_[N];
    T rank_[N];
    T size_[N];

public:
    UnionFind();
    T Root(T idx);
    bool IsSame(T x, T y);
    void Unite(T x, T y);
    T GetSize(T idx);
};

template <typename T, T N>
UnionFind<T, N>::UnionFind()
{
    for (T i = 0; i < N; i++) {
        parent_[i] = i;
        rank_[i] = 0;
        size_[i] = 1;
    }
}

template <typename T, T N>
T UnionFind<T, N>::Root(T idx)
{
    return parent_[idx] == idx ? idx : parent_[idx] = Root(parent_[idx]);
}

template <typename T, T N>
bool UnionFind<T, N>::IsSame(T x, T y)
{
    return Root(x) == Root(y);
}

template <typename T, T N>
void UnionFind<T, N>::Unite(T x, T y)
{
    x = Root(x);
    y = Root(y);
    if (x == y) {
        return;
    }

    if (rank_[x] < rank_[y]) {
        parent_[x] = y;
        size_[y] += size_[x];
    } else {
        parent_[y] = x;
        size_[x] += size_[y];
        if (rank_[x] == rank_[y]) {
            rank_[x]++;
        }
    }
}

template <typename T, T N>
T UnionFind<T, N>::GetSize(T idx)
{
    return size_[Root(idx)];
}

long long pow_mod(long long n, long long p, long long m)
{
    if (p == 0) {
        return 1;
    }

    if (p % 2 == 0) {
        long long t = pow_mod(n, p / 2, m);
        return t * t % m;
    }
    return n * pow_mod(n, p - 1, m) % mod;
}

long long nCr_mod(long long n, long long r)
{
    long long x = 1;
    for (long long i = n - r + 1; i <= n; i++) {
        x *= i;
        x %= mod;
    }

    long long a = 1;
    for (long long i = 1; i <= r; i++) {
        a *= i;
        a %= mod;
    }

    long long y = pow_mod(a, mod - 2, mod) % mod;

    return x * y % mod;
}

bool is_prime(long long n)
{
    if (n == 1) {
        return false;
    }

    for (long long i = 2; i * i <= n; i++) {
        if (n % i == 0) {
            return false;
        }
    }

    return true;
}

bool is_leap(long long y)
{
    return (y % 4 == 0 && y % 100 != 0) || y % 400 == 0;
}

void next_day(long long& y, long long& m, long long& d)
{
    d++;

    if (d > last_days[m - 1] + (m == 2 && is_leap(y) ? 1 : 0)) {
        d = 1;
        m++;
    }

    if (m > 12) {
        y++;
        m = 1;
    }
}

int main()
{
    long long h, w;
    std::cin >> h >> w;

    if (h % 2 == 0 && w % 2 == 0) {
        std::cout << h * w / 2 << std::endl;
        return 0;
    }

    if (h % 2 == 0) {
        std::cout << h * (w / 2) + h / 2 << std::endl;
        return 0;
    }

    if (w % 2 == 0) {
        std::cout << w * (h / 2) + w / 2 << std::endl;
        return 0;
    }

    std::cout << h * (w / 2) + h / 2 + 1 << std::endl;
}