142B - Help General - CodeForces Solution

// #pragma GCC optimize("Ofast,unroll-loops")
// #pragma GCC target("avx2,popcnt,lzcnt,abm,bmi,bmi2,fma,tune=native")
// #pragma expected_value
// #pragma isolated_call
// #pragma disjoint

#include <bits/stdc++.h>
using namespace std;

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using ordered_set = tree< int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>;

#define ll long long
#define ull unsigned long long
#define ld long double
#define pb push_back
#define setpre(i) setprecision(i)<<fixed
#define foru(i, a, b) for(int i=a;i<=b;++i)
#define ford(i, a, b) for(int i=a;i>=b;--i)
#define mp make_pair
#define moo cout<<"moo "
#define fi first
#define se second

ll lmod = 1e9 + 7;
typedef vector<int> vi;
typedef vector< vector<int> > vvi;
typedef pair<int, int> ii;
typedef pair<double, double> dd;

int a[] = {0, 0, 1, 4};

int main(){
    #ifdef LOCAL
    freopen("input.inp", "r", stdin);
    freopen("output.out", "w", stdout);
    #endif
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

    int n, m; cin >> n >> m;
    if(n > m) swap(n, m);
    if(n == 1){
        cout << m << endl;
        return 0;
    }

    if(n == 2){
        cout << m / 4 * 4 + min((m % 4), 2) * 2 << endl;
        return 0;
    }

    if(n == 3){
        cout << max({(n / 2 + (n % 2)) * (m / 2 + (m % 2)), m * (1 + (n - 1) / 3), n * (1 + (m - 1) / 3), 4 + (m - 2) / 4 * 5 + a[(m - 2) % 4], (n * m) - (n * m) / 2}) << endl;
        return 0;
    }

    cout << max({(n / 2 + (n % 2)) * (m / 2 + (m % 2)), m * (1 + (n - 1) / 3), n * (1 + (m - 1) / 3), n * m - (n * m) / 2}) << endl;

    return 0;
}