#include<bits/stdc++.h>
using namespace std;
int mai()
{
	long long n,m;
	cin>>n>>m;
	cout<<n*m/2;
}

using namespace std;
int main()
{
	long long  n,m;
	cin>>n>>m;
	cout<<n*m/(1*2);
}

#include<bits/stdc++.h>//万能头
using namespace std;
int main()
{
    long long n,m,s;//定义
    cin>>n>>m;//输入
    s=(n*m)/(1*2);//拿n与m的体积去除以1*2算出能切几块
    cout<<s;//输出
    return 0;
}

#include<bits/stdc++.h> using namespace std;//l long long n,m;//j int main(){//m cin>>n>>m;//m cout<<n*m/2;//m return 0; }

#include<bits/stdc++.h>
#define ll long long
#define un unsigned
#define db double
#define st string
#define int long long
#define ct const
#define xh(a,b,c) for(int a=b;a<=c;a++)
#define wx while(1)
#define dn(a,b,c) for(int a=b;a>=c;a--)
using namespace std;
int n,m;
signed main(){
	cin>>n>>m;
	cout<<n*m/2;
	return 0;
}

#include<bits/stdc++.h>
using namespace std;
long n,m;
int main()
{
	cin>>n>>m;
	cout<<n*m/2; 
	return 0;
}

只需n与m的积/2即可（不管奇偶性） #include<bits/stdc++.h> using namespace std; long long n,m; int main() { freopen("a.in","r",stdin); freopen("a.out","w",stdout); cin>>n>>m; cout<<n*m/2; return 0; }