#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; }