没什么好讲的，这里放一个bitset的做法

const ll N = 1e6+10;
ll n, x, y;
bitset <N> f(1);
void work()
{
  cin>>n;
  for (ll i=1; i<=n; i++) {
    cin>>x>>y;
    f = (f<<(x*x))|(f<<(y*y));
  }
  cout<<f.count();
}