令所有的 \(w<h\) ，一个集合的 \(res=min(w)\times min(h)\)
按 \(w\) 排序，假设一开始的分组为 \([1,n],[n+1,2n]\)，每次考虑加一个数到后一个区间内，就要维护后缀区间的第 \(n+1\) 大和第 \(n\) 大的数，考虑用 \(2\) 个multiset 维护，并预处理前缀 \(h\) 的 \(min\)
// Author: xiaruize
#ifndef ONLINE_JUDGE
#define debug(x) cerr << "On Line:" << __LINE__ << #x << "=" << x << endl
bool start_of_memory_use;
#else
#define debug(x)
#endif
#include <bits/stdc++.h>
using namespace std;
#ifndef ONLINE_JUDGE
clock_t start_clock = clock();
#endif
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define mms(arr, n) memset(arr, n, sizeof(arr))
#define rep(i, a, n) for (int i = (a); i <= (n); ++i)
#define per(i, n, a) for (int i = (n); i >= (a); --i)
int max(int a, int b)
{
if (a > b)
return a;
return b;
}
int min(int a, int b)
{
if (a < b)
return a;
return b;
}
const int INF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e5 + 10;
int n, m;
pii a[N];
int x[N], y[N];
int XA, XB, XC;
int YA, YB, YC;
int mi[N];
multiset<int> s, t;
bool cmp(pii a, pii b)
{
return a.second < b.second;
}
void solve()
{
cin >> n;
n *= 2;
cin >> m;
rep(i, 1, m) cin >> x[i];
cin >> m;
rep(i, 1, m) cin >> y[i];
cin >> XA >> XB >> XC >> YA >> YB >> YC;
rep(i, m + 1, n)
{
x[i] = (x[i - 1] * XA + XB) % XC + 1;
y[i] = (y[i - 1] * YA + YB) % YC + 1;
}
int maxn = 0;
rep(i, 1, n)
{
if (x[i] > y[i])
swap(x[i], y[i]);
a[i] = {x[i], y[i]};
}
sort(a + 1, a + n + 1);
mi[0] = INF;
rep(i, 1, n) mi[i] = min(mi[i - 1], a[i].second);
rep(i, n / 2 + 1, n)
{
s.insert(a[i].second);
t.insert(a[i].second);
}
int res = a[1].first * mi[n / 2] + a[n / 2 + 1].first * (*t.begin());
per(i, n / 2, 1)
{
s.insert(a[i].second);
while (s.size() > n / 2 + 1)
s.erase(prev(s.end()));
t.insert(a[i].second);
while (t.size() > n / 2)
t.erase(t.begin());
res = max(res, a[1].first * min(mi[i - 1], *s.rbegin()) + a[i].first * (*s.begin()));
res = max(res, a[1].first * min(mi[i - 1], *s.begin()) + a[i].first * (*t.begin()));
}
cout << res << endl;
}
#ifndef ONLINE_JUDGE
bool end_of_memory_use;
#endif
signed main()
{
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int testcase = 1;
// cin >> testcase;
while (testcase--)
solve();
#ifndef ONLINE_JUDGE
cerr << "Memory use:" << (&end_of_memory_use - &start_of_memory_use) / 1024.0 / 1024.0 << "MiB" << endl;
cerr << "Time use:" << (double)clock() / CLOCKS_PER_SEC * 1000.0 << "ms" << endl;
#endif
return 0;
}