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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#642082 | #8577. 평균 최대화 | user10086 | 0 | 0ms | 42112kb | C++17 | 4.9kb | 2024-10-15 09:52:53 | 2024-10-15 09:52:55 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 3e5 + 10, inf = 1e13;
int n, idx, l[N], r[N], a[N], s[N], sz[N], rx[N], rs[N];
#define __int128 int
struct F
{
int i, j;
F()
{
i = j = 0;
}
F(int a, int b)
{
int d = __gcd(a, b);
i = a / d, j = b / d;
}
bool operator> (const F& f2) const
{
return (__int128)i * f2.j > (__int128)j * f2.i;
}
bool operator< (const F& f2) const
{
return (__int128)i * f2.j < (__int128)j * f2.i;
}
F operator + (const F& f2) const
{
__int128 a = (__int128)i * f2.j + (__int128)j * f2.i, b = (__int128)j * f2.j;
auto d = __gcd(a, b);
return {a / d, b / d};
}
void reduce()
{
auto d = __gcd(i, j);
i /= d, j /= d;
}
}ans[N];
struct DS
{
vector<F> d;
void insert(F v)
{
d.insert(lower_bound(d.begin(), d.end(), v), v);
}
void assign(int pos, F v)
{
assert(pos <= d.size());
for (int i = 0; i < pos; i++) d[i] = v;
}
F operator[] (int x) const
{
F s = {0, 1};
for (int i = 0; i < x; i++) s = s + d[i];
return s;
}
void print() const
{
for (auto v : d) printf("%lld/%lld ", v.i, v.j); printf("\n");
}
};
DS ds[N];
DS* dp[N];
vector<int> son[N];
vector<int> rg[N];
map<array<int, 2>, int> mp;
int build(int l, int r)
{
idx++, ::l[idx] = l, ::r[idx] = r, mp[{l, r}] = idx;
sz[idx] = 1;
int x = idx; rx[x] = r - l + 1, rs[x] = s[r] - s[l - 1];
for (int i = l; i <= r; i++)
{
if (rg[i].empty()) continue;
int j = rg[i].back(); rg[i].pop_back();
int y = build(i, j);
son[x].push_back(y);
rx[x] -= (j - i + 1), rs[x] -= (s[j] - s[i - 1]);
i = j;
}
return x;
}
template<class T>
void chkmax(T &x, T y)
{
if (y > x) x = y;
}
DS* merge(DS* a, DS* b)
{
if (a->d.size() < b->d.size()) swap(a, b);
for (F v : b->d) a->insert(v);
return a;
}
pair<int, F> getp(const DS* ds, int a, int b)
{
// max (y - b) / (x - a)
assert(!ds->d.empty());
int l = 0, r = ds->d.size();
auto getres = [&](int x)
{
F y1 = (*ds)[x];
y1.i -= b * y1.j, y1.j *= (x - a);
y1.reduce();
return y1;
};
while (l + 1 < r)
{
int mid1 = l + (r - l + 1) / 3, mid2 = r - (r - l) / 3;
F y1 = getres(mid1), y2 = getres(mid2);
if (y1 < y2) l = mid1 + 1;
else if (y1 > y2) r = mid2 - 1;
else l = mid1, r = mid2;
}
// printf("res(l) = (%lld, %lld/%lld), res(r) = (%lld, %lld/%lld)\n", l, getres(l).i, getres(l).j, r, getres(r).i, getres(r).j);
if (getres(l) > getres(r)) return {l, getres(l)};
else return {r, getres(r)};
}
void dfs(int x)
{
// printf("dfs(%lld)\n", x);
dp[x] = &ds[x];
if (son[x].empty())
{
if (rx[x]) for (int i = 1; i <= rx[x]; i++) dp[x]->insert({rs[x], rx[x]});
}
else
{
dp[x] = &ds[x];
for (int y : son[x]) dfs(y), dp[x] = merge(dp[x], dp[y]);
// dp[x]->print();
// (-rx[x], -rs[x])
auto res = getp(dp[x], -rx[x], -rs[x]);
// printf("proc: res = {%lld, %lld/%lld}", res.first, res.second.i, res.second.j);
int p = res.first;
F v = res.second;
dp[x]->assign(p, v);
for (int i = 1; i <= rx[x]; i++) dp[x]->insert(v);
}
// dp[x]->print();
auto res = getp(dp[x], -2, -(a[l[x] - 1] + a[r[x] + 1]));
ans[x] = res.second;
// printf("%lld: (%lld, %lld/%lld)\n", x, res.first, ans[x].i, ans[x].j);
}
void initialize(vector<signed> A)
{
n = (int)(A.size());
for (int i = 1; i <= n; i++) a[i] = A[i - 1];
for (int i = 1; i <= n; i++) s[i] = a[i] + s[i - 1];
vector<int> s;
for (int i = n; i >= 1; i--)
{
while (!s.empty() && a[s.back()] > a[i])
{
int x = s.back(); s.pop_back();
if (x != i + 1) rg[i + 1].push_back(x - 1);
}
if (!s.empty() && s.back() != i + 1) rg[i + 1].push_back(s.back() - 1);
if (!s.empty() && a[s.back()] >= a[i]) s.pop_back();
s.push_back(i);
}
// for (int i = 1; i <= n; i++)
// for (int j : rg[i]) printf("[%lld, %lld]\n", i, j);
build(2, n - 1);
// for (int i = 1; i <= n; i++)
// printf("rx[%lld] = %lld, rs[%lld] = %lld\n", i, rx[i], i, rs[i]);
dfs(1);
// for (int i = 1; i <= n; i++)
// for (int j = 0; j <= sz[i]; j++)
// printf("dp[%lld][%lld] = %lld\n", i, j, dp[i][j]);
// for (int i = 1; i <= n; i++)
// for (int j : son[i]) printf("[%lld, %lld] -> [%lld, %lld]\n", l[i], r[i], l[j], r[j]);
}
array<long long, 2> maximum_average(signed i, signed j)
{
i++, j++;
if (j == i + 1) return {a[i] + a[j], 2};
assert(mp.find({i + 1, j - 1}) != mp.end());
int id = mp[{i + 1, j - 1}];
return {ans[id].i, ans[id].j};
}
//signed main()
//{
// int n; cin >> n;
// vector<signed> A;
// for (int i = 1, ai; i <= n; i++) cin >> ai, A.push_back(ai);
// initialize(A);
// while (1)
// {
// int l, r; cin >> l >> r;
// auto res = maximum_average(l, r);
// cout << res[0] << ' ' << res[1] << endl;
// }
//}
Details
Tip: Click on the bar to expand more detailed information
Subtask #1:
score: 0
Wrong Answer
Test #1:
score: 0
Wrong Answer
time: 0ms
memory: 42112kb
input:
10 2 4 3 9 9 9 9 9 9 1 2 0 2 0 9
output:
3 1 32 5
result:
wrong answer Wrong Answer on query #2: 32/5 != 60/9
Subtask #2:
score: 0
Skipped
Dependency #1:
0%
Subtask #3:
score: 0
Skipped
Dependency #2:
0%
Subtask #4:
score: 0
Time Limit Exceeded
Test #15:
score: 0
Time Limit Exceeded
input:
300000 1 2 4 4 4 4 3 2 4 4 3 4 4 4 4 4 4 4 4 4 3 4 3 4 4 4 4 4 4 4 4 3 3 4 4 4 3 4 3 4 4 4 4 4 4 4 4 4 4 3 3 4 4 4 3 4 4 3 4 4 4 4 4 4 4 3 2 4 4 4 4 4 4 4 4 4 4 4 4 4 3 4 4 4 4 4 4 4 4 4 4 4 2 4 4 2 4 4 3 4 4 4 2 3 4 4 4 4 4 4 3 2 4 4 4 2 4 4 4 4 4 4 4 4 4 4 4 2 4 4 4 4 4 4 3 4 4 3 4 4 4 4 4 4 4 4 4...
output:
Unauthorized output
result:
Subtask #5:
score: 0
Skipped
Dependency #3:
0%
Subtask #6:
score: 0
Time Limit Exceeded
Test #28:
score: 0
Time Limit Exceeded
input:
300000 1 300000 300001 299999 300003 299998 300005 299997 300007 299996 300009 299995 300011 299994 300013 299993 300015 299992 300017 299991 300019 299990 300021 299989 300023 299988 300025 299987 300027 299986 300029 299985 300031 299984 300033 299983 300035 299982 300037 299981 300039 299980 3000...
output:
Unauthorized output
result:
Subtask #7:
score: 0
Skipped
Dependency #4:
0%
Subtask #8:
score: 0
Skipped
Dependency #1:
0%