QOJ.ac
QOJ
ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#670734 | #7627. Phony | TLE_Automat | WA | 1ms | 3760kb | C++20 | 7.8kb | 2024-10-23 23:46:20 | 2024-10-23 23:46:21 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
#define SZ(x) ((int)((x).size()))
#define lb(x) ((x) & (-(x)))
#define bp(x) __builtin_popcount(x)
#define bpll(x) __builtin_popcountll(x)
#define mkp make_pair
#define pb push_back
#define fi first
#define se second
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, int> pli;
typedef pair<ll, ll> pll;
typedef pair<double, int> pdi;
mt19937 mrnd(std::chrono::steady_clock::now().time_since_epoch().count());
int rnd(int l, int r) {
return mrnd() % (r - l + 1) + l;
}
double rnd01() {
return mrnd() * 1.0 / UINT32_MAX;
}
namespace Fhq_Treap {
const int MAXN = 5e5 + 10;
struct Node {
int lson, rson;
int pri, sz;
ll val;
};
int cnt_id;
Node tree[MAXN << 4];
queue<int> rub;
#define ls(x) tree[x].lson
#define rs(x) tree[x].rson
struct Tree {
int rt = 0;
void clear(int x) {
tree[x].val = tree[x].pri = tree[x].sz = 0;
rub.push(x);
}
void all_clear() {
auto vec = iter_id();
for (auto x : vec) {
clear(x);
}
}
vector<int> iter_id() { // 中序遍历
vector<int> res;
auto dfs = [&](auto &&self, int x) -> void {
if (ls(x)) {
self(self, ls(x));
}
res.push_back(x);
if (rs(x)) {
self(self, rs(x));
}
};
dfs(dfs, rt);
return res;
}
vector<ll> iter_val() {
vector<ll> res;
auto dfs = [&](auto &&self, int x) -> void {
if (ls(x)) {
self(self, ls(x));
}
res.push_back(tree[x].val);
if (rs(x)) {
self(self, rs(x));
}
};
dfs(dfs, rt);
return res;
}
int newp(ll val) {
int cur;
if (!rub.empty()) {
cur = rub.front();
rub.pop();
} else {
cur = ++cnt_id;
}
tree[cur].val = val;
tree[cur].pri = rnd(1, 114514);
tree[cur].sz = 1;
return cur;
}
void push_up(int x) {
tree[x].sz = tree[ls(x)].sz + tree[rs(x)].sz + 1;
}
void split(int x, int &l, int &r, ll val) {
if (!x) {
l = r = 0;
return ;
}
if (tree[x].val <= val) {
l = x;
split(rs(x), rs(l), r, val);
} else {
r = x;
split(ls(x), l, ls(r), val);
}
push_up(x);
}
int merge(int x, int y) {
if (!x || !y) {
return x ^ y;
}
if (tree[x].pri <= tree[y].pri) {
rs(x) = merge(rs(x), y);
return push_up(x), x;
} else {
ls(y) = merge(x, ls(y));
return push_up(y), y;
}
}
void insert(ll val) {
int l_rt, r_rt;
split(rt, l_rt, r_rt, val);
rt = merge(merge(l_rt, newp(val)), r_rt);
}
void del(ll val) {
int l_rt, r_rt, mid_rt;
split(rt, l_rt, r_rt, val);
split(l_rt, l_rt, mid_rt, val - 1);
mid_rt = merge(ls(mid_rt), rs(mid_rt));
rt = merge(merge(l_rt, mid_rt), r_rt);
}
int size() {
return tree[rt].sz;
}
int greater_equal_cnt(ll val) {
int l_rt, r_rt;
split(rt, l_rt, r_rt, val - 1);
int res = tree[r_rt].sz;
rt = merge(l_rt, r_rt);
return res;
}
ll kth_mx(int k) {
int x = rt;
while (true) {
if (tree[rs(x)].sz + 1 < k) {
k -= tree[rs(x)].sz + 1;
x = ls(x);
}
else if (tree[rs(x)].sz + 1 > k) {
x = rs(x);
}
else {
return tree[x].val;
}
}
}
};
}
using Fhq_Treap::Tree;
void solve() {
int n, m; ll k;
cin >> n >> m >> k;
vector a(n + 1, 0ll);
const ll D = 1e18;
for (int i = 1; i <= n; i++) {
cin >> a[i];
a[i] += D;
}
sort(a.begin() + 1, a.begin() + n + 1);
map<ll, int> rid;
map<int, ll> pt;
map<int, int> pos;
int mxid = 0, secid = 0;
for (int i = 1; i <= n; i++) {
ll p = a[i] / k;
if (!rid[p]) {
rid[p] = ++mxid;
pt[mxid] = p;
pos[mxid] = i - 1;
}
}
secid = mxid - 1;
vector tr(n + 1, Tree());
for (int i = 1; i <= n; i++) {
ll p = a[i] / k, q = a[i] % k;
tr[rid[p]].insert(q);
}
int mxdc = 0;
auto merge = [&](int x, int y) -> int {
if (tr[x].size() > tr[y].size()) {
swap(x, y);
}
auto vec = tr[x].iter_val();
for (auto cur : vec) {
tr[y].insert(cur);
}
tr[x].all_clear();
return y;
};
while (m--) {
char op;
cin >> op;
if (op == 'C') {
ll t;
cin >> t;
while (true) {
int tot = tr[mxid].size();
if (mxdc + t < tot) {
mxdc += t;
break;
} else if (secid == 0) {
t -= (tot - mxdc);
pt[mxid] = pt[mxid] - 1 - (t / tot);
mxdc = t % tot;
} else {
t -= (tot - mxdc);
mxdc = 0;
if ((pt[mxid] - 1 - pt[secid]) * tot > t) {
pt[mxid] = pt[mxid] - 1 - (t / tot);
mxdc = t % tot;
break;
} else {
t -= (pt[mxid] - 1 - pt[secid]) * tot;
pt[mxid] = pt[secid];
mxid = merge(mxid, secid);
secid--;
}
}
}
} else {
int x;
cin >> x;
int tot = tr[mxid].size();
if (x <= tot - mxdc) {
cout << tr[mxid].kth_mx(mxdc + x) + k * pt[mxid] - D << '\n';
} else {
int sectot = tr[secid].size();
if (x <= tot + sectot) {
x -= tot - mxdc;
ll l = 0, r = 2e18, ans = -1;
while (l <= r) {
ll mid = (l + r) >> 1;
int gecnt = tr[secid].greater_equal_cnt(mid - pt[secid] * k) +
min(tr[mxid].greater_equal_cnt(mid + k - pt[mxid] * k), mxdc);
if (gecnt >= x) {
ans = mid;
l = mid + 1;
} else {
r = mid - 1;
}
}
cout << ans - D << '\n';
} else {
x -= (tot + sectot);
cout << a[pos[secid] - x + 1] - D << '\n';
}
}
}
}
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
int T = 1;
while (T--) solve();
return 0;
}
詳細信息
Test #1:
score: 100
Accepted
time: 1ms
memory: 3572kb
input:
3 5 5 7 3 9 A 3 C 1 A 2 C 2 A 3
output:
3 4 -1
result:
ok 3 lines
Test #2:
score: -100
Wrong Answer
time: 0ms
memory: 3760kb
input:
5 8 8 294 928 293 392 719 A 4 C 200 A 5 C 10 A 2 C 120 A 1 A 3
output:
294 -640 -689 -5584 -5586
result:
wrong answer 2nd lines differ - expected: '200', found: '-640'