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 |
---|---|---|---|---|---|---|---|---|---|
#486710 | #7108. Couleur | lwenj | RE | 1ms | 7876kb | C++14 | 3.1kb | 2024-07-21 23:23:49 | 2024-07-21 23:23:49 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 10;
int n, idx;
int arr[N], root[N];
struct node {
int l, r;
int cnt;
}tr[N * 4];
map<int, long long> mp;
multiset<long long> ms;
int query(int cur, int l, int r, int ql, int qr) {
if (ql > qr)
return 0;
if (l > r) return 0;
if (l >= ql && r <= qr) return tr[cur].cnt;
int mid = (l + r) >> 1, sum = 0;
if (ql <= mid)
sum += query(tr[cur].l, l, mid, ql, qr);
if (qr > mid)
sum += query(tr[cur].r, mid + 1, r, ql, qr);
return sum;
}
int build(int l, int r) {
int cur = ++idx;
if (l == r) {
tr[cur].cnt = 0;
return cur;
}
int mid = (l + r) >> 1;
tr[cur].l = build(l, mid), tr[cur].r = build(mid + 1, r);
return cur;
}
int add(int p, int l, int r, int x) {
int cur = ++idx;
tr[cur] = tr[p];
if (l == r) {
tr[cur].cnt++;
return cur;
}
int mid = (l + r) / 2;
if (x <= mid) tr[cur].l = add(tr[p].l, l, mid, x);
else tr[cur].r = add(tr[p].r, mid + 1, r, x);
tr[cur].cnt = tr[tr[cur].l].cnt + tr[tr[cur].r].cnt;
return cur;
}
void split(int L, int R, int X) {
long long old = mp[L - 1]; ms.erase(ms.find(old));
long long base = query(root[R], 1, n, 1, arr[X] - 1) - query(root[X], 1, n, 1, arr[X] - 1);
base += query(root[X - 1], 1, n, arr[X] + 1, n) - query(root[L - 1], 1, n, arr[X] + 1, n);
if (X - L < R - X) {
long long a = 0, b = base;
for (int i = L; i < X; i++) {
a += query(root[i - 1], 1, n, arr[i] + 1, n) - query(root[L - 1], 1, n, arr[i] + 1, n);
b += query(root[R], 1, n, 1, arr[i] - 1) - query(root[X], 1, n, 1, arr[i] - 1);
}
mp[L - 1] = a; ms.insert(mp[L - 1]);
mp[X] = old - a - b; ms.insert(mp[X]);
} else {
long long a = 0, b = base;
for (int i = X + 1; i <= R; i++) {
a += query(root[i - 1], 1, n, arr[i] + 1, n) - query(root[X], 1, n, arr[i] + 1, n);
b += query(root[X - 1], 1, n, arr[i] + 1, n) - query(root[L - 1], 1, n, arr[i] + 1, n);
}
mp[L - 1] = old - a - b; ms.insert(mp[L - 1]);
mp[X] = a; ms.insert(mp[X]);
}
}
void solve() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) scanf("%d", &arr[i]);
root[0] = build(1, n);
for (int i = 1; i <= n; i++) root[i] = add(root[i - 1], 1, n, arr[i]);
long long tmp = 0;
for (int i = 1; i <= n; i++) tmp += query(root[i - 1], 1, n, arr[i] + 1, n);
mp.clear(); ms.clear();
mp[0] = tmp; ms.insert(tmp);
mp[n + 1] = 0; ms.insert(0);
long long ans = *prev(ms.end());
for (int i = 1; i <= n; i++) {
printf("%lld%c", ans, "\n "[i < n]);
long long x; scanf("%lld", &x);
x ^= ans;
auto it = prev(mp.lower_bound(x));
split(it->first + 1, next(it)->first - 1, x);
ans = *prev(ms.end());
}
}
int main()
{
int t; scanf("%d", &t);
while (t--) solve();
return 0;
}
// 1
// 5
// 4 3 1 1 1
// 5 4 5 3 1
详细
Test #1:
score: 100
Accepted
time: 1ms
memory: 7876kb
input:
3 5 4 3 1 1 1 5 4 5 3 1 10 9 7 1 4 7 8 5 7 4 8 21 8 15 5 9 2 4 5 10 6 15 4 8 8 1 12 1 10 14 7 14 2 9 13 10 3 37 19 23 15 7 2 10 15 2 13 4 5 8 7 10
output:
7 0 0 0 0 20 11 7 2 0 0 0 0 0 0 42 31 21 14 14 4 1 1 1 0 0 0 0 0 0
result:
ok 3 lines
Test #2:
score: -100
Runtime Error
input:
11116 10 10 5 10 3 6 4 8 5 9 8 31 27 24 11 12 3 0 2 3 1 10 8 2 7 2 8 10 1 10 9 10 6 5 2 13 2 1 0 1 3 1 10 7 10 7 6 1 3 10 6 7 9 21 18 10 1 6 5 4 8 9 10 10 2 10 4 8 8 5 7 2 6 7 20 10 9 1 15 0 4 2 9 7 10 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 10 1 2 3 4 5 6 7 8 9 10 6 3 5 2 7 10 9 1 4 8 10 1 10 1 3...
output:
21 18 16 12 10 6 4 1 1 0 12 12 10 10 4 4 4 2 1 0 20 16 9 5 3 3 3 0 0 0 22 14 8 8 5 5 2 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 19 12 7 4 4 2 2 1 0 0 20 18 8 3 1 1 0 0 0 0 45 21 21 10 3 3 3 0 0 0 17 11 8 2 1 1 1 0 0 0 13 4 1 0 0 0 0 0 0 0 29 27 22 15 9 7 4 3 1 0 26 16 9 2 1 1 1 1 1 0 0 0 0 0 0 ...