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 |
---|---|---|---|---|---|---|---|---|---|
#230969 | #7108. Couleur | UFRJ# | RE | 0ms | 3828kb | C++20 | 4.4kb | 2023-10-28 22:38:42 | 2023-10-28 22:38:43 |
Judging History
answer
#include<bits/stdc++.h>
#include<bits/extc++.h>
using namespace std;
using lint = int64_t;
template<typename K, typename V, typename Comp = less<K>>
using ordered_map = __gnu_pbds::tree<K, V, Comp, __gnu_pbds::rb_tree_tag, __gnu_pbds::tree_order_statistics_node_update>;
template<typename K, typename Comp = less<K>>
using ordered_set = ordered_map<K, __gnu_pbds::null_type, Comp>;
template<typename T> struct FT {
vector<T> s;
FT(int n=0) : s(n) {}
FT(const vector<T> &A) : s(int(A.size())) {
const int N = int(A.size());
for(int pos = 0; pos < N; ++pos) {
s[pos] += A[pos];
int nxt = (pos | (pos + 1));
if(nxt < N) s[nxt] += s[pos];
}
}
void update(int pos, T dif) {
for(; pos < (int) s.size(); pos |= pos + 1) s[pos] += dif;
}
T query(int pos) {
T res = 0;
for(; pos > 0; pos &= pos - 1) res += s[pos-1];
return res;
}
};
FT<lint> ft;
vector<lint> a;
lint count_inv(int l, int r){
if(l >= r) return 0;
lint t = 0;
for(int i = r; i>=l; i--){
t += ft.query(a[i]);
ft.update(a[i], 1);
}
for(int i = r; i>=l; i--){
ft.update(a[i], -1);
}
return t;
}
using pii = pair<int, int>;
const int inf = 1e7;
void solve(){
int n;
cin>>n;
a.resize(n);
ft = FT<lint>(n+2);
for(auto &x : a) cin>>x;
map<pair<int, int>, lint> inv;
map<lint, int> freq;
inv[{0, n-1}] = count_inv(0, n-1);
freq[inv[{0, n-1}]]++;
lint z = inv[{0, n-1}];
for(int _ = 0; _ < n; _++){
assert(!freq.empty());
z = freq.rbegin()->first;
cout<<z;
if(_ != n-1) cout<<" ";
int p; cin>>p;
p = p ^ z;
p--;
auto it = inv.upper_bound({p+1, -1});
assert(it != inv.begin());
it--;
auto [interval, T] = *it;
auto [l, r] = interval;
//cout<<"querying "<<p<<endl;
//cout<<"interval "<<l<<" "<<r<<" "<<T<<endl;
freq[T]--;
if(freq[T] == 0) freq.erase(T);
inv.erase({l, r});
if(l == r) continue;
if(p == l){
lint inv_right = T;
for(int i = p+1; i<=r; i++){
if(a[p] > a[i]) inv_right--;
}
inv[{l+1, r}] = inv_right;
freq[inv_right]++;
//cout<<"p == l"<<endl;
continue;
}
if(p == r){
lint inv_lef = T;
for(int i = l; i<p; i++){
if(a[p] < a[i]) inv_lef--;
}
inv[{l, r-1}] = inv_lef;
freq[inv_lef]++;
//cout<<"p == r"<<endl;
continue;
}
int sz_left = p - l;
int sz_right = r - (l-1) + 1;
if(sz_left < sz_right){
lint inv_lef = count_inv(l, p-1);
lint inv_right = T - inv_lef;
ordered_set<pii> S;
for(int i = l; i<p; i++){
S.insert({a[i], i});
}
int L = S.size();
inv_right -= L - S.order_of_key({a[p], inf});
for(int i = p+1; i<=r; i++){
inv_right -= L - S.order_of_key({a[i], inf});
if(a[i] < a[p])
inv_right--;
}
inv[{l, p-1}] = inv_lef;
inv[{p+1, r}] = inv_right;
freq[inv_lef]++;
freq[inv_right]++;
} else {
lint inv_right = count_inv(p+1, r);
lint inv_lef = T - inv_right;
ordered_set<pii> S;
for(int i = p+1; i<=r; i++){
S.insert({a[i], i});
}
inv_lef -= S.order_of_key({a[p], -1});
for(int i = l; i<p; i++){
inv_lef -= S.order_of_key({a[i], -1});
if(a[i] > a[p])
inv_lef--;
}
inv[{l, p-1}] = inv_lef;
inv[{p+1, r}] = inv_right;
freq[inv_lef]++;
freq[inv_right]++;
}
// for(auto [interval, T] : inv){
// //cout<<"( " <<interval.first<<", "<<interval.second<<") "<<T<<endl;
// }
//cout<<endl;
}
cout<<"\n";
//cout<<"-----"<<endl;
}
int main() {
cin.tie(nullptr)->sync_with_stdio(false);
int t; cin>>t;
while(t--){
solve();
}
return 0;
}
详细
Test #1:
score: 100
Accepted
time: 0ms
memory: 3828kb
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 ...