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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 |
---|---|---|---|---|---|---|---|---|---|
#160764 | #7108. Couleur | ucup-team134# | AC ✓ | 3285ms | 28728kb | C++14 | 5.5kb | 2023-09-02 21:21:18 | 2023-09-02 21:21:19 |
Judging History
answer
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>
#define ll long long
#define pb push_back
#define f first
#define s second
#define sz(x) (int)(x).size()
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define ios ios_base::sync_with_stdio(false);cin.tie(NULL)
#define ld long double
#define li __int128
using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>; ///find_by_order(),order_of_key()
template<int D, typename T>struct vec : public vector<vec<D - 1, T>> {template<typename... Args>vec(int n = 0, Args... args) : vector<vec<D - 1, T>>(n, vec<D - 1, T>(args...)) {}};
template<typename T>struct vec<1, T> : public vector<T> {vec(int n = 0, T val = T()) : vector<T>(n, val) {}};
template<class T1,class T2> ostream& operator<<(ostream& os, const pair<T1,T2>& a) { os << '{' << a.f << ", " << a.s << '}'; return os; }
template<class T> ostream& operator<<(ostream& os, const vector<T>& a){os << '{';for(int i=0;i<sz(a);i++){if(i>0&&i<sz(a))os << ", ";os << a[i];}os<<'}';return os;}
template<class T> ostream& operator<<(ostream& os, const deque<T>& a){os << '{';for(int i=0;i<sz(a);i++){if(i>0&&i<sz(a))os << ", ";os << a[i];}os<<'}';return os;}
template<class T> ostream& operator<<(ostream& os, const set<T>& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}
template<class T> ostream& operator<<(ostream& os, const set<T,greater<T> >& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}
template<class T> ostream& operator<<(ostream& os, const multiset<T>& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}
template<class T> ostream& operator<<(ostream& os, const multiset<T,greater<T> >& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}
template<class T1,class T2> ostream& operator<<(ostream& os, const map<T1,T2>& a) {os << '{';int i=0;for(auto p:a){if(i>0&&i<sz(a))os << ", ";os << p;i++;}os << '}';return os;}
int ri(){int x;scanf("%i",&x);return x;}
void rd(int&x){scanf("%i",&x);}
void rd(long long&x){scanf("%lld",&x);}
void rd(double&x){scanf("%lf",&x);}
void rd(long double&x){scanf("%Lf",&x);}
void rd(string&x){cin>>x;}
void rd(char*x){scanf("%s",x);}
template<typename T1,typename T2>void rd(pair<T1,T2>&x){rd(x.first);rd(x.second);}
template<typename T>void rd(vector<T>&x){for(T&p:x)rd(p);}
template<typename C,typename...T>void rd(C&a,T&...args){rd(a);rd(args...);}
//istream& operator>>(istream& is,__int128& a){string s;is>>s;a=0;int i=0;bool neg=false;if(s[0]=='-')neg=true,i++;for(;i<s.size();i++)a=a*10+s[i]-'0';if(neg)a*=-1;return is;}
//ostream& operator<<(ostream& os,__int128 a){bool neg=false;if(a<0)neg=true,a*=-1;ll high=(a/(__int128)1e18);ll low=(a-(__int128)1e18*high);string res;if(neg)res+='-';if(high>0){res+=to_string(high);string temp=to_string(low);res+=string(18-temp.size(),'0');res+=temp;}else res+=to_string(low);os<<res;return os;}
void test(){
int n=ri();
vector<int> a(n);
rd(a);
set<pair<pair<int,int>,int>> in;
vector<pair<ordered_set<pair<int,int>>,ll>> sets;
in.insert({{0,n-1},0});
multiset<ll> cnts;
ll cnt=0;
ordered_set<pair<int,int>> st;
for(int i=0;i<n;i++){
cnt+=sz(st)-st.order_of_key({a[i]+1,-1});
st.insert({a[i],i});
}
sets.pb({st,cnt});
cnts.insert(cnt);
for(int i=0;i<n;i++){
ll br=*cnts.rbegin();
printf("%lld",br);
if(i!=n-1){
printf(" ");
}
/*for(auto p:sets){
printf("[");
for(auto d:p.f){
printf("(%i %i) ",d.f,d.s);
}
printf("], %lld\n",p.s);
}*/
ll p;
rd(p);
p^=br;
p--;
//printf("Brisem %i\n",p);
auto it=in.lower_bound({{p,n},n});
it--;
int l=(*it).f.f,r=(*it).f.s,ind=(*it).s;
in.erase(it);
//printf("%i %i %i %i\n",l,r,ind,p);
int levi=p-l,desni=r-p;
cnts.erase(cnts.find(sets[ind].s));
if(levi<desni){
for(int i=l;i<=p;i++){
sets[ind].f.erase({a[i],i});
}
for(int i=l;i<=p;i++){
sets[ind].s-=sets[ind].f.order_of_key({a[i],-1});
}
if(l!=p){
int newInd=sz(sets);
ordered_set<pair<int,int>> novi;
ll noviCnt=0;
for(int i=l;i<p;i++){
noviCnt+=sz(novi)-novi.order_of_key({a[i]+1,-1});
novi.insert({a[i],i});
}
sets[ind].s-=noviCnt;
cnts.insert(noviCnt);
cnt+=noviCnt;
sets[ind].s-=sz(novi)-novi.order_of_key({a[p]+1,-1});
sets.pb({novi,noviCnt});
in.insert({{l,p-1},newInd});
}
if(r!=p){
in.insert({{p+1,r},ind});
cnts.insert(sets[ind].s);
}
}
else{
for(int i=p;i<=r;i++){
sets[ind].f.erase({a[i],i});
}
for(int i=p;i<=r;i++){
sets[ind].s-=sz(sets[ind].f)-sets[ind].f.order_of_key({a[i]+1,-1});
}
if(r!=p){
int newInd=sz(sets);
ordered_set<pair<int,int>> novi;
ll noviCnt=0;
for(int i=p+1;i<=r;i++){
noviCnt+=(int)novi.size()-novi.order_of_key({a[i]+1,-1});
novi.insert({a[i],i});
}
sets[ind].s-=noviCnt;
cnts.insert(noviCnt);
sets[ind].s-=novi.order_of_key({a[p],-1});
sets.pb({novi,noviCnt});
in.insert({{p+1,r},newInd});
}
if(l!=p) {
in.insert({{l,p-1},ind});
cnts.insert(sets[ind].s);
}
}
}
printf("\n");
}
int main()
{
int t=ri();
while(t--){
test();
}
return 0;
}
这程序好像有点Bug,我给组数据试试?
详细
Test #1:
score: 100
Accepted
time: 1ms
memory: 3996kb
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: 0
Accepted
time: 3285ms
memory: 28728kb
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 ...
result:
ok 11116 lines
Extra Test:
score: 0
Extra Test Passed