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
#511374#5449. 楼梯ewrilan#0 7ms5744kbC++204.7kb2024-08-09 19:57:412024-08-09 19:57:42

Judging History

你现在查看的是最新测评结果

  • [2024-08-09 19:57:42]
  • 评测
  • 测评结果:0
  • 用时:7ms
  • 内存:5744kb
  • [2024-08-09 19:57:41]
  • 提交

answer

//
#ifndef __SIZEOF_INT128__
  #define __SIZEOF_INT128__
#endif
#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace chrono;
using namespace __gnu_pbds;
template <typename T> using oset =  tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
#define rep(i, p, k) for(int i(p); i < (k); ++i)
#define per(i, p, k) for(int i(p); i > (k); --i)
#define sz(x) (int)(x).size()
#define sc static_cast
typedef long long ll;
typedef long double ld;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef __int128_t lll;
//#define int ll
template <typename T = int> using par = std::pair <T, T>;
#define fi first
#define se second
#define test int _number_of_tests(in()); while(_number_of_tests--)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define pb emplace_back
struct Timer {
    string name{""};
    time_point<high_resolution_clock> end, start{high_resolution_clock::now()};
    duration<float, std::milli> dur;
    Timer() = default;
    Timer(string nm): name(nm) {}
    ~Timer() {
        end = high_resolution_clock::now(); dur= end - start;
        cout << "@" << name << "> " << dur.count() << " ms" << '\n';
    }
};
template <typename T = int> inline T in()
{
    static T x;
    std::cin >> x;
    return x;
}
std::string yn(bool b)
{
    if(b) return "YES\n";
    else return "NO\n";
}
template <typename F, typename S> std::ostream& operator<<(std::ostream& out, const std::pair <F, S>& par);
template <typename T> std::ostream& operator<< (std::ostream& out, const std::vector <T>& wek)
{
    for(const auto& i : wek)out << i << ' ';
    return out;
}
template <typename F, typename S> std::ostream& operator<<(std::ostream& out, const std::pair <F, S>& par)
{
    out << '{'<<par.first<<", "<<par.second<<"}";
    return out;
}
#define show(x) cerr << #x << " = " << x << '\n';
std::int32_t main()
{
    std::cout.tie(nullptr); //for luck
    std::cin.tie(nullptr); std::ios_base::sync_with_stdio(0);
    int m(in());
    if(1 || m <= 1000)
    {
        map <int, ll> ar;
        stack <pair <int, int>> dd; 
        int a, b;
        rep(i, 0, m){
            ar[0] = 0;
            switch(in<char>())
            {
                case '+':
                    cin >> a >> b;
                    ar[a] += b;
                    dd.push({a, b});
                    break;
                case 'R':
                    cin >> a;
                    rep(j, 0, a){
                        auto [a, b] = dd.top(); dd.pop();
                        if(a == -1)continue;
                        ar[a] -= b;
                        if(!ar[a])ar.erase(a);
                    }
                    break;
                case '-':
                    while(sz(dd))dd.pop();
                    cin >> a >> b;
                    ar[a-1] += b;
                    while(b){
                        if(ar.rbegin()->second > b){
                            ar.rbegin()->second -= b;
                            b = 0;
                        }
                        else{
                            b -= ar.rbegin()->second;
                            ar.erase(prev(ar.end()));
                        }
                    }
                    break;
                case '?':
                    dd.push({-1, -1});
                    cin >> a;
                    auto p(ar.rbegin()), k(ar.rbegin());
                    ll s(1);
                    ll px(p->second);
                    while(next(k) != ar.rend()){
                        while(p != k && s > a){
                            s -= p->first - next(p)->first;
                            s -= next(p)->second;
                            ++p;
                            px += p->second;
                        }
                        if(s <= a && a <= s + p->second-1){
                            cout << k->first << ' ' << px - (a-s) << '\n';
                            break;
                        }
                        if(s + p->second <= a && a < s + p->second-1 + k->first - next(k)->first ){
                            cout << k->first - (a - s - p->second+1) << ' ' << px - (p->second-1) << '\n';
                            break;
                        }
                        ++k;
                        s += k->second;
                        s += prev(k)->first - k->first;
                    }
                    break;
            }
            // for(auto i: ar)cout << i << ' ';
            // cout << '\n';
        }
        return 0;
    }
    return 0;
}

詳細信息

Subtask #1:

score: 0
Wrong Answer

Test #1:

score: 0
Wrong Answer
time: 0ms
memory: 3568kb

input:

1000
- 1 999995
? 770771330
? 688406105220012703
? 466054413
? 1466199
? 940610359316496655
? 310504014100463831
? 765557590
? 419614901
? 830584303
? 85199513
? 768715778674812284
? 742663363105169125
? 859012516258231128
? 168409807482711289
? 842755243
? 618667253264707663
? 957265852
+ 1 1
+ 1 1...

output:


result:

wrong output format Unexpected end of file - int64 expected

Subtask #2:

score: 0
Skipped

Dependency #1:

0%

Subtask #3:

score: 0
Skipped

Dependency #1:

0%

Subtask #4:

score: 0
Skipped

Dependency #1:

0%

Subtask #5:

score: 0
Skipped

Dependency #1:

0%

Subtask #6:

score: 0
Wrong Answer

Test #111:

score: 0
Wrong Answer
time: 7ms
memory: 5744kb

input:

300000
? 308230551
? 154394919
? 77796824
? 523232316
? 601650936815206607
? 986805724
? 283169431815882827
? 781223930
? 785380179984309713
? 36818911074958611
? 452850684
? 392810692
? 812929344
? 9753139
? 236758865441063408
? 448106017
? 382652997142237763
? 667762111
? 201388730
? 433119061
? 6...

output:


result:

wrong output format Unexpected end of file - int64 expected

Subtask #7:

score: 0
Skipped

Dependency #1:

0%