제출 #1216443

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1216443		steveonalex	A Plus B (IOI23_aplusb)	C++20	100 / 100	32 ms	3516 KiB

aplusb

#include <bits/stdc++.h>
 
using namespace std;
 
typedef long long ll;
typedef unsigned long long ull;
 
#define MASK(i) (1LL << (i))
#define GETBIT(mask, i) (((mask) >> (i)) & 1)
#define ALL(v) (v).begin(), (v).end()
#define block_of_code if(true)
 
ll max(ll a, ll b){return (a > b) ? a : b;}
ll min(ll a, ll b){return (a < b) ? a : b;}
ll gcd(ll a, ll b){return __gcd(a, b);}
ll lcm(ll a, ll b){return a / gcd(a, b) * b;}
 
ll LASTBIT(ll mask){return (mask) & (-mask);}
int pop_cnt(ll mask){return __builtin_popcountll(mask);}
int ctz(ull mask){return __builtin_ctzll(mask);}
int logOf(ull mask){return 63 - __builtin_clzll(mask);}
 
mt19937_64 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
ll rngesus(ll l, ll r){return l + (ull) rng() % (r - l + 1);}
double rngesus_d(double l, double r){
    double cur = rngesus(0, MASK(60) - 1);
    cur /= MASK(60) - 1;
    return l + cur * (r - l);
}
 
template <class T1, class T2>
    bool maximize(T1 &a, T2 b){
        if (a < b) {a = b; return true;}
        return false;
    }
 
template <class T1, class T2>
    bool minimize(T1 &a, T2 b){
        if (a > b) {a = b; return true;}
        return false;
    }
 
template <class T>
    void printArr(T container, string separator = " ", string finish = "\n", ostream &out = cout){
        for(auto item: container) out << item << separator;
        out << finish;
    }
 
template <class T>
    void remove_dup(vector<T> &a){
        sort(ALL(a));
        a.resize(unique(ALL(a)) - a.begin());
    }

#include "aplusb.h"

vector<int> smallest_sums(int n, vector<int> a, vector<int> b){
    vector<int> ans;
    #define pii pair<int,int>
    priority_queue<pii, vector<pii>, greater<pii>> pq;
    for(int i = 0; i < n; ++i) pq.push(make_pair(a[0] + b[i], 0));

    for(int it = 0; it < n; ++it){
        pii cur = pq.top(); pq.pop();
        ans.push_back(cur.first);
        if (cur.second + 1 < n) {
            cur.first += a[cur.second + 1] - a[cur.second];
            cur.second++;
            pq.push(cur);
        }
    }
    return ans;
}

• Subtask 1: 10 / 10
• Subtask 2: 20 / 20
• Subtask 3: 30 / 30
• Subtask 4: 40 / 40