Submission #691534

#	Time	Username	Problem	Language	Result	Execution time	Memory
691534		fatemetmhr	Measures (CEOI22_measures)	C++17	59 / 100	196 ms	5068 KiB

Main

// hmmm... na dige chizi baraye goftan namoonde
 
 
// ~ Be Name Khoda ~

#include <bits/stdc++.h>
//#pragma GCC optimize ("Ofast")
//#pragma GCC target("avx2")
//#pragma GCC optimize("unroll-loops,O3")

using namespace std;

typedef long long ll;

#define pb       push_back
#define mp       make_pair
#define all(x)   x.begin(), x.end()
#define fi       first
#define se       second

const int maxn  =  1e6   + 10;
const int maxn5 =  5e5   + 10;
const int maxnt =  1.2e6 + 10;
const int maxn3 =  1e3   + 10;
const int mod   =  1e9   +  7;
const ll  inf   =  1e18;

ll d;
ll a[maxn5];

inline ll solve(int n){
    //cout << "ok " << n << endl;
    sort(a, a + n);
    ll lastpos = a[0], tim = 0;
    for(int i = 1; i < n; i++){
        if(a[i] - lastpos >= d)
            lastpos = max(lastpos + d, a[i] - tim);
        else if(a[i] + tim - lastpos >= d)
            lastpos += d;
        else{
            ll need = d - (a[i] - lastpos) - tim;
            ll x = a[i] + tim;
            //cout << "here " << x << ' ' << need << ' ' << tim << endl;
            need /= 2;
            tim += need;
            lastpos = x + need;
        }
        //cout << i << ' ' << lastpos << ' ' << tim << endl;
    }
    return tim;
}

 
int main(){
    ios_base::sync_with_stdio(false); cin.tie(0);
 
    int n, m;
    cin >> n >> m >> d;
    d *= 2;
    for(int i = 0; i < n; i++){
        cin >> a[i];
        a[i] *= 2;
    }
    sort(a, a + n);
    ll lastpos, tim;
    for(int i = 0; i < m; i++){
        cin >> a[i + n];
        a[i + n] *= 2;
        ll ans;
        if(m > 10){
            if(i == 0)
                lastpos = a[0], tim = 0;
            else{
                if(a[i] - lastpos >= d)
                    lastpos = max(lastpos + d, a[i] - tim);
                else if(a[i] + tim - lastpos >= d)
                    lastpos += d;
                else{
                    ll need = d - (a[i] - lastpos) - tim;
                    ll x = a[i] + tim;
                    //cout << "here " << x << ' ' << need << ' ' << tim << endl;
                    need /= 2;
                    tim += need;
                    lastpos = x + need;
                }
            }
            ans = tim;
        }
        else
            ans = solve(i + n + 1);
        cout << ans / 2 << (ans & 1 ? ".5 " : " ");
    }
    cout << endl;

}

• Subtask 1: 10 / 10
• Subtask 2: 14 / 14
• Subtask 3: 35 / 35
• Subtask 4: 0 / 41