Submission #1270438

#	Time	Username	Problem	Language	Result	Execution time	Memory
1270438		witek_cpp	Snowball (JOI21_ho_t2)	C++20	0 / 100	1 ms	580 KiB

Main

#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
 
using namespace std;
typedef long long ll;
 
#define st first
#define nd second
#define f(a, c, b) for (int a = c; b > a; a++)
#define pb push_back
#define all(a) a.begin(), a.end()
#define sz(a) int(a.size())

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n, q;
    cin >> n >> q;
    vector<ll> a(n);
    f(i, 0, n) cin >> a[i];
    vector<ll> w(q);
    f(i, 0, q) cin >> w[i];
    ll mksl = 0;
    ll mksr = 0;
    ll pos = 0;
    vector<pair<ll, pair<ll, ll>>> bs = {{0, {0, 0}}};
    f(i, 0, q) {
        pos += w[i];
        if (-pos > mksl) {
            mksl = -pos;
            bs.pb({mksr + mksl, {mksl, mksr}});
        }
        if (pos > mksr) {
            mksr = pos;
            bs.pb({mksr + mksl, {mksl, mksr}});
        }
    }
    /*for (pair<ll, pair<ll, ll>> ele: bs) {
        cout << ele.st << " " << ele.nd.st << " " << ele.nd.nd << "\n";
    }*/
    vector<ll> ans(n, 0);
    ans[0] += mksl;
    ans.back() += mksr;
    f(i, 0, n - 1) {
        if ((a[i + 1] - a[i]) >= (mksr + mksl)) {
            ans[i] += mksr;
            ans[i + 1] += mksl;
        }
        int l, r, mid;
        l = 0;
        r = sz(bs) - 2; //chcemy znalelx maksymalnie ind na prawo w ktorym sie zmiescimy
        ll odl = a[i + 1] - a[i];
        while (r > l) {
            mid = (r + l + 1)/2;
            if (bs[mid].st <= odl) l = mid;
            else r = mid - 1;
        }
        //cout << "ind " << i << " bs mi skonczyl na " << r << "\n";
        ll pl = bs[r + 1].nd.st - bs[r].nd.st;
        ll pr = bs[r + 1].nd.nd - bs[r].nd.nd;
        ans[i] += bs[r].nd.nd;
        ans[i + 1] += bs[r].nd.st;
        ll s = bs[r].st;
        if (pl) {
            ans[i + 1] += (odl - s);
        } else {
            ans[i] += (odl - s);
        }
    }
    f(i, 0, n) cout << ans[i] << "\n";
    return 0;
}

• Subtask 1: 0 / 33
• Subtask 2: 0 / 67