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Submission #639921

# Submission time Handle Problem Language Result Execution time Memory
639921 2022-09-12T17:18:40 Z Wunka Hacker (BOI15_hac) C++17
40 / 100
 1000 ms 4336 KB 
#include<bits/stdc++.h>
using namespace std;
 
#define ll long long 
 
struct segtree {
    vector<ll> tree; 
    int size = 1; 
 
    void init(int n) {
        while(size < n) size *= 2; 
        tree.assign(2 * size - 1, 0);
    }
 
    void set(int i, ll v, int x, int lx, int rx) {
        if(rx - lx == 1) {
            tree[x] = v;
            return; 
        } 
        
        int mid = (lx + rx) / 2; 
        if(i < mid) {
            set(i, v, 2 * x + 1, lx, mid); 
        }
        else{
            set(i, v, 2 * x + 2, mid, rx); 
        }
        tree[x] = max(tree[2 * x + 1], tree[2 * x + 2]); 
    }
    void set(int i, ll v) {
        set(i, v, 0, 0, size); 
    }
 
 
    ll get_max(int l, int r, int x, int lx, int rx) {
        if(lx >= r || rx <= l) return 0; 
        if(lx >= l && rx <= r) return tree[x]; 
        int mid = (lx + rx) / 2; 
 
        ll m1 = get_max(l, r, 2 * x + 1, lx, mid);
        ll m2 = get_max(l, r, 2 * x + 2, mid, rx); 
        return max(m1, m2);  
    }
    ll get_max(int l, int r) {
        return get_max(l, r, 0, 0, size); 
    }
}; 
 
// NOTE: Pentru al doilea jucator marimea lantului o sa fie floor(n / 2), de aceea o sa fac i + (n / 2)
 
vector<ll> computeNsums(vector<ll>& a) {
    int n = a.size() / 2;
    vector<ll> nsums(n); 
    vector<ll> pref(2 * n + 1);
    for(int i = 0; i < 2 * n; i++) {
        pref[i + 1] = pref[i] + a[i]; 
    }
 
    for(int i = 0; i < n; i++) {
        nsums[i] = pref[i + (n / 2)] - pref[i]; 
    }
 
    return nsums;
}
 
int main() {
    ios_base::sync_with_stdio(false); 
    cin.tie(NULL);
 
    int n; 
    cin >> n; 
    vector<ll> a(2 * n); 
    ll S = 0; 
    for(int i = 0; i < n; i++) {
        cin >> a[i]; 
        a[i + n] = a[i]; 
        S += a[i]; 
    }
    
    vector<ll> b = computeNsums(a); 
    //cout << "our b array: "; 
    //for(int i = 0; i < n; i++) cout << b[i] << ' '; 
    //cout << '\n'; 
    /*
    segtree st; 
    st.init(n); 
    for(int i = 0; i < n; i++) {
        st.set(i, b[i]); 
    }
    //cout << "THIS IS OUR SUM:" << S << '\n'; 
    //cout << '\n'; 
    ll ans = 0; 
    for(int i = 0; i < n; i++) {
        cout << "we are in position " << i << '\n'; 
        int l = (i + 1) % n, r = ((i - (n + 1) / 2) + n) % n; 
        int mn = min(l, r), mx = max(l, r); 
        cout << "currently our admissible interval:" << mn << ' ' << mx << '\n'; 
        //ll res = st.get_max(mn, mx + 1); 
        //ans = max(ans, S - st.get_max(mn, mx + 1)); 
        //cout << '\n'; 
        ll m1 = st.get_max(0, mn + 1), m2 = st.get_max((i + 1) % n, mx + 1); 
        ll res = max(m1, m2); 
        ans = max(ans, S - res); 
        cout << "comparing " << ans << " with " << res << " which is "<< S - res << '\n';
        cout << '\n'; 
    }
    
    // (L; R) - de la care element putem sa formam sirul, cu care element se incepe ultimul sir. 
    for(int i = 0; i < n; i++) {
        int L = (i + 1) % n, R = ((i - (n + 1) / 2) + n) % n; 
        // L should be greater than R
        if(R > L) {
            swap(R, L); 
        }
        ll m1 = st.get_max(L, n + 1), m2 = st.get_max(0, R + 1); 
        ll res = max(m1, m2); 
        ans = max(ans, S - res);
 
    }
    */ 
    // este o modalitate cum se poate de rezolvat problema fara de segment tree : observam ca noi ca si cum miscam un interval de marimea k si in el noi scoatem cel mai din stanga elemente si adaugam cel mai din dreapta element, atunci cand miscam intervalul dat. aceasta ne da ideea de sliding window.  
    ll ans = 0; 
    ll L = 1, R = ((0 - (n) / 2) + n) % n; 
    multiset<ll> window; 
    for(int i = L; i <= R; i++){
        window.insert(b[i]); 
    } 
 
    for(int i = 0; i < n; i++) {
        ll res = *max_element(window.begin(), window.end()); 
        ans = max(ans, S - res); 
        window.erase(find(window.begin(), window.end(), b[L])); 
        L = (L + 1) % n; 
        R = (R + 1) % n; 
        window.insert(b[R]); 
    }
    //cout << "========================================" << '\n'; 
    cout << ans << '\n'; 
}
 
/*
    Diferenta intre v1 si v2. 
    -- Daca ambii jucaturi incearca sa imi maximizeze scorul sau, aceasta nu inseamna faptul ca raspunsul o sa fie obligatoriu ca primul o sa aiba mai mare valoarea decat al doilea jucator SAU ca in asemenea situatie cand diferenta intre scoruri o sa fie minimala --> noi o sa avem cel mai mare raspuns pentru X. MOtivul pentru aceasta consta intr-aceea, ca asemenea ipoteze nu obligatoriu asuma ca al doilea jucator are strategie optimala, de aceea nu putem sa spunem ca ele sunt corecte. 
 
 
    Observam ca dupa prima miscare noi deja stim cum o sa arate primul sir si noi stim cum o sa arata sirul la al doilea jucator. De ce?
    Pentru ca sirul primului jucator este format unic - obligatoriu o sa se invarte intre intervalele de marimea k, care contin primul varf ales, al doilea jucator o sa incearca sa faca asa un pas - astfel incat valoarea sirului primului jucator, o sa fie minimal posibila. Cum procedeaza al doilea jucator? Noi precomputam pentru fiecare varf X, care este valoarea sirului care porneste din X si are marimea k, daca sirul merge in ordinea acelor ceasornicului. 
 
    Dupa aceasta, noi incercam din toate sirurile acestea, care nu il contin pe X - sa luam intervalul cu sma maximala. SUma_sirului1 = S - suma_sirului2. Problema s-a redus la cautarea banala valorii maximale a suma_sirului1. 
 
*/

Subtask #1
20.0 / 20.0

# Verdict Execution time Memory Grader output
1 Correct 0 ms 212 KB Output is correct
2 Correct 1 ms 212 KB Output is correct
3 Correct 1 ms 212 KB Output is correct
4 Correct 1 ms 212 KB Output is correct
5 Correct 1 ms 212 KB Output is correct
6 Correct 1 ms 212 KB Output is correct
7 Correct 1 ms 320 KB Output is correct
8 Correct 1 ms 260 KB Output is correct
9 Correct 1 ms 212 KB Output is correct
10 Correct 1 ms 212 KB Output is correct
11 Correct 1 ms 212 KB Output is correct
12 Correct 1 ms 212 KB Output is correct
13 Correct 1 ms 212 KB Output is correct
14 Correct 0 ms 212 KB Output is correct
15 Correct 0 ms 212 KB Output is correct
16 Correct 1 ms 212 KB Output is correct
17 Correct 1 ms 212 KB Output is correct
18 Correct 1 ms 212 KB Output is correct

Subtask #2
20.0 / 20.0

# Verdict Execution time Memory Grader output
1 Correct 0 ms 212 KB Output is correct
2 Correct 1 ms 212 KB Output is correct
3 Correct 1 ms 212 KB Output is correct
4 Correct 1 ms 212 KB Output is correct
5 Correct 1 ms 212 KB Output is correct
6 Correct 1 ms 212 KB Output is correct
7 Correct 1 ms 320 KB Output is correct
8 Correct 1 ms 260 KB Output is correct
9 Correct 1 ms 212 KB Output is correct
10 Correct 1 ms 212 KB Output is correct
11 Correct 1 ms 212 KB Output is correct
12 Correct 1 ms 212 KB Output is correct
13 Correct 1 ms 212 KB Output is correct
14 Correct 0 ms 212 KB Output is correct
15 Correct 0 ms 212 KB Output is correct
16 Correct 1 ms 212 KB Output is correct
17 Correct 1 ms 212 KB Output is correct
18 Correct 1 ms 212 KB Output is correct
19 Correct 2 ms 340 KB Output is correct
20 Correct 5 ms 340 KB Output is correct
21 Correct 4 ms 320 KB Output is correct
22 Correct 52 ms 460 KB Output is correct
23 Correct 141 ms 560 KB Output is correct
24 Correct 48 ms 468 KB Output is correct
25 Correct 136 ms 556 KB Output is correct
26 Correct 153 ms 572 KB Output is correct
27 Correct 1 ms 212 KB Output is correct
28 Correct 1 ms 324 KB Output is correct
29 Correct 1 ms 212 KB Output is correct
30 Correct 120 ms 552 KB Output is correct
31 Correct 120 ms 552 KB Output is correct

Subtask #3
0 / 20.0

# Verdict Execution time Memory Grader output
1 Correct 1 ms 320 KB Output is correct
2 Correct 2 ms 340 KB Output is correct
3 Correct 145 ms 572 KB Output is correct
4 Execution timed out 1083 ms 4336 KB Time limit exceeded
5 Halted 0 ms 0 KB -

Subtask #4
0 / 40.0

# Verdict Execution time Memory Grader output
1 Correct 0 ms 212 KB Output is correct
2 Correct 1 ms 212 KB Output is correct
3 Correct 1 ms 212 KB Output is correct
4 Correct 1 ms 212 KB Output is correct
5 Correct 1 ms 212 KB Output is correct
6 Correct 1 ms 212 KB Output is correct
7 Correct 1 ms 320 KB Output is correct
8 Correct 1 ms 260 KB Output is correct
9 Correct 1 ms 212 KB Output is correct
10 Correct 1 ms 212 KB Output is correct
11 Correct 1 ms 212 KB Output is correct
12 Correct 1 ms 212 KB Output is correct
13 Correct 1 ms 212 KB Output is correct
14 Correct 0 ms 212 KB Output is correct
15 Correct 0 ms 212 KB Output is correct
16 Correct 1 ms 212 KB Output is correct
17 Correct 1 ms 212 KB Output is correct
18 Correct 1 ms 212 KB Output is correct
19 Correct 2 ms 340 KB Output is correct
20 Correct 5 ms 340 KB Output is correct
21 Correct 4 ms 320 KB Output is correct
22 Correct 52 ms 460 KB Output is correct
23 Correct 141 ms 560 KB Output is correct
24 Correct 48 ms 468 KB Output is correct
25 Correct 136 ms 556 KB Output is correct
26 Correct 153 ms 572 KB Output is correct
27 Correct 1 ms 212 KB Output is correct
28 Correct 1 ms 324 KB Output is correct
29 Correct 1 ms 212 KB Output is correct
30 Correct 120 ms 552 KB Output is correct
31 Correct 120 ms 552 KB Output is correct
32 Correct 1 ms 320 KB Output is correct
33 Correct 2 ms 340 KB Output is correct
34 Correct 145 ms 572 KB Output is correct
35 Execution timed out 1083 ms 4336 KB Time limit exceeded
36 Halted 0 ms 0 KB -