Submission #64434

#	Time	Username	Problem	Language	Result	Execution time	Memory
64434		SpeedOfMagic	Hacker (BOI15_hac)	C++17	0 / 100	4 ms	620 KiB

hac

/** MIT License Copyright (c) 2018 Vasilyev Daniil **/
#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")
template<typename T> using v = vector<T>;
#define int long long
typedef long double ld;
typedef string str;
typedef vector<int> vint;
#define rep(a, l, r) for(int a = (l); a < (r); a++)
#define pb push_back
#define sz(a) ((int) a.size())
const long long inf = 4611686018427387903; //2^62 - 1
#if 0  //FileIO
const string fileName = "";
ifstream fin ((fileName == "" ? "input.txt"  : fileName + ".in" ));
ofstream fout((fileName == "" ? "output.txt" : fileName + ".out"));
#define get fin>>
#define put fout<<
#else
#define get cin>>
#define put cout<<
#endif
#define eol put endl
void read() {}     template<typename Arg,typename... Args> void read (Arg& arg,Args&... args){get (arg)     ;read(args...) ;}
void print(){}     template<typename Arg,typename... Args> void print(Arg  arg,Args...  args){put (arg)<<" ";print(args...);}
void debug(){eol;} template<typename Arg,typename... Args> void debug(Arg  arg,Args...  args){put (arg)<<" ";debug(args...);}
char curFlag = 'A'; void flag() {put curFlag++ << endl;}
int getInt(){int a; get a; return a;}
//code goes here
int f(int a, int b) { //f - query you need to calculate, but for two values
    return max(a, b);
}
const int nothing = 0; //nothing means that f(x, nothing)=x
struct segTreeNode {
    int val = nothing, lazy = 0;

    segTreeNode(){}

    void update(int s) { //function that updates single node from lazy
        this -> val += s;
    }
};

v<segTreeNode> segTree;
int n, tot;

void update(int l, int r, int s, int cur = 1, int ll = 1, int rr = tot) {
    if(l > r)
        return;

    if(segTree[cur].lazy != 0) {
        segTree[cur].update(segTree[cur].lazy);
        if(cur < tot) {
            segTree[cur * 2].lazy += segTree[cur].lazy;
            segTree[cur * 2 + 1].lazy += segTree[cur].lazy;
        }
        segTree[cur].lazy = 0;
    }

    if(l == ll && r == rr) {
        segTree[cur].update(s);
        if(cur < tot) {
            segTree[cur * 2].lazy += s;
            segTree[cur * 2 + 1].lazy += s;
        }
        return;
    }

    int mid = (ll + rr) / 2;
    update(l, min(r, mid), s, cur * 2, ll, mid);
    update(max(l, mid + 1), r, s, cur * 2 + 1, mid + 1, rr);
    segTree[cur].val = f(segTree[cur * 2].val, segTree[cur * 2 + 1].val);
}

int query(int l, int r, int cur = 1, int ll = 1, int rr = tot) {
    assert(l >= ll && r <= rr);
    if(l > r)
        return nothing;

    if(segTree[cur].lazy != 0) {
        segTree[cur].update(segTree[cur].lazy);
        if(cur < tot) {
            segTree[cur * 2].lazy += segTree[cur].lazy;
            segTree[cur * 2 + 1].lazy += segTree[cur].lazy;
        }
        segTree[cur].lazy = 0;
    }

    if(l == ll && r == rr)
        return segTree[cur].val;

    int mid = (ll + rr) / 2;

    int res = f(query(l, min(r, mid), cur * 2, ll, mid), query(max(l, mid + 1), r, cur * 2 + 1, mid + 1, rr));
    segTree[cur].val = f(segTree[cur * 2].val, segTree[cur * 2 + 1].val);
    return res;
}

void init(vector<int> vals) {
    n = vals.size();
    tot = n;
    while(tot & (tot - 1))
        tot++;

    segTree = vector<segTreeNode>(2 * tot);
    rep(i, tot, tot + n)
        segTree[i].val = vals[i - tot];

    for(int i = tot - 1; i; i--)
        segTree[i].val = f(segTree[i * 2].val, segTree[i * 2 + 1].val);
}

void run() {
    get n;
    int v[n];
    rep(i, 0, n)
        get v[i];
    init(vint(n, 0));
    vint d[n];
    int s = 0;
    rep(j, 0, n / 2 + n % 2)
        s += v[j];
    rep(i, 0, n) {
        if (i)
            s = s - v[i - 1] + v[(i + n / 2 + n % 2 - 1 + n) % n];
        d[i] = {s, i, (i + n / 2 + n % 2 - 1 + n) % n};
    }


    sort(d, d + n);
    for (int i = n - 1; i >= 0; i--) {
        //debug(d[i][0], d[i][1] + 1, d[i][2] + 1);
        if (d[i][1] > d[i][2]) {
            update(d[i][1] + 1, n, 1);
            update(1, d[i][2] + 1, 1);
        } else {
            update(d[i][1] + 1, d[i][2] + 1, 1);
        }
        
        if (query(1, n) == n / 2 + n % 2) {
            put d[i][0];
            return;
        }
    }
}

int32_t main() {srand(time(0)); ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); put fixed; put setprecision(15); run(); return 0;}

• Subtask 1: 0 / 20
• Subtask 2: 0 / 20
• Subtask 3: 0 / 20
• Subtask 4: 0 / 40