Problem:
Score : 1200 points

Problem Statement

#nck {
width: 30px;
height: auto;
}

Imagine a game played on a line. Initially, the player is located at position 0 with N candies in his possession, and the exit is at position E. There are also N bears in the game. The i-th bear is located at x_i. The maximum moving speed of the player is 1 while the bears do not move at all.

When the player gives a candy to a bear, it will provide a coin after T units of time. More specifically, if the i-th bear is given a candy at time t, it will put a coin at its position at time t+T. The purpose of this game is to give candies to all the bears, pick up all the coins, and go to the exit. Note that the player can only give a candy to a bear if the player is at the exact same position of the bear. Also, each bear will only produce a coin once. If the player visits the position of a coin after or at the exact same time that the coin is put down, the player can pick up the coin. Coins do not disappear until collected by the player.

Shik is an expert of this game. He can give candies to bears and pick up coins instantly. You are given the configuration of the game. Please calculate the minimum time Shik needs to collect all the coins and go to the exit.

Constraints

1 \leq N \leq 100,000

1 \leq T, E \leq 10^9

0 < x_i < E

x_i < x_{i+1} for 1 \leq i < N

All input values are integers.

Partial Scores

In test cases worth 600 points, N \leq 2,000.

Input

The input is given from Standard Input in the following format:

N E T
x_1 x_2 ... x_N

Output

Print an integer denoting the answer.

Sample Input 1

3 9 1
1 3 8

Sample Output 1

12

The optimal strategy is to wait for the coin after treating each bear. The total time spent on waiting is 3 and moving is 9. So the answer is 3 + 9 = 12.

Sample Input 2

3 9 3
1 3 8

Sample Output 2

16

Sample Input 3

2 1000000000 1000000000
1 999999999

Sample Output 3

2999999996

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define FOR(i, begin, end) for(int i=(begin);i<(end);i++)
#define REP(i, n) FOR(i,0,n)
#define IFOR(i, begin, end) for(int i=(end)-1;i>=(begin);i--)
#define IREP(i, n) IFOR(i,0,n)
#define Sort(v) sort(v.begin(), v.end())
#define Reverse(v) reverse(v.begin(), v.end())
#define all(v) v.begin(),v.end()
#define SZ(v) ((int)v.size())
#define Lower_bound(v, x) distance(v.begin(), lower_bound(v.begin(), v.end(), x))
#define Upper_bound(v, x) distance(v.begin(), upper_bound(v.begin(), v.end(), x))
#define Max(a, b) a = max(a, b)
#define Min(a, b) a = min(a, b)
#define bit(n) (1LL<<(n))
#define bit_exist(x, n) ((x >> n) & 1)
#define debug(x) cout << #x << "=" << x << endl;
#define vdebug(v) cout << #v << "=" << endl; REP(i, v.size()){ cout << v[i] << ","; } cout << endl;
#define mdebug(m) cout << #m << "=" << endl; REP(i, m.size()){ REP(j, m[i].size()){ cout << m[i][j] << ","; } cout << endl;}
#define pb push_back
#define f first
#define s second
#define int long long
#define INF 1000000000000000000
template<typename T> istream &operator>>(istream &is, vector<T> &v){ for (auto &x : v) is >> x; return is; }
template<typename T> ostream &operator<<(ostream &os, vector<T> &v){ for(int i = 0; i < v.size(); i++) { cout << v[i]; if(i != v.size() - 1) cout << endl; }; return os; }
template<typename T> void Out(T x) { cout << x << endl; }
template<typename T1, typename T2> void Ans(bool f, T1 y, T2 n) { if(f) Out(y); else Out(n); }

using vec = vector<int>;
using mat = vector<vec>;
using Pii = pair<int, int>;
using PiP = pair<int, Pii>;
using PPi = pair<Pii, int>;
using bools = vector<bool>;
using pairs = vector<Pii>;

//int dx[4] = {1,0,-1,0};
//int dy[4] = {0,1,0,-1};
//char d[4] = {'D','R','U','L'};

const int mod = 1000000007;
//const int mod = 998244353;
#define Add(x, y) x = (x + (y)) % mod
#define Mult(x, y) x = (x * (y)) % mod

struct LazySegmentTree
{
    //A: data, B: 作用素
    using T_A = int;
    using T_B = int;
    T_A id_A = INF;
    T_B id_B = INF;
    int N;
    vector<T_A> dat;
    vector<T_B> ope;

    T_A F_AA(T_A u, T_A v){
        return min(u, v);
    }

    T_A F_AB(T_A u, T_B v, int sz){
        return min(u, v);
    }

    T_B F_BB(T_B u, T_B v){
        return min(u, v);
    }

    LazySegmentTree(int n){
        N = 1;
        while(n > N) N = N << 1;
        dat = vector<T_A>(2 * N - 1, id_A);
        ope = vector<T_B>(2 * N - 1, id_B);
    }

    LazySegmentTree(int n, vector<T_A> v){
        N = 1;
        while(n > N) N = N << 1;
        dat = vector<T_A>(2 * N - 1, id_A);
        ope = vector<T_B>(2 * N - 1, id_B);
        REP(i, n) dat[i + N - 1] = v[i];
        IREP(i, N - 1) dat[i] = F_AA(dat[i * 2 + 1], dat[i * 2 + 2]); 
    }

    //遅延評価
    void eval(int k, int l, int r){
        if(ope[k] != id_B){
            dat[k] = F_AB(dat[k], ope[k], r - l);
            if(r - l > 1){
                ope[k * 2 + 1] = F_BB(ope[k * 2 + 1], ope[k]);
                ope[k * 2 + 2] = F_BB(ope[k * 2 + 2], ope[k]);
            }
            ope[k] = id_B;
        }
    }

    void update(int a, int b, T_B x, int k, int l, int r){

        eval(k, l, r);
        if(b <= l || r <= a) return;

        if(a <= l && r <= b){
            ope[k] = F_BB(ope[k], x);
            eval(k, l, r);
        }else{
            update(a, b, x, k * 2 + 1, l, (l + r) / 2);
            update(a, b, x, k * 2 + 2, (l + r) / 2, r);
            dat[k] = F_AA(dat[k * 2 + 1], dat[k * 2 + 2]);
        }

    }

    T_A get(int a, int b, int k, int l, int r){
        if(r <= a || b <= l) return id_A;

        eval(k, l, r);
        if(a <= l && r <= b) return dat[k];
        else{
            T_A vl = get(a, b, k * 2 + 1, l, (l + r) / 2);
            T_A vr = get(a, b, k * 2 + 2, (l + r) / 2, r);
            return F_AA(vl, vr);
        }
    }

    //[a,b)の区間更新
    void update(int a, int b, T_B x){ update(a, b, x, 0, 0, N); }

    //[a, b)の演算結果
    T_A get(int a, int b){ return get(a, b, 0, 0, N); }

    T_A val(int k){ return get(k, k + 1); }
};

signed main(){

    int N, E, T; cin >> N >> E >> T;
    vec x(N); cin >> x;

    LazySegmentTree ST1(N), ST2(N);
    int t0 = 0;
    REP(i, N){
        int t;
        if(i == 0) t = t0 + x[i];
        else t = t0 + x[i] - x[i - 1];

        int i0 = Upper_bound(x, x[i] + T / 2);
        ST1.update(i, i0, t + T - x[i]);
        ST2.update(i0, N, t - 3 * x[i]);

        int v1 = ST1.val(i) + x[i];
        int v2 = ST2.val(i) + 3 * x[i];
        t0 = min(v1, v2);
    }
    int ans = t0 + E - x[N - 1];
    Out(ans);

    return 0;
}