Submission #69871

#	Time	Username	Problem	Language	Result	Execution time	Memory
69871		Benq	Sparklers (JOI17_sparklers)	C++14	100 / 100	210 ms	31120 KiB

sparklers

#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/rope>

using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;

int N,K,T,mid;
vi X;

bool good(vd x, vd y) {
    if (x[0] > y[0]) return 0;
    pi L = {0,0}, R = {0,0};
    bool ok = 1;
    while (ok) {
        ok = 0;
        while (L.s < sz(x)-1 && x[L.s+1] <= y[R.f]) {
            ok = 1;
            if (x[++L.s] < x[L.f]) L.f = L.s;
        }
        while (R.s < sz(y)-1 && x[L.f] <= y[R.s+1]) {
            ok = 1;
            if (y[++R.s] > y[R.f]) R.f = R.s;
        }
    }
    return L.s == sz(x)-1 && R.s == sz(y)-1;
}

bool OK() {
    vd x,y;
    F0Rd(i,K+1) x.pb((ld)2*mid*T*i-X[i]);
    FOR(i,K,N) y.pb((ld)2*mid*T*i-X[i]);
    if (!good(x,y)) return 0;
    reverse(all(x)), reverse(all(y));
    if (!good(x,y)) return 0;
    return 1;
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> N >> K >> T; X.resize(N); K--;
    F0R(i,N) cin >> X[i];
    /*mid = 25303;
    cout << OK();
    exit(0);*/
    int lo = 0, hi = 1000000000;
    while (lo < hi) {
        mid = (lo+hi)/2;
        if (OK()) hi = mid;
        else lo = mid+1;
    }
    cout << lo;
}

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
* if you have no idea just guess the appropriate well-known algo instead of doing nothing :/
*/

• Subtask 1: 30 / 30
• Subtask 2: 20 / 20
• Subtask 3: 50 / 50