Submission #1295298

#	Time	Username	Problem	Language	Result	Execution time	Memory
1295298		ender_shayan	Sparklers (JOI17_sparklers)	C++20	0 / 100	5 ms	572 KiB

sparklers

#include <bits/stdc++.h>

using namespace std;

// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;

typedef long long ll;
typedef long double	ld;
typedef pair<int, int>	pii  ;
typedef pair<ll, ll>	pll  ;
typedef vector<pii>     vii  ;
typedef vector<int>     veci ;
typedef vector<pll>     vll  ;
typedef vector<ll>      vecll;

// find_by_order             order_of_key

//#pragma GCC optimize("O3,unroll-loops")
//#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define F		        first
#define S		        second
#define pb		        push_back
#define endl            '\n'
#define Mp		        make_pair
#define all(x)          x.begin(), x.end()
#define debug(x)        cerr << #x << " = " << x << endl
#define set_dec(x)	    cout << fixed << setprecision(x);
#define fast_io         ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define file_io         freopen("in.txt" , "r" , stdin) ; freopen("out.txt" , "w" , stdout);
#define lb              lower_bound
#define ub              upper_bound
#define for1(n)         for(int i=1;i<=n;i++)
#define for0(n)         for(int i=0;i<n;i++)
#define forn(n)         for(int i=n;i>0;i--)
#define pq              priority_queue <pii, vector<pii>, greater<pii>>


const ll mod = 2e9 ;// 998244353 ;// 1e9+9;

ll inf=1e18;
const int N=1e6+100,L=21,bs=701,NN=1e6;
int B[N],C[N],D[N],E[N],n,m,q;
vector<int>g[N];
ll A[N],t,k;
ll mxx[N],mxxx[N];
bitset<N>is,iss;
bool check(ll t){
    vector<pll>vec1,vec2,vec3,vec4,vec5,vec6;
    vec3.pb({-inf,-inf*2});vec4.pb({-inf,-inf*2});
    for1(k-1){
        pll p={A[i+1]-A[i],t*2-(A[i+1]-A[i])};
        while(vec3.back().S>=0 && p.S<0){
            p.F=max(p.F,vec3.back().F-p.S);
            p.S+=vec3.back().S;
            vec3.pop_back();
        }
        vec3.pb(p);
    }
    for(int i=n-1;i>=k;i--){
        pll p={A[i+1]-A[i],t*2-(A[i+1]-A[i])};
        
        while(vec4.back().S>=0 && p.S<0){
            p.F=max(p.F,vec4.back().F-p.S);
            p.S+=vec4.back().S;
            vec4.pop_back();
        }
        vec4.pb(p);
    }
    ll p=t*2;
    while(vec1.size()!=1){
        if(vec3.back().F<=p && vec3.back().S>0){
            p+=vec3.back().S;
            vec3.pop_back();
        }
        else if(vec4.back().F<=p && vec4.back().S>0){
            p+=vec4.back().S;
            vec4.pop_back();
        }
        else if(vec3.back().F<=p && vec4.back().F<=p+vec3.back().S){
            p+=vec3.back().S;
            vec3.pop_back();
        }
        else if(vec4.back().F<=p && vec3.back().F<=p+vec4.back().S){
            p+=vec4.back().S;
            vec4.pop_back();
        }
        else
            break;
    }
    // cout<<vec3.size()<<" "<<vec4.size()<<endl;
    return(vec3.size()==1 && vec4.size()==1);

}
int main(){
    fast_io
    cin>>n>>k>>t;
    for1(n){
        cin>>A[i];
    }
    ll l=-1,r=mod/t+3;
    while(r-l!=1){
        ll mid=(l+r)>>1;
        if(check(t*mid)==0)l=mid;
        else r=mid;
    }
    cout<<l+1<<endl;


}

• Subtask 1: 0 / 30
• Subtask 2: 0 / 20
• Subtask 3: 0 / 50