이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#include "peru.h"
//#include<ext/pb_ds/assoc_container.hpp>
//#include<ext/pb_ds/tree_policy.hpp>
using namespace std;
//using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef unsigned int ui;
#define ii pair<int,int>
#define pll pair<ll,ll>
#define vi vector<int>
#define vii vector<ii>
#define vll vector<ll>
#define vpll vector<pll>
#define matrix vector<vi>
#define matrixLL vector<vll>
#define vs vector<string>
#define vui vector<ui>
#define msi multiset<int>
#define mss multiset<string>
#define si set<int>
#define ss set<string>
#define PB push_back
#define PF push_front
#define PPB pop_back
#define PPF pop_front
#define X first
#define Y second
#define MP make_pair
#define FOR(i, a, b) for (int i = int(a); i < int(b); i++)
#define REP(i, n) FOR(i, 0, n)
#define all(x) (x).begin(), (x).end()
const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, -1, 1};
const int dxx[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dyy[] = {0, 0, -1, 1, -1, 1, -1, 1};
const string abc="abcdefghijklmnopqrstuvwxyz";
const string ABC="ABCDEFGHIJKLMNOPQRSTUVWXYZ";
const ld pi = 3.14159265;
const int mod = 1e9 + 7;
const int MOD = 1e9 + 7;
const int MAXN = 3e6 + 7;
const int inf = mod;
const ll INF = 1e18;
const ll zero = ll(0);
const int logo = 20;
const int off = 1 << logo;
const int trsz = off << 1;
ll dp[MAXN];
stack<ll> l, r;
struct node{
ll opt;
int arr, ind;
node(ll opt, int arr, int ind){
this->opt = opt;
this->arr = arr;
this->ind = ind;
}
};
deque<node> d;
int add(ll a, ll b){
return (a + b) % mod;
}
int mul(ll a, ll b){
return (a * b) % mod;
}
ll get_min(){
ll ret = INF;
if(!l.empty()) ret = min(ret, l.top());
if(!r.empty()) ret = min(ret, r.top());
return ret;
}
void push(stack<ll> &s, ll val){
if(s.empty()) s.push(val);
else s.push(min(val, s.top()));
}
void rebuild(){
while(!l.empty()) l.pop();
while(!r.empty()) r.pop();
int sz = d.size();
int line = sz / 2;
vector<node> v;
REP(i, line) v.PB(d.front()), d.PPF();
while(!v.empty()){
push(l, v.back().opt);
d.PF(v.back());
v.PPB();
}
REP(i, sz - line) v.PB(d.back()), d.PPB();
while(!v.empty()){
push(r, v.back().opt);
d.PB(v.back());
v.PPB();
}
}
int solve(int n, int k, int *arr){
int ret = 0;
for(int i = 1; i <= n; i++){
if(i == 1){
dp[i] = arr[i - 1];
d.PB(node(dp[i], arr[i - 1], i));
push(r, arr[i - 1]);
ret = add(mul(ret, 23), dp[i]);
continue;
}
//prvi veci u stacku(1 indeks.)
int dpind = i;
while(!d.empty() and d.back().arr <= arr[i - 1]){
dpind = d.back().ind;
d.PPB();
if(r.empty()) rebuild();
else r.pop();
}
d.PB(node(dp[dpind - 1] + arr[i - 1], arr[i - 1], dpind));
push(r, dp[dpind - 1] + arr[i - 1]);
if(i > k){
ll arrmax = d.front().arr;
d.pop_front();
if(l.empty()) rebuild();
else l.pop();
//ak sam popo maksimum i maks je zadni ubacen
if(d.empty()){
d.push_front(node(dp[i - k] + arrmax, arrmax, i - k + 1));
push(l, dp[i - k] + arrmax);
}else if(d.front().ind > i - k + 1){
//ako onaj kojeg sam popo nije najlijeviji ga vrati nazad
d.push_front(node(dp[i - k] + arrmax, arrmax, i - k + 1));
push(l, dp[i - k] + arrmax);
}
}
dp[i] = get_min();
//cout << dp[i] << " ";
ret = add(mul(ret, 23), dp[i]);
}
return ret;
}
/*
void solv(){
int n, k;
cin >> n >> k;
vi arr; arr.clear();
REP(i, n){
int x;
cin >> x;
arr.PB(x);
}
cout << solve(n, k, arr);
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int t=1;
//cin >> t;
while(t--)solv();
return 0;
}
*/
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