Submission #855757

#	Time	Username	Problem	Language	Result	Execution time	Memory
855757		TimDee	Peru (RMI20_peru)	C++17	49 / 100	1049 ms	174308 KiB

peru

//  Esti <3

//\
     šťastia pre nás :)
//   you're already the best
//             _
//   ^ ^      //
// >(O_O)<___//
//   \ __ __  \
//    \\ \\ \\\\
 
#include <bits/stdc++.h>
using namespace std;
 
//#pragma GCC optimize("O3","unroll-loops")
//#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
 
#pragma GCC optimize("O3")
#pragma GCC target("popcnt")

using ll = long long;
#define int long long
#define forn(i,n) for(int i=0; i<(n); ++i)
#define pb push_back
#define pi pair<int,int>
#define f first
#define s second 
#define vii(a,n) vector<int> a(n); forn(i,n) cin>>a[i];
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
 
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
const ll inf = 1e18;
const ll mod = 1e9+7;//998244853;

// \
\
:smiling_face_with_3_hearts: :smiling_face_with_3_hearts:  :smiling_face_with_3_hearts:  
 
//vidime sa veľmi skoro, moje slnko

//const int sz = 1<<22;
int sz=1;
struct node {
	int ans=inf, mina=inf;
	int32_t maxb=0, minb=0;
};

//node t[sz];
//int32_t lazy[sz];
//int dp[sz];
//int32_t b[sz];
vector<node> t;
vector<int32_t> lazy, b;
vector<int> dp;
void merge(int v) {
	if (2*v+2 >= sz-1) {
		t[v].ans = min(dp[2*v+1-sz+1]+b[2*v+1-sz+1], dp[2*v+2-sz+1]+b[2*v+2-sz+1]);
		t[v].mina = min(dp[2*v+1-sz+1], dp[2*v+2-sz+1]);
		t[v].minb = min(b[2*v+1-sz+1], b[2*v+2-sz+1]);
		t[v].maxb = max(b[2*v+1-sz+1], b[2*v+2-sz+1]);
		if (t[v].minb == t[v].maxb) t[v].ans=min(t[v].ans, t[v].maxb + t[v].mina);
		return;
	}
	t[v].ans = min(t[2*v+1].ans, t[2*v+2].ans);
	t[v].mina = min(t[2*v+1].mina, t[2*v+2].mina);
	t[v].maxb = max(t[2*v+1].maxb, t[2*v+2].maxb);
	t[v].minb = min(t[2*v+1].minb, t[2*v+2].minb);
	if (t[v].minb == t[v].maxb) t[v].ans=min(t[v].ans, t[v].maxb + t[v].mina);
}
int32_t x;
void push(int v) {
	x = lazy[v];
	lazy[v]=0;
	if (2*v+2<sz-1) {
		lazy[2*v+1] = max(lazy[2*v+1],x);
		lazy[2*v+2] = max(lazy[2*v+2],x);
	} else {
		b[2*v+1-sz+1]=max(b[2*v+1-sz+1],x);
		b[2*v+2-sz+1]=max(b[2*v+2-sz+1],x);
	}
	t[v].minb = t[v].maxb = x;
	t[v].ans = t[v].mina + x;
}

void Set(int i) {
	int v = (sz+i-2)>>1;
	while (v > 0) {
		merge(v);
		v = (v-1)>>1;
	}
	merge(0);
}
void upd(int v, int l, int r, int lx, int rx, int x) {
	if (r-l==1) {
		if (rx<=l || r<=lx) return;
		if (b[v-sz+1] >= x) return;
		b[v-sz+1] = x;
		return;
	}
	if (lazy[v]) push(v);
	if (t[v].minb >= x) return;
	if (rx<=l || r<=lx) return;
	if (lx<=l && r<=rx) {
		if (t[v].maxb <= x) {
			lazy[v] = x;
			push(v);
			return;
		}
	}
	int m=(l+r)>>1;
	upd(2*v+1,l,m,lx,rx,x);
	upd(2*v+2,m,r,lx,rx,x);
	merge(v);
}
void upd(int l, int r, int x) {
	upd(0,0,sz,l,r,x);
}
int query(int v, int l, int r, int lx, int rx) {
	if (r-l==1) {
		if (rx<=l || r<=lx) return inf;
		return dp[v-sz+1]+b[v-sz+1];
	}
	if (lazy[v]) push(v);
	if (rx<=l || r<=lx) return inf;
	if (lx<=l && r<=rx) return t[v].ans;
	int m=(l+r)>>1;
	int lq = query(2*v+1,l,m,lx,rx);
	int rq = query(2*v+2,m,r,lx,rx);
	return min(lq,rq);
}
int query(int l, int r) {
	return query(0,0,sz,l,r);
}

int32_t solve(int32_t n, int32_t k, int32_t* a) {
	if (n==1) return a[0]%mod;

	while (sz<n) sz<<=1;
	t.resize(sz);
	b.resize(sz);
	dp.resize(sz);
	lazy.resize(sz);
   	
	forn(i,n+1) dp[i]=inf;
	dp[0]=0;
	Set(0);
	for (int i=1; i<=n; ++i) {
		upd(0,i,a[i-1]);
		int x = query(max(i-k,0ll),i);
		dp[i] = x;
		Set(i);
	}

	int ans=0, p=1;
	for (int i=n; i; --i) {
		dp[i]%=mod;
		ans=(ans+p*dp[i])%mod;
		p=(p*23)%mod;
	}
	return ans;

}

Compilation message (stderr)

peru.cpp:3:1: warning: multi-line comment [-Wcomment]
    3 | //\
      | ^
peru.cpp:9:1: warning: multi-line comment [-Wcomment]
    9 | //   \ __ __  \
      | ^
peru.cpp:36:1: warning: multi-line comment [-Wcomment]
   36 | // \
      | ^

• Subtask 1: 18 / 18
• Subtask 2: 31 / 31
• Subtask 3: 0 / 51