Submission #855746

#	Submission time	Handle	Problem	Language	Result	Execution time	Memory
855746	2023-10-01T17:36:48 Z	TimDee	Peru (RMI20_peru)	C++17	Compilation error	0 ms	0 KB

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;
struct node {
	int ans=inf, mina=0;
	bool full=1;
	int32_t maxb=0, minb=0;
};

node t[2*sz];
int32_t lazy[2*sz];
void merge(node&ret, node a, node b) {
	ret.ans = min(a.ans, b.ans);
	ret.full = (a.full && b.full) && (a.maxb == b.maxb);
	ret.mina = min(a.mina, b.mina);
	ret.maxb = max(a.maxb, b.maxb);
	ret.minb = min(a.minb, b.minb);
	if (ret.full) ret.ans=min(ret.ans, ret.maxb + ret.mina);
}
int32_t x;
void push(int v) {
	x = lazy[v];
	lazy[v]=0;
	if (2*v+2<2*sz) {
		lazy[2*v+1] = max(lazy[2*v+1],x);
		lazy[2*v+2] = max(lazy[2*v+2],x);
	}
	t[v].minb = t[v].maxb = x;
	t[v].ans = t[v].mina + x;
	t[v].full = 1;
}

void Set(int i, int x) {
	t[sz-1+i].mina = x;
	t[sz-1+i].ans = x + t[sz-1+i].maxb;
	int v = (sz+i-2)>>1;
	while (v > 0) {
		merge(t[v],t[2*v+1],t[2*v+2]);
		v = (v-1)>>1;
	}
	merge(t[0],t[1],t[2]);
}
void upd(int v, int l, int r, int lx, int rx, int x) {
	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;
		}
		assert(r-l > 1);
	}
	int m=(l+r)>>1;
	upd(2*v+1,l,m,lx,rx,x);
	upd(2*v+2,m,r,lx,rx,x);
	merge(t[v],t[2*v+1],t[2*v+2]);
}
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 (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;
   	
   	Set(0,0);
	vector<int> dp(n+1,inf);
	dp[0]=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,x);
	}

	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

Compilation timeout while compiling peru