Submission #855725

#	Time	Username	Problem	Language	Result	Execution time	Memory
855725		TimDee	Peru (RMI20_peru)	C++17	49 / 100	235 ms	262144 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

struct sgt {
	struct node {
		int ans=inf, full=1, mina=0, maxb=0, minb=0;
	};
	node merge(node a, node b) {
		node ret;
		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);

		return ret;
	}
	void push(int v) {
		int x = lazy[v];
		lazy[v]=0;
		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;
		//cout<<"? "<<v<<' '<<t[v].mina<<' '<<t[v].ans<<' '<<x<<'\n';
		t[v].ans = t[v].mina + x;
		t[v].full = 1;
	}

	vector<node> t; int sz=1;
	vector<int> lazy;

	sgt(int n) {
		while (sz<n) sz<<=1;
		t.resize(2*sz);
		lazy.resize(4*sz);
	}
	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) {
			t[v] = merge(t[2*v+1],t[2*v+2]);
			v = (v-1)>>1;
		}
		t[0] = merge(t[1],t[2]);
	}
	void upd(int v, int l, int r, int lx, int rx, int x) {
		//cout<<"Upd "<<v<<' '<<l<<' '<<r<<' '<<lx<<' '<<rx<<' '<<x<<'\n';
		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);
				//cout<<"! "<<t[v].ans<<' '<<t[v].mina<<' '<<t[v].full<<' '<<t[v].minb<<' '<<t[v].maxb<<'\n';
				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);
		t[v]=merge(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;
   	
   	sgt st(n+1);
   	st.set(0,0);
	vector<int> dp(n+1,inf);
	dp[0]=0;
	for (int i=1; i<=n; ++i) {
		st.upd(0,i,a[i-1]);
		int x = st.query(max(i-k,0ll),i);
		dp[i] = x;
		st.set(i,x);
		//cout<<dp[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