# |
Submission time |
Handle |
Problem |
Language |
Result |
Execution time |
Memory |
855754 |
2023-10-01T17:58:08 Z |
TimDee |
Peru (RMI20_peru) |
C++17 |
|
0 ms |
0 KB |
// 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=inf;
int32_t maxb=0, minb=0;
};
node t[sz];
int32_t lazy[sz];
int dp[sz];
int32_t b[sz];
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;
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
Compilation timeout while compiling peru