Submission #918739

#	Time	Username	Problem	Language	Result	Execution time	Memory
918739		Abito	Real Mountains (CCO23_day1problem2)	C++17	0 / 25	66 ms	188320 KiB

Main

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#define F first
#define S second
#define pb push_back
#define ppb pop_back
#define ep insert
#define endl '\n'
#define elif else if
#define pow pwr
#define sqrt sqrtt
//#define int long long
#define ll long long
#define y1 YONE
#define free freeee
#define lcm llcm
typedef unsigned long long ull;
using namespace std;
const int N=1e6+5,M=1e6+3;
int a[N],n,p,mx,b[N];
multiset<int> seg[4*N+5];
void build(int x,int l,int r){
    for (int i=l;i<=r;i++) seg[x].ep(a[i]);
    if (l==r) return;
    int mid=(l+r)/2;
    build(x*2,l,mid);
    build(x*2+1,mid+1,r);
    return;
}
void edit(int x,int l,int r,int i,int val){
    seg[x].erase(seg[x].find(val));
    seg[x].ep(val+1);
    if (l==r) return;
    int mid=(l+r)/2;
    if (i<=mid) edit(x*2,l,mid,i,val);
    else edit(x*2+1,mid+1,r,i,val);
    return;
}
int query(int x,int l,int r,int lx,int rx,int val){
    if (l>rx || r<lx) return INT_MAX;
    if (l>=lx && r<=rx) return *seg[x].upper_bound(val);
    int mid=(l+r)/2;
    return min(query(x*2,l,mid,lx,rx,val),query(x*2+1,mid+1,r,lx,rx,val));
}
int32_t main(){
    ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
    cin>>n;
    for (int i=1;i<=n;i++){
        cin>>a[i];
        if (a[i]<=mx) continue;
        p=i;
        mx=a[i];
    }
    for (int i=1;i<=p;i++) b[i]=max(b[i-1],a[i]);
    for (int i=n;i>=p;i--) b[i]=max(b[i+1],a[i]);
    set<pair<int,int>> s;
    for (int i=1;i<=n;i++) if (a[i]!=b[i]) s.ep({a[i],i});
    build(1,1,n);
    int ans=0;
    while (!s.empty()){
        pair<int,int> x=*s.begin();
        s.erase(x);
        int l=query(1,1,n,1,x.S-1,x.F),r=query(1,1,n,x.S+1,n,x.F);
        ans=(ans+x.F+l+r)%M;
        edit(1,1,n,x.S,x.F);
        a[x.S]++;
        if (a[x.S]!=b[x.S]) s.ep({a[x.S],x.S});
    }cout<<ans<<endl;
    return 0;
}

• Subtask 1: 0 / 3
• Subtask 2: 0 / 3
• Subtask 3: 0 / 3
• Subtask 4: 0 / 3
• Subtask 5: 0 / 4
• Subtask 6: 0 / 5
• Subtask 7: 0 / 4