@ding870 @tangtianyao0123 ovo

Hope I can solve DE.

Hell!I was ill!But hope I can solve EF(in this awful way)

Can you solve E

There are N cards and N piles of cards.

in question D, it should be Q piles of cards

It is hard for me to belive that 1k+ solved F. It didn't feel that easy to me.

Felt like D was in my hand, but it slipped through a hole

Easy A-F, Truly Hard G

Speed really matters. Those who solve A-F got rank 10 to rank 1040.

How to solve D ?

I solved G problem first part with a datastructure based solution. (non random)
Submission: https://atcoder.jp/contests/abc455/submissions/75273702

1st Subproblem:

• Count the number of beautiful sequences in which all elements have frequency = V
• So we can iterate on L pointer, from backside,(i.e) N to 1. and maintain all bad intervals for R pointers.
• Bad R pointers for a element a[i] = [first occurence , V th occurence) + [(V+1) th occurence , N]
• So this could be visualized as , for every new L pointer, we remove atmost 2 and add atmost 2 bad intervals.
• Now the problem boils down to, interval addition and removal, and finding points covered by no intervals. (good R pointers)
• This can be solved by LazySegTree, maintain (minimum , frequency of minimum) values in segment tree node, and perform range addition.
• The frequency of 0, in Segment tree, would give me the number of good R pointers on iterating for each L pointer.
• Idea: https://mirror.codeforces.com/blog/entry/137300

how's my code wrong at C? can somebody help me pls

#include<bits/stdc++.h>
using namespace std;
map<int,int>mp;
int main(){
	int n,k;
	cin>>n>>k;
	vector<int>a(n);
	int cols=0;
	long long sum=0; 
	vector<pair<long long,int> >p;
	for(int i=0;i<n;i++){
		cin>>a[i];
		sum+=a[i];
		if(mp[a[i]]==0){
			cols++;
			p.push_back(make_pair(0,a[i]));
		}
		mp[a[i]]++;
	}
	for(int i=0;i<cols;i++){
		p[i].first=(long long)mp[p[i].second]*p[i].second;
	}
	sort(p.rbegin(),p.rend());
	for(int i=0;i<min(k,(int)p.size());i++){
		sum-=p[i].first;
	}
	cout<<sum<<endl;
	return 0;
}