Submission #156819

#	Time	Username	Problem	Language	Result	Execution time	Memory
156819		jhnah917	Split the sequence (APIO14_sequence)	C++14	0 / 100	18 ms	4840 KiB

sequence

#include <bits/stdc++.h>
#define x first
#define y second
#define all(v) v.begin(), v.end()
using namespace std;

typedef long long ll;
typedef pair<ll, ll> p;

struct Line{
	ll a, b, c;
	ll get(ll x){
		return a*x+b;
	}
};

struct CHT{
	vector<Line> stk;
	int idx = 1;
	
	void insert(Line v){
		while(stk.size() >= 2){
			if(v.a == stk.back().a && v.b >= stk.back().b) stk.pop_back();
			else if((stk.back().b - v.b)*(stk.back().a - stk[stk.size()-2].a) < (v.a - stk.back().a)*(stk[stk.size()-2].b - v.b)){
				stk.pop_back();
			}else break;
		}
		if(stk.back().a == v.a && stk.back().b <= v.b && v.c) return;
		stk.push_back(v);
	}
	
	p query(ll x){
		while(idx+1 < stk.size() && (stk[idx].b - stk[idx+1].b) < (stk[idx+1].a - stk[idx].a) * x) idx++;
		return {stk[idx].get(x), stk[idx].c};
	}
};

ll arr[101010];
ll sum[101010];
ll dp[2][101010];
int track[222][101010];
vector<int> ans;

int main(){
	ios_base::sync_with_stdio(0); cin.tie(0);
	int n, k; cin >> n >> k;
	for(int i=1; i<=n; i++) cin >> arr[i], sum[i] = sum[i-1] + arr[i];
	for(int i=1; i<=k; i++){
		CHT cht; cht.stk.reserve(101010); cht.insert({0, 0, 0});
		for(int j=1; j<=n; j++){
			auto now = cht.query(sum[j]);
			dp[i&1][j] = now.x;
			track[i][j] = now.y;
			cht.insert({sum[j], dp[(i-1)&1][j] - sum[j]*sum[j], j});
		}
	}
	
	cout << dp[k&1][n] << "\n";
	
	int now = n;
	ans.push_back(-1);
	for(int i=1; i<=k; i++){
		ans.push_back(track[k-i+1][now]);
		now = track[k-i+1][now];
	}
	sort(all(ans));
	for(int i=1; i<=k; i++){
		if(!ans[i]) ans[i] = 1;
		if(ans[i] <= ans[i-1]) ans[i] = ans[i-1] + 1;
	}
	for(int i=1; i<=k; i++) cout << ans[i] << " ";
}

Compilation message (stderr)

sequence.cpp: In member function 'p CHT::query(ll)':
sequence.cpp:33:15: warning: comparison between signed and unsigned integer expressions [-Wsign-compare]
   while(idx+1 < stk.size() && (stk[idx].b - stk[idx+1].b) < (stk[idx+1].a - stk[idx].a) * x) idx++;
         ~~~~~~^~~~~~~~~~~~

• Subtask 1: 0 / 11
• Subtask 2: 0 / 11
• Subtask 3: 0 / 11
• Subtask 4: 0 / 17
• Subtask 5: 0 / 21
• Subtask 6: 0 / 29