Submission #23639

#	Time	Username	Problem	Language	Result	Execution time	Memory
23639		SYury	Split the sequence (APIO14_sequence)	C++11	0 / 100	16 ms	81876 KiB

sequence

#include<bits/stdc++.h>

using namespace std;
typedef long long lint;
typedef long double ldb;
typedef unsigned long long uli;

#define X first
#define Y second
#define F(i, l, r) for(int i = l; i != r; i++)
#define Df(i, l, r) for(int i = l; i != r; i--)
#define I(i, a) for(auto i : a)
#define pb push_back
#define rs resize
#define mk make_pair
#define asg assign
#define all(x) x.begin(),x.end()
#define ret return
#define cont continue
#define brk break
#define ins insert
#define era erase
#define fi0(x) memset(x, 0, sizeof(x))
#define finf(x) memset(x, 127, sizeof(x))
#define acpy(y, x) memcpy(y, x, sizeof(y))
#define y1 adjf
#define tm dhgdg

const int MAXN = 1e5 + 3;
const int MAXK = 2e2 + 2;
const ldb eps = 1e-16;

int pr[MAXN][MAXK];
int a[MAXN];
int pref[MAXN];
int n, k;

struct line{
	lint b;
	int id;
	line(lint pb, int pid):b(pb),id(pid){}
	lint f(int x){
		ret pref[id] * 1ll * x + b;
	}
	lint k(){
		ret pref[id];
	}
};

struct convex_hull{
	vector<line> l;
	int ptr = 0;
	void add_line(line curr){
		while(!l.empty() && l.back().k() == curr.k() && l.back().b <= curr.b)l.pop_back();
		if(!l.empty() && l.back().k() == curr.k() && l.back().b > curr.b){
			ptr = min(ptr, (int)l.size() - 1);
			ret;
		}
		while(l.size() >= 2){
			line l1 = l[(int)l.size() - 1], l2 = l[(int)l.size() - 2];
			//lint f1 = (l1.k - l2.k) * (l1.b - curr.b), f2 = (l2.b - l1.b) * (curr.k - l1.k);
			ldb f2 = (l2.b - l1.b)/((ldb)l1.k() - l2.k()), f1 = (l1.b - curr.b)/((ldb)curr.k() - l1.k());
			if(f2 < f1 - eps)brk;
			l.pop_back();
		}
		l.pb(curr);
		ptr = min(ptr, (int)l.size() - 1);
	}
	pair<int, lint> get(int x){
		if(l.empty())ret mk(-1, 0);
		while(ptr < (int)l.size() - 1 && l[ptr].f(x) <= l[ptr + 1].f(x))ptr++;
		ret mk(l[ptr].id, l[ptr].f(x));
	}
};

convex_hull ch[MAXK];

void restore(int i, int j){
	if(i < 0)ret;
	printf("%d ", i + 1);
	restore(pr[i][j], j - 1);
}

lint dcp[MAXK];

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	scanf("%d%d", &n, &k);
	F(i, 0, n)scanf("%d", &a[i]);
	F(i, 0, n)pref[i] = a[i] + ((i == 0) ? 0 : pref[i - 1]);
	k++;
	F(i, 0, n)
		F(j, 1, k + 1)
			pr[i][j] = -2;
	pr[0][1] = -1;
	ch[1].add_line(line(-pref[0] * 1ll * pref[0], 0));
	F(i, 1, n){
		dcp[0] = 0;
		pr[i][1] = -1;
		F(j, 2, k + 1){
			pair<int, lint> tmp = ch[j - 1].get(pref[i]);
			int pi = tmp.X;
			if(pi == -1)cont;
			pr[i][j] = pi;
			dcp[j] = tmp.Y;
			if(j > 2 && pr[i][j] != -2 && pr[i][j - 1] != -2 && pr[i][j] != pr[i][j - 1])ret 0;
		}
		F(j, 1, k + 1)if(pr[i][j] != -2)ch[j].add_line(line(dcp[j] - pref[i] * 1ll * pref[i], i));	
	}
	printf("%lld\n", dcp[k]);
	restore(pr[n - 1][k], k - 1);
	ret 0;
}

Compilation message (stderr)

sequence.cpp: In function 'int main()':
sequence.cpp:89:23: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
  scanf("%d%d", &n, &k);
                       ^
sequence.cpp:90:30: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
  F(i, 0, n)scanf("%d", &a[i]);
                              ^

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