Submission #312264

#	Time	Username	Problem	Language	Result	Execution time	Memory
312264		arwaeystoamneg	Split the sequence (APIO14_sequence)	C++17	0 / 100	1703 ms	22008 KiB

sequence

/*
ID: awesome35
LANG: C++14
TASK: vans
*/
#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
#include<unordered_set>
#include<unordered_map>
#include<chrono>

using namespace std;
typedef pair<int, int> pii;
typedef long long ll;
typedef pair<ll, ll> pll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<pair<int, int>> vpi;
typedef vector<pair<ll, ll>> vpll;

#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define trav(a,x) for (auto& a: x)

#define pb push_back
#define mp make_pair
#define rsz resize
#define sz(x) int(x.size())
#define all(x) x.begin(),x.end()
#define f first
#define s second
#define cont continue
#define endl '\n'
#define ednl '\n'
#define test int testc;cin>>testc;while(testc--)
#define pq priority_queue
const int dx[4] = { 1,0,-1,0 }, dy[4] = { 0,1,0,-1 }; // for every grid problem!!
const ll linf = 4000000000000000000LL;
const ll inf = 1000000007;//998244353
const ld pi = 3.1415926535;

void pv(vi a) { trav(x, a)cout << x << " "; cout << endl; }void pv(vll a) { trav(x, a)cout << x << " "; cout << endl; }void pv(vector<vi>a) {
	F0R(i, sz(a)) { cout << i << endl; pv(a[i]); cout << endl; }
}void pv(vector<vll>a) { F0R(i, sz(a)) { cout << i << endl; pv(a[i]); }cout << endl; }void pv(vector<string>a) { trav(x, a)cout << x << endl; cout << endl; }
void build_primes(vi& primes, int size)
{
	vi visited;
	visited.rsz(size, 0);
	FOR(i, 2, size)
	{
		if (visited[i] == 0)
		{
			primes.pb(i);
			int a = i;
			while (a < size)
			{
				visited[a] = 1;
				a += i;
			}
		}
	}
}
vector<vector<ll>> matrix_mult(vector<vector<ll>>& a, vector<vector<ll>>& b)
{
	int n = a.size();
	vector<vector<ll>> answer;
	answer.resize(n);
	for (int i = 0; i < n; i++) answer[i].resize(n, 0);
	for (int i = 0; i < n; i++)
	{
		for (int j = 0; j < n; j++) // calculate answer[i][j]
		{
			for (int k = 0; k < n; k++)
				answer[i][j] = (answer[i][j] + a[i][k] * b[k][j]) % inf;
		}
	}
	return answer;
}
int modInverse(int a, int m)
{
	int m0 = m;
	int y = 0, x = 1;
	if (m == 1)
		return 0;
	while (a > 1)
	{
		// q is quotient 
		int q = a / m;
		int t = m;
		// m is remainder now, process same as 
		// Euclid's algo 
		m = a % m, a = t;
		t = y;
		// Update y and x 
		y = x - q * y;
		x = t;
	}
	// Make x positive 
	if (x < 0)
		x += m0;

	return x;
}
ll power(ll x, ll y)
{
	ll k = 1LL << 60;
	ll z = 1;
	while (k != 0)
	{
		z *= z;
		z %= inf;
		if (y >= k)
		{
			z *= x;
			z %= inf;
			y -= k;
		}
		k >>= 1;
	}
	return z;
}
struct point
{
	ld x, y;
	bool operator<(const point& rhs)const
	{
		if (x == rhs.x)return y < rhs.y;
		return x < rhs.x;
	}
};
struct pt
{
	ll x, y;
	bool operator<(const point& rhs)const
	{
		if (x == rhs.x)return y < rhs.y;
		return x < rhs.x;
	}
};
// remember that you need to take abs
long double area(point x, point y, point z)
{
	return (x.y * y.x + y.y * z.x + z.y * x.x - x.x * y.y - y.x * z.y - z.x * x.y) / 2.0;
}
bool clockwise(point x, point y, point z)
{
	return area(x, y, z) > 0;
}
// remember that you need to take abs
long double area(pt x, pt y, pt z)
{
	return (x.y * y.x + y.y * z.x + z.y * x.x - x.x * y.y - y.x * z.y - z.x * x.y) / 2.0;
}
bool clockwise(pt x, pt y, pt z)
{
	return area(x, y, z) > 0;
}
ll gcd(ll a, ll b)
{
	if (a > b)swap(a, b);
	if (a == 0)return b;
	return gcd(a, b % a);
}
int popcount(ll a)
{
	int count = 0;
	while (a)
	{
		count += (a & 1);
		a >>= 1;
	}
	return count;
}
ll choose(ll n, ll r)
{
	ll p = 1, k = 1;
	if (n - r < r)
		r = n - r;
	if (r != 0) {
		while (r) {
			p *= n;
			k *= r;
			long long m = gcd(p, k);
			p /= m;
			k /= m;
			n--;
			r--;
		}
	}
	else
		p = 1;
	return p;
}
vll prefix_hash(string& a, vll& powers)
{
	int n = sz(a);
	vll prefix(n + 1);
	F0R(i, n)prefix[i + 1] = (prefix[i] + powers[i] * (a[i] - 'a' + 1)) % inf;
	return prefix;
}
struct custom_hash {
	static uint64_t splitmix64(uint64_t x) {
		// http://xorshift.di.unimi.it/splitmix64.c
		x += 0x9e3779b97f4a7c15;
		x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
		x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
		return x ^ (x >> 31);
	}
	//the return type was size_t. But isnt that problematic?
	uint64_t operator()(uint64_t x) const {
		static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
		return splitmix64(x + FIXED_RANDOM);
	}
};
struct custom_hash_fast {
	//the return type was size_t. But isnt that problematic?
	uint64_t operator()(uint64_t x) const {
		static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
		x ^= FIXED_RANDOM;
		return x ^ (x >> 16);
	}
};
void setIO(string s) {
	ios_base::sync_with_stdio(0); cin.tie(0);
	//freopen((s + ".in").c_str(), "r", stdin);
	//freopen((s + ".out").c_str(), "w", stdout);
}
const int mxN = 1e5;
int n, ptr, last;
ll a[mxN + 1], b[mxN + 1], c[mxN + 1];
int check(ll ai, ll bi)
{
	if (a[last - 1] == ai)
	{
		return 1;
	}
	return (b[last - 1] - b[last - 2]) / (a[last - 2] - a[last - 1]) >= (bi - b[last - 1]) / (a[last - 1] - ai);
}
void al(ll ai, ll bi, ll index) {
	while (last - ptr >= 2 && check(ai, bi))
		--last;
	a[last] = ai, b[last] = bi, c[last] = index;
	++last;
}
ll intersect(ll xi, ll ptr)
{
	return a[ptr] * xi + b[ptr];
}
pair<ll,ll> query(ll xi) {
	while (last - ptr >= 2 && intersect(xi, ptr) <= intersect(xi, ptr + 1))
		++ptr;
	return { a[ptr] * xi + b[ptr],c[ptr] };
}
int main()
{
	setIO("cbarn");
	int n, k;
	cin >> n >> k;
	vll a(n);
	F0R(i, n)cin >> a[i];
	vll prefix(n + 1);
	F0R(i, n)prefix[i + 1] = prefix[i] + a[i];
	vector<vi>ans(k + 1, vi(n));
	vector<vll>dp(2, vll(n));
	F0R(i, k)
	{
		FOR(j, i + 1, n)
		{
			al(prefix[j], dp[0][j - 1] - prefix[j] * prefix[j], j);
			pair<ll,ll>t = query(prefix[j + 1]);
			dp[1][j] = t.f;
			ans[i + 1][j] = t.s;
		}
		dp[0] = dp[1];
	}
	//pv(dp);
	cout << dp[0][n - 1] << endl;
	vi b;
	int i = k, j = n - 1;
	while (i)
	{
		j = ans[i][j];
		b.pb(j--);
		i--;
	}
	reverse(all(b));
	pv(b);
}

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