Разбор Educational Codeforces Round 142

№	Пользователь	Рейтинг
1	tourist	4009
2	jiangly	3823
3	Benq	3738
4	Radewoosh	3633
5	jqdai0815	3620
6	orzdevinwang	3529
7	ecnerwala	3446
8	Um_nik	3396
9	ksun48	3390
10	gamegame	3386

№	Пользователь	Вклад
1	cry	166
2	maomao90	163
2	Um_nik	163
4	atcoder_official	161
5	adamant	160
6	-is-this-fft-	158
7	awoo	157
8	TheScrasse	154
9	nor	153
9	Dominater069	153

Разбор

Решение (Neon)

#include <bits/stdc++.h>

using namespace std;

int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int cnt1 = 0;
    for (int i = 0; i < n; ++i) {
      int x;
      cin >> x;
      cnt1 += (x == 1);
    }
    cout << n - cnt1 / 2 << '\n';
  }
}

1792B - Стендап комик

Идея: BledDest

Разбор

Решение (awoo)

for _ in range(int(input())):
    a1, a2, a3, a4 = map(int, input().split())
    if a1 == 0:
        print(1)
    else:
        print(a1 + min(a2, a3) * 2 + min(a1 + 1, abs(a2 - a3) + a4))

1792C - Min Max сортировка

Идея: BledDest

Разбор

Решение (Neon)

#include <bits/stdc++.h>

using namespace std;

int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> pos(n + 1);
    for (int i = 0; i < n; ++i) {
      int x;
      cin >> x;
      pos[x] = i;
    }
    int l = (n + 1) / 2, r = (n + 2) / 2;
    while (l > 0 && (l == r || (pos[l] < pos[l + 1] && pos[r - 1] < pos[r]))) {
      --l;
      ++r;
    }
    cout << (n - r + l + 1) / 2 << '\n';
  }
}

1792D - Перестановки с неподвижным префиксом

Идея: BledDest

Разбор

Решение (awoo)

#include <bits/stdc++.h>
 
using namespace std;
 
#define forn(i, n) for(int i = 0; i < int(n); i++) 

int get(const vector<int> &a, const vector<int> &b){
	int res = 0;
	while (res < int(a.size()) && a[res] == b[res])
		++res;
	return res;
}

int main(){
	int t;
	scanf("%d", &t);
	while (t--){
		int n, m;
		scanf("%d%d", &n, &m);
		vector<vector<int>> a(n, vector<int>(m));
		forn(i, n) forn(j, m){
			scanf("%d", &a[i][j]);
			--a[i][j];
		}
		vector<vector<int>> b(n, vector<int>(m));
		forn(i, n) forn(j, m) b[i][a[i][j]] = j;
		sort(b.begin(), b.end());
		forn(i, n){
			int j = lower_bound(b.begin(), b.end(), a[i]) - b.begin();
			int ans = 0;
			if (j > 0) ans = max(ans, get(a[i], b[j - 1]));
			if (j < n) ans = max(ans, get(a[i], b[j]));
			printf("%d ", ans);
		}
		puts("");
	}
}

1792E - Таблица и делители

Идея: adedalic

Разбор

Решение (adedalic)

#include<bits/stdc++.h>

using namespace std;

#define fore(i, l, r) for(int i = int(l); i < int(r); i++)
#define sz(a) int((a).size())

#define x first
#define y second

typedef long long li;
typedef long double ld;
typedef pair<int, int> pt;

template<class A, class B> ostream& operator <<(ostream& out, const pair<A, B> &p) {
	return out << "(" << p.x << ", " << p.y << ")";
}
template<class A> ostream& operator <<(ostream& out, const vector<A> &v) {
	fore(i, 0, sz(v)) {
		if(i) out << " ";
		out << v[i];
	}
	return out;
}

const int INF = int(1e9);
const li INF64 = li(1e18);
const ld EPS = 1e-9;

int n;
li m1, m2;

inline bool read() {
	if(!(cin >> n >> m1 >> m2))
		return false;
	return true;
}

vector<pt> mFact;
vector<li> divs;

void factM(li m1, li m2) {
	mFact.clear();

	for (int d = 2; d * d <= m1 || d * d <= m2; d++) {
		int cnt = 0;
		while (m1 % d == 0) {
			m1 /= d;
			cnt++;
		}
		while (m2 % d == 0) {
			m2 /= d;
			cnt++;
		}
		if (cnt > 0)
			mFact.push_back({d, cnt});
	}
	if (m1 > m2)
		swap(m1, m2);
	if (m1 > 1)
		mFact.push_back({m1, 1});
	if (m2 > 1) {
		if (m2 == m1)
			mFact.back().y++;
		else
			mFact.push_back({m2, 1});
	}
}

void genDivisors(int pos, li val) {
	if (pos >= sz(mFact)) {
		divs.push_back(val);
		return;
	}
	li cur = val;
	fore (pw, 0, mFact[pos].y + 1) {
		genDivisors(pos + 1, cur);
		if (pw < mFact[pos].y)
			cur *= mFact[pos].x;
	}
}

inline void solve() {
	factM(m1, m2);

	divs.clear();
	genDivisors(0, 1);
	sort(divs.begin(), divs.end());

	vector<int> ans(sz(divs), 0);

	vector<li> dp(sz(divs), -1);
	fore (id, 0, sz(divs)) {
		if (divs[id] <= n)
			dp[id] = divs[id];
		for (auto [p, pw] : mFact) {
			if (divs[id] % p != 0)
				continue;
			
			int pos = int(lower_bound(divs.begin(), divs.end(), divs[id] / p) - divs.begin());
			dp[id] = max(dp[id], dp[pos]);
		}

		if (divs[id] / dp[id] <= n)
			ans[id] = divs[id] / dp[id];
	}

	int cnt = 0;
	int xorSum = 0;
	fore (i, 0, sz(ans)) {
		cnt += ans[i] > 0;
		xorSum ^= ans[i];
	}
	
//	cout << sz(ans) << endl;
//	cout << ans << endl;
	cout << cnt << " " << xorSum << endl;
}

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
	int tt = clock();
#endif
	ios_base::sync_with_stdio(false);
	cin.tie(0), cout.tie(0);
	cout << fixed << setprecision(15);

	int t; cin >> t;
	
	while (t--) {
		
		read();
		solve();
		
#ifdef _DEBUG
		cerr << "TIME = " << clock() - tt << endl;
		tt = clock();
#endif
	}
	return 0;
}

1792F1 - Покраска графа (простая версия)

Идея: BledDest

Разбор

Решение (BledDest)

#include<bits/stdc++.h>

using namespace std;

const int MOD = 998244353;

int add(int x, int y)
{
    x += y;
    while(x >= MOD) x -= MOD;
    while(x < 0) x += MOD;
    return x;
}

int mul(int x, int y)
{
    return (x * 1ll * y) % MOD;
}

int varMul(int x)
{
    return x;
}

template<typename... Args>
int varMul(int x, Args... args)
{
    return mul(x, varMul(args...));
}

int binpow(int x, int y)
{
    int z = 1;
    while(y)
    {
        if(y & 1) z = mul(z, x);
        x = mul(x, x);
        y /= 2;
    }
    return z;
}

vector<int> fact, rfact;
vector<int> dp;
int n;

void precalc()
{
    fact.resize(n + 1);
    rfact.resize(n + 1);
    fact[0] = 1;
    for(int i = 1; i <= n; i++)
        fact[i] = mul(i, fact[i - 1]);
    for(int i = 0; i <= n; i++)
        rfact[i] = binpow(fact[i], MOD - 2);
    dp.resize(n + 1, -1);
}

int C(int n, int k)
{
    if(n < 0 || n < k || k < 0) return 0;
    return varMul(fact[n], rfact[k], rfact[n - k]);
}

int calc(int x)
{
    if(dp[x] != -1) return dp[x];
    if(x == 1) return dp[x] = 1;
    if(x == 2) return dp[x] = 1;
    dp[x] = 0;
    int& d = dp[x];
    for(int i = 1; i < x; i++)
    {
        d = add(d, varMul(calc(i), (i == x - 1 ? (MOD + 1) / 2 : calc(x - i)), 2, C(x - 1, i - 1)));
    }   
    return d;
}

int main()
{
    cin >> n;
    precalc();
    cout << add(mul(calc(n), 2), -2) << endl;
}

1792F2 - Покраска графа (сложная версия)

Идея: BledDest

Разбор

Решение (BledDest)

#include<bits/stdc++.h>

using namespace std;

const int LOGN = 18;
const int N = (1 << LOGN);
const int MOD = 998244353;
const int g = 3;

#define forn(i, n) for(int i = 0; i < int(n); i++)

inline int mul(int a, int b)
{
    return (a * 1ll * b) % MOD;
}

inline int norm(int a) 
{
    while(a >= MOD)
        a -= MOD;
    while(a < 0)
        a += MOD;
    return a;
}

inline int binPow(int a, int k) 
{
    int ans = 1;
    while(k > 0) 
    {
        if(k & 1)
            ans = mul(ans, a);
        a = mul(a, a);
        k >>= 1;
    }
    return ans;
}

inline int inv(int a) 
{
    return binPow(a, MOD - 2);
}

vector<int> w[LOGN];
vector<int> iw[LOGN];
vector<int> rv[LOGN];

void precalc() 
{
    int wb = binPow(g, (MOD - 1) / (1 << LOGN));
    
    for(int st = 0; st < LOGN; st++) 
    {
        w[st].assign(1 << st, 1);
        iw[st].assign(1 << st, 1);
        
        int bw = binPow(wb, 1 << (LOGN - st - 1));
        int ibw = inv(bw);
        
        int cw = 1;
        int icw = 1;
        
        for(int k = 0; k < (1 << st); k++) 
        {
            w[st][k] = cw;
            iw[st][k] = icw;
            
            cw = mul(cw, bw);
            icw = mul(icw, ibw);
        }
        
        rv[st].assign(1 << st, 0);
        
        if(st == 0) 
        {
            rv[st][0] = 0;
            continue;
        }
        int h = (1 << (st - 1));
        for(int k = 0; k < (1 << st); k++)
            rv[st][k] = (rv[st - 1][k & (h - 1)] << 1) | (k >= h);
    }
}

inline void fft(int a[N], int n, int ln, bool inverse) 
{   
    for(int i = 0; i < n; i++) 
    {
        int ni = rv[ln][i];
        if(i < ni)
            swap(a[i], a[ni]);
    }
    
    for(int st = 0; (1 << st) < n; st++) 
    {
        int len = (1 << st);
        for(int k = 0; k < n; k += (len << 1)) 
        {
            for(int pos = k; pos < k + len; pos++) 
            {
                int l = a[pos];
                int r = mul(a[pos + len], (inverse ? iw[st][pos - k] : w[st][pos - k]));
                
                a[pos] = norm(l + r);
                a[pos + len] = norm(l - r);
            }
        }
    }
    
    if(inverse) 
    {
        int in = inv(n);
        for(int i = 0; i < n; i++)
            a[i] = mul(a[i], in);
    }
}

int aa[N], bb[N], cc[N];

vector<int> multiply(vector<int> a, vector<int> b) 
{
    int sza = a.size();
    int szb = b.size();
    int n = 1, ln = 0;
    while(n < (sza + szb))
        n <<= 1, ln++;
    for(int i = 0; i < n; i++)
        aa[i] = (i < sza ? a[i] : 0);
    for(int i = 0; i < n; i++)
        bb[i] = (i < szb ? b[i] : 0);
        
    fft(aa, n, ln, false);
    fft(bb, n, ln, false);
    
    for(int i = 0; i < n; i++)
        cc[i] = mul(aa[i], bb[i]);
        
    fft(cc, n, ln, true);
    
    int szc = n;
    vector<int> c(szc);
    szc = n;
    for(int i = 0; i < n; i++)
        c[i] = cc[i];
    return c;
}                    

int main()
{
    int n;
    cin >> n;
    vector<int> fact(n + 1);
    fact[0] = 1;
    for(int i = 0; i < n; i++)
        fact[i + 1] = mul(fact[i], i + 1);
    precalc();
    vector<int> A = {0, 1, 2};
    vector<int> B = {0, 1, 1};
    vector<int> C = {0, 1, 1};
    vector<int> D = {0, 1, 1};
    vector<int> conv;
    const int K = 2000;
    int last_conv = -1e9;
    while(A.size() <= n)
    {
        int cur = A.size();
        if(cur - last_conv >= K)
        {
            last_conv = cur - 1;
            conv = multiply(C, D);
        }
        /*for(auto x : conv) cerr << x << " ";
        cerr << endl;*/
        int val_A;
        if(last_conv * 2 >= cur)
        {
            val_A = conv[cur];
            // [cur - last_conv, last_conv] are already used
            for(int i = 1; i < (cur - last_conv); i++)
            {
                val_A = norm(val_A + mul(C[i], D[cur - i]));
            }
            for(int i = last_conv + 1; i < cur; i++)
            {
                val_A = norm(val_A + mul(C[i], D[cur - i]));
            }
        }
        else
        {
            val_A = 0;
            for(int i = 1; i <= cur - 1; i++)
            {
                val_A = norm(val_A + mul(C[i], D[cur - i]));
            }
        }
        val_A = mul(val_A, fact[cur - 1]);
        val_A = mul(val_A, 2);
        A.push_back(val_A);
        B.push_back(mul(val_A, inv(2)));
        C.push_back(mul(val_A, inv(fact[cur])));
        D.push_back(mul(B.back(), inv(fact[cur - 1])));
    }
    cout << norm(A[n] - 2) << endl;
}

Комментарии (56)

Показать архивные | Написать комментарий?

hrabesim

22 месяца назад, # |

Isn't it possible to have an O(n) solution for problem C (Min Max Sort)?

→ Ответить

steveonalex

22 месяца назад, # ^ |

yeah, there's an easy $$$O(n)$$$ solution for problem C, involving using the array $$$pos[i]$$$, to get the position of index $$$i$$$ in the permutation, then start at the middle value and finding the $$$LIS$$$ to both side. My submission: 190340250

primeprogrammer2021

+13

maybe you read only the last line.

read the third last line.

Indeed I did. Thanks :)

At_Atharva

can somebody explain how to come up with the equation in problem B.

simply_jo

← Rev. 3 →

+10

Here, the Optimal strategy to solve the problem is by first saying all the jokes that both Alice and bob likes. (type 1 jokes) then by saying jokes alternately that (Alice likes,bob doesn't) and vice versa.

Equation: a1 + min(a2,a3) x 2

Final strategy is to say remaining jokes : first by saying remaining type 2, type 3 jokes and then by saying the jokes that both doesn't like (Type 4). this should be compared with the points that he already acquired hence a1 is taken. We take a1 + 1 as the judges goes when points become negative. Hence we add 1 to make it -1. Finally, We take the minimum of both these values to form the final equation.

Equation: a1 + min(a2,a3).2 + min(a1 + 1,abs(a2 - a3) + a4)

pantw

Does anyone have an idea on proving the efficiency of bisearch-on-divisors approach for prob. E? Never expected it will be this fast = =

Can somebody explain what is the meaning of run binary search on k in problem C tutorial?

go_emon

← Rev. 2 →

Remove value out of [k, n + k — 1] from array. If the remaining numbers are increasing （from [k, n + k — 1]), then the number k is appropriate.
You can use binary search to find k from [1, n].
You can see this submission, for more details:
https://mirror.codeforces.com/contest/1792/submission/190334713

irkstepanov

+62

For whatever it's worth, here's my $$$O(n \log n)$$$ solution for problem F.

First, we notice that the property for the set $$$S$$$ is the same as having a one-colored cut in $$$S$$$ (in other words, the set of vertices can be split into $$$S_1$$$ and $$$S_2$$$ so that all edges between $$$S_1$$$ and $$$S_2$$$ have the same color). The only issue is that a given set $$$S$$$ can possibly be split into $$$S_1$$$ and $$$S_2$$$ in several ways. So I introduce $$$f(n)$$$ to be equal to the number of graphs on $$$n$$$ vertices where the global cut is blue. Then the answer is $$$2 f(n) - 2$$$, since the global cut may be red, but $$$2$$$ extra cases arise when all edges have the same color.

To find $$$f(n)$$$ we need to partition the set of $$$n$$$ vertices into $$$k \geq 2$$$ subsets so that vertices from different subsets are connected by blue edges, but the subsets themselves obey the rules recursively (they have global cuts of red edges). Since $$$f_{\text{red}} = f_{\text{blue}}$$$, we have

$$$ f(n) = \sum\limits_{k \geq 2}~ \sum\limits_{a_1 + \ldots + a_k = n} C(a_1, \ldots, a_k) \cdot f(a_1) \cdot \ldots \cdot f(a_k),$$$

where $$$C(a_1, \ldots, a_k)$$$ roughly means the number of ways to choose subsets of sizes $$$a_1, \ldots, a_k$$$ from a set of size $$$n = a_1 + \ldots + a_k$$$. If $$$b_1$$$ occurs $$$c_1$$$ times, $$$\ldots$$$, $$$b_m$$$ occurs $$$c_m$$$ times within the multiset $$${ a_1, \ldots, a_k}$$$, then

$$$C(a_1, \ldots, a_k) = \frac{n!}{a_1! \cdot \ldots \cdot a_k! \cdot c_1! \cdot \ldots \cdot c_m!}.$$$

Define $h(n) = \frac{f(n)}{n!}.$ The base values are $$$h(0) = 0, h(1) = 1$$$. Via $$$H(x)$$$ we denote the generating function of $$$h$$$: $$$H(x) = h(0) + h(1) \cdot x + h(2) \cdot x^2 + \ldots$$$. After a careful examination (I don't know how to prove rigorously it though) we obtain

$$$ e^{H(x)} = 2H(x) + 1 - x.$$$

Everything else is the standard approach of how to solve these recurrences: if we know $$$H(x) \bmod{x^{m}}$$$, then from the equality above we can obtain $$$H(x) \bmod{x^{2m}}$$$. Underneath we need FFT (NTT) and an exponential generating function. Each step from $$$m$$$ to $$$2m$$$ takes $$$O(m \log m)$$$ time, so the overall complexity if $$$O(n \log n)$$$.

ivan100sic

I also found this solution, thanks for sharing it!

David713

I don't get why we have to make sure that no set of vertices is connected by both colors in F1. Doesn't the lemma proved it impossible to connect a set of vertices with both colors?

BledDest

The lemma proves the other thing: it is impossible to have a set of vertices which is neither blue-connected nor red-connected.

Thanks. I made an oversight while reading the lemma.

OMG_wc

A-F1 video editorial for Chinese:

BiliBili

delusion_loss_vanity

+18

In fact, we can calculate the convolution-like sequences (such as those in problem F) in $$$O(n \log^2 n)$$$ or $$$O(n \log^2 n / \log \log n)$$$ or even faster. One can find the approach from Elegia's report. The implementations usually has lower constant factor in time than those of the Newton iteration (if exist).

Here is my implementation, which imitates Elegia's implementations for other problems.

(but it's just as fast as a brute force lol)

mmaxio

Do you have this report as a PDF?

Elegia

+24

Here (sorry for Chinese only)

Do you know how to find ref. [5] (罗煜翔。(2020)。浅谈 Nimber 和多项式算法。IOI2020 中国国家集训队论⽂集。) ? I wanna see how to do semi-relaxed multiplication in $$$ \frac{n\log ^2 n}{\log \log n} $$$(if that's what it says). And what nimbers have to do with it xd.

Let's discuss about it here, actually in that report, most contents are just putting the general algorithm into the nimber framework. Nimber does not play an important role in the relaxed convolution.

__hermit__

"The array may large" in E made me realise how I don't trust my gut feeling at all.

HHY_zZhu

Where does the log(divs(m)) of the complexity in question E come from Isn't it only need O(divs(m)⋅z(m)) to calculate the dp array?

oh i get it,because you can't find the element within O(1)

then why not use hash

fishy15

18 месяцев назад, # ^ |

constant factor is worse, I converted hash -> binary search + vector and it ran within time limit

economist

can somebody explain why in D if we use lower_bound ans EITHER result or the previous one? why not just result? thanks in advance

awoo

It depends on the value immediately after the longest common prefix. While searching for $$$p$$$, you find some inverse $$$q_1, q_2, \dots, q_m$$$ that starts with $$$p_1, p_2, \dots, p_k$$$ for some $$$k$$$. The next value is different. If $$$q_{k+1} > p_{k+1}$$$, then lower_bound will point at $$$q$$$ (or one of inverses with such prefix if there are multiple). However, if $$$q_{k+1} < p_{k+1}$$$, then $$$q$$$ will be smaller than $$$p$$$, and lower_bound will jump over it. The next inverse has to be greater than $$$p$$$, so you only have to look one step behind.

Yaria

For D, I thought while $$$n=5\times 10^4$$$, the number of subsets of $$$\{1,...,m\}$$$ is $$$10^3$$$.

So I constructed a map $$$M$$$ mapping each subset $$$L\subset \{1,...,m\}$$$ to $$$\{(a_i[p_1], ..., a_i[p_t])\ |\ i\in \{1,...,n\},\ \{p_1<...<p_t\}=L\}$$$. Then I iterated over every permutation and every $$$k$$$ to see whether the corresponding $$$k$$$ locations have the desired values.

For constructing $$$M$$$, it seemed that I can iterate over all subsets of $$$\{1,...,m\}$$$ and for each subset iterate over all permutations. This 2-layered loop is about $$$5\times 10^7$$$. But my actual implementation requires $$$O(2^m\times n\times m\times \log(n))\approx 8\times 10^9$$$ and thus got TLE. Not sure whether this is optimizable.

jeal0uspengu1n

I thought of the same thing during the contest but midway realized that its gonna give TLE :(

Edit: This video has a similar approach

SakurakoujiRuna

You can implement it with Trie in $$$O(2^m\times n \times m)$$$. I'm not sure whether it is fast enough.

nayeem2021

21 месяц назад, # ^ |

It is not necessary to store all 2^m subsets. Let a permutation be 3 2 4 1. For this we need to store 0 0 0 1, 0 2 0 1, 3 2 0 1, 3 2 4 1. So the complexity is O(n*m)

SNX03

20 месяцев назад, # ^ |

Why don't we activate the numbers randomly, something like '0 2 0 0'?

luanmenglei

I have some questions in problem F1

Why iterate k — the number of vertices whick are in the same 'red' component as 1 but not iterate the blue component, is this because you are counting the blue component?
What about the edges between vertices in the same componet as 1 and the rest vertices, are they must be the same color? how to proof?

Hope someone can help me, thx a lot

We are considering the case when the whole graph is blue-connected, so there's no need to iterate on the blue component. The case when the graph is red-connected is symmetric to it.
It's easy to see that all these edges are blue, since any red edge between any vertex from the "red" component of vertex 1 and any vertex outside this component means that we haven't picked the whole component

I understood!

Thanks a lot

elEctroWolf

Can someone give a test case on which my code is failing? I couldn't view the 156th item of 2nd test case. Is there any way to view it?

Click the "Click to see detail" button in the bottom of the page to see the detailed test cases

tushar_kumar

has anyone solved problem C with binary search ?

yadavDipesh

Code

~~~~~

int l = -1;
    int r = n + 1;
    while(r - l > 1){
        int m = (l+r) >>> 1;

        int min = m;
        int max = n - m + 1;

        List<Integer> list = new ArrayList<>();
        for(int i = 0 ; i < n ; i++){
           if(arr[i] > min && arr[i] < max){
             list.add(arr[i]);
           }
        }

        boolean cc = true;
        for(int i = 0 ; i < list.size() - 1 ; i++){
           int a = list.get(i);
           int b = list.get(i+1);
           if(a > b) cc = false;
        }

        if(cc) r = m;
        else l = m;
    }

    out.println(r);

Socrates1232

NEED PROVEMENT FOR PROBLEM F

Nevermind

F1's solution is flawed I guess. Consider case with n=3, when A1=B1=1, A2=2, B2=1, you get B3=4 which is incorrect.

So what's wrong with the $$$B_3=4$$$?

A3 which is the answer for n=3 is 6 in the pretest case, but surely 6 is not 4 * 2 :P

Well, maybe the tutorial forget something or i just didn't see.

But don't forget the limitation 1 and 2.

The $$$A_3=8$$$ contains the case that all blue and all red.

I noticed that too. I'm just saying that define "An" as "the answer for n" is not very strict and could lead to someone's misunderstanding(like me)

另外看兄弟id八成也是国人，直接说汉语就好了吧:)

哈哈

Sorry, I initially stated the problem without the constraints "at least one edge should be red" and "at least one edge should be blue". I wrote the editorial for that version, but then decided to introduce these constraints. So, the actual answer is $$$A_n-2$$$ since we need to discard the case "all edges are red" and the case "all edges are blue".

cockatooo

19 месяцев назад, # |

Time for problem E should be more strict. The following brute force solution passes:

Generate an array of increasing divisors. Make a copy and maintain it in std::set<long long>. For each divisor $$$d_i$$$ in the original array, find the number of divisors that has $$$d_i$$$ as its first row, i.e. start from divs_set.lower_bound(d_i * d_i) and iterate until the value is greater than $$$d_i * n$$$, counting the number of divisors that are a multiple of $$$d_i$$$, and remove them from the set afterwards.

the_Incharge

18 месяцев назад, # |

Can someone give some intuition why way 2 in E works

adityagamer

13 месяцев назад, # |

In editorial of problem E,it is mentioned that " There is also a way to get rid of extra log(divs(m)) factor if you iterate through dp in a smart way ".

How to do so? Could not find anything in comments and figure out neither.

sharath__chandra

9 месяцев назад, # |

is it possible to solve problem 'D' using trie?

Byte-Ghifo

8 месяцев назад, # ^ |

Yes, it is

I just solved it using Trie: 251308798

you insert in the $$$trie$$$ the inverse of arrays $$$a_i$$$

and then for every array you look for the longest common prefix.

thanks

Ylandolsi

4 месяца назад, # |

D can be solved with string hashing...

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