Thanks for participating in the round, we hope you liked the problems!
| Handle | A | B | C | D | E1 | E2 | F |
|---|---|---|---|---|---|---|---|
| Vladithur | 16K | 8K | 3K | 900 | 500 | 50 | 4 |
| GlowCheese | 14K | 9K | 5K | 500 | ? | ? | ? |
| thanhchauns2 | 14K | 8K | 2K | 1K | 500 | ? | ? |
| QuangBuiCPP | 14K | 7K | 2K | 1K | 500 | 24 | 5 |
| welleyth | 16K | 10K | 4K | 1.5K | 600 | 130 | 2 |
| Kon567889 | 14K | 10K | 5K | 2K | ? | ? | 5 |
The smallest possible sum is $$$1 + 2 + \ldots + k$$$.
#include <bits/stdc++.h>
#define all(x) (x).begin(), (x).end()
#define allr(x) (x).rbegin(), (x).rend()
#define gsize(x) (int)((x).size())
const char nl = '\n';
typedef long long ll;
typedef long double ld;
using namespace std;
int main() {
ios::sync_with_stdio(0); cin.tie(0);
int T;
cin >> T;
while (T--) {
int n, k;
cin >> n >> k;
vector<int> a(n);
for (int i = 0; i < n; i++) cin >> a[i];
vector<int> b = a;
sort(all(b));
int x = b[k - 1];
int ans = 0;
for (int i = 0; i < k; i++) {
if (a[i] > x) ans++;
}
cout << ans << nl;
}
}
Bonus: solve for every $$$k$$$ from $$$1$$$ to $$$n$$$ for $$$n \le 10^5$$$.
In an optimal answer, for all $$$1 \le i \le n$$$, $$$\operatorname{lcm}(i, p_i) = i \cdot p_i$$$.
$$$\operatorname{lcm}(x, x + 1) = x \cdot (x + 1)$$$.
$$$\operatorname{lcm}(x, x + 1) + \operatorname{lcm}(x + 1, x) = x^2 + (x + 1)^2 - 1$$$.
#include <bits/stdc++.h>
#define all(x) (x).begin(), (x).end()
#define allr(x) (x).rbegin(), (x).rend()
#define gsize(x) (int)((x).size())
const char nl = '\n';
typedef long long ll;
typedef long double ld;
using namespace std;
int main() {
ios::sync_with_stdio(0); cin.tie(0);
int T;
cin >> T;
while (T--) {
int n;
cin >> n;
if (n % 2 == 0) {
for (int i = 1; i <= n; i += 2) {
cout << i + 1 << ' ' << i << ' ';
}
cout << nl;
} else {
cout << 1 << ' ';
for (int i = 2; i <= n; i += 2) {
cout << i + 1 << ' ' << i << ' ';
}
cout << nl;
}
}
}
Bonus: try to prove the solution without the editorial!
The operation only decreases numbers.
An array is sorted if there is no index $$$i$$$ such that $$$a_i > a_{i + 1}$$$.
Suppose there is an index $$$i$$$ such that $$$a_i > a_{i + 1}$$$. What operations do you need to perform to make $$$a_i \le a_{i + 1}$$$?
#include <bits/stdc++.h>
#define all(x) (x).begin(), (x).end()
#define allr(x) (x).rbegin(), (x).rend()
#define gsize(x) (int)((x).size())
const char nl = '\n';
typedef long long ll;
typedef long double ld;
using namespace std;
int main() {
ios::sync_with_stdio(0); cin.tie(0);
int T;
cin >> T;
while (T--) {
int n;
cin >> n;
vector<int> a(n);
map<int, vector<int>> p;
for (int i = 0; i < n; i++) {
cin >> a[i];
p[a[i]].push_back(i);
}
int ans = 0;
set<int> ts;
for (int i = 0; i < n - 1; i++) {
if (a[i] > a[i + 1]) ts.insert(i);
}
while (!ts.empty()) {
int i = *ts.begin();
int x;
if (a[i] > 0) {
x = a[i];
} else {
x = a[i + 1];
}
for (int j: p[x]) {
a[j] = 0;
ts.erase(j - 1);
ts.erase(j);
if (j > 0 && a[j - 1] > a[j]) ts.insert(j - 1);
if (j + 1 < n && a[j] > a[j + 1]) ts.insert(j);
}
ans++;
}
cout << ans << nl;
}
}
Bonus: solve for when $$$a_i$$$ can also be negative.
$$$\operatorname{d}(u; v) \le 2 \cdot \min(a_1 \ldots a_n)$$$
diameter $$$\le 2 \cdot \min(a_1 \ldots a_n)$$$
diameter $$$\le \max\limits_{1 \le i \le n - 1}{\min(a_i,a_{i+1})}$$$
Binary search on the answer or do a clever greedy.
#include <bits/stdc++.h>
#define all(x) (x).begin(), (x).end()
#define allr(x) (x).rbegin(), (x).rend()
#define gsize(x) (int)((x).size())
const char nl = '\n';
typedef long long ll;
typedef long double ld;
using namespace std;
const int inf = 1000000000;
bool f(vector<int> a, int k, int ans) {
int n = gsize(a);
int dk = 0;
for (int i = 0; i < n; i++) {
if (a[i] * 2 < ans) {
a[i] = inf;
k--;
dk++;
}
}
if (k < 0) return false;
if ((k > 0 && dk > 0) || k >= 2) return true;
if (k == 1 && dk == 0) {
return *max_element(all(a)) >= ans;
}
for (int i = 0; i < n - 1; i++) {
if (min(a[i], a[i + 1]) >= ans) return true;
}
return false;
}
void solve() {
int n, k;
cin >> n >> k;
vector<int> a(n);
for (int i = 0; i < n; i++) cin >> a[i];
int lb = 0, ub = inf, tans = 0;
while (lb <= ub) {
int tm = (lb + ub) / 2;
if (f(a, k, tm)) {
tans = tm;
lb = tm + 1;
} else {
ub = tm - 1;
}
}
cout << tans << nl;
}
int main() {
ios::sync_with_stdio(0); cin.tie(0);
int T;
cin >> T;
while (T--) solve();
}
Bonus: solve for every $$$k$$$ from $$$1$$$ to $$$n$$$.
1712E2 - LCM Sum (hard version)
Count the number of "bad" triplets such that $$$\operatorname{lcm}(i, j, k) < i + j + k$$$.
$$$\operatorname{lcm}(i, j, k) \le 2 \cdot k$$$ in a bad triplet.
A triplet is bad iff $$$\operatorname{lcm}(i, j, k) = k$$$ or ($$$\operatorname{lcm}(i, j, k) = 2 \cdot k$$$ and $$$i + j > k$$$).
$$$\operatorname{lcm}(i, j, k) = x$$$ implies $$$i$$$, $$$j$$$, and $$$k$$$ are all divisors of $$$x$$$.
For every $$$k$$$, iterate over all pairs of divisors of $$$2 \cdot k$$$.
Think of the "bad" triplets as of segments $$$[i; k]$$$ with some weight.
Solve in offline.
#include <bits/stdc++.h>
#define all(x) (x).begin(), (x).end()
#define allr(x) (x).rbegin(), (x).rend()
#define gsize(x) (int)((x).size())
const char nl = '\n';
typedef long long ll;
typedef long double ld;
using namespace std;
const int maxn = 200000;
int t[(maxn + 1) * 4];
void update(int v, int tl, int tr, int i, int x) {
t[v] += x;
if (tl != tr - 1) {
int tm = (tl + tr) / 2;
if (i < tm) update(2*v + 1, tl, tm, i, x);
else update(2*v + 2, tm, tr, i, x);
}
}
int query(int v, int tl, int tr, int l, int r) {
if (l >= r) return 0;
if (tl == l && tr == r) return t[v];
else {
int tm = (tl + tr) / 2;
int r1 = query(2*v + 1, tl, tm, l, min(r, tm));
int r2 = query(2*v + 2, tm, tr, max(l, tm), r);
return r1 + r2;
}
}
int main() {
ios::sync_with_stdio(0); cin.tie(0);
vector<vector<int>> d(2*maxn + 1);
for (int i = 1; i <= maxn; i++) {
for (int j = i; j <= 2*maxn; j += i) {
d[j].push_back(i);
}
}
vector<vector<pair<int, int>>> segs(maxn + 1);
for (int k = 1; k <= maxn; k++) {
for (int ti = 0; ti < gsize(d[2*k]); ti++) {
int i = d[2*k][ti], ct = 0;
for (int tj = ti + 1; tj < gsize(d[2*k]); tj++) {
int j = d[2*k][tj];
if (j >= k) break;
if (i + j > k || (k % i == 0 && k % j == 0)) ct++;
}
if (ct > 0) {
segs[i].emplace_back(k, ct);
}
}
}
int T;
cin >> T;
vector<ll> ans(T);
vector<array<int, 3>> queries;
for (int i = 0; i < T; i++) {
ll l, r;
cin >> l >> r;
queries.push_back({l, r, i});
ans[i] = (r - l + 1) * (r - l) * (r - l - 1) / 6;
}
sort(all(queries));
int ti = T - 1;
for (int i = maxn; i >= 1; i--) {
// Update ST
for (auto [r, ct]: segs[i]) {
update(0, 0, maxn + 1, r, ct);
}
// Get Answers
while (ti >= 0 && queries[ti][0] == i) {
ans[queries[ti][2]] -= query(0, 0, maxn + 1, i, queries[ti][1] + 1);
ti--;
}
}
for (int i = 0; i < T; i++) cout << ans[i] << nl;
}
Bonus: solve the problem in $$$\mathcal{O}((n + t) \log n)$$$ or better.
Let $$$f_v$$$ be the distance to the closest vertex $$$\in L$$$ to $$$v$$$. You can calculate it for every vertex in $$$O(n)$$$.
The distance in the new graph $$$\operatorname{d'}(u, v)$$$ is equal to $$$\min(\operatorname{d}(u, v), f_u + f_v + x)$$$
Suppose there is a vertex $$$v$$$ with $$$f_v > 0$$$. Then there is always a vertex $$$u$$$ in the subtree of $$$v$$$ such that $$$f_u < f_v$$$ and $$$depth_u$$$ > $$$depth_v$$$.
Think small-to-large merging.
Maintain the largest value of $$$depth_v$$$ over all $$$v$$$ with $$$f_v = i$$$ for all needed $$$i$$$.
In this problem, small-to-large merging works in $$$O(n)$$$.
The diameter is $$$\le n$$$. So you can increase it by $$$1$$$ at most $$$n$$$ times.
#include <bits/stdc++.h>
#define all(x) (x).begin(), (x).end()
#define allr(x) (x).rbegin(), (x).rend()
#define gsize(x) (int)((x).size())
const char nl = '\n';
typedef long long ll;
typedef long double ld;
using namespace std;
const int maxn = 1000000;
int f[maxn], d[maxn];
vector<int> g[maxn], g2[maxn], val[maxn];
int n, q;
vector<int> ans, x;
void solve() {
for (int v = n - 1; v >= 0; v--) {
for (int u: g[v]) {
if (gsize(val[u]) > gsize(val[v])) swap(val[u], val[v]);
for (int fu = 0; fu < gsize(val[u]); fu++) {
for (int qi = 0; qi < q; qi++) {
while (true) {
int fv = max(0, ans[qi] - fu - x[qi]);
if (fv < gsize(val[v]) &&
val[v][fv] + val[u][fu] - 2*d[v] >= ans[qi]) ans[qi]++;
else break;
}
}
}
for (int fu = 0; fu < gsize(val[u]); fu++) {
val[v][fu] = max(val[v][fu], val[u][fu]);
}
val[u].clear();
val[u].shrink_to_fit();
}
{
int fu = f[v];
for (int qi = 0; qi < q; qi++) {
while (true) {
int fv = max(0, ans[qi] - fu - x[qi]);
if (fv < gsize(val[v]) &&
val[v][fv] - d[v] >= ans[qi]) ans[qi]++;
else break;
}
}
if (fu == gsize(val[v])) {
val[v].push_back(d[v]);
}
}
}
for (int i = 0; i < q; i++) {
ans[i]--;
}
}
int main() {
ios::sync_with_stdio(0); cin.tie(0);
cin >> n;
vector<int> deg(n);
for (int i = 1; i < n; i++) {
int p;
cin >> p;
d[i] = d[p - 1] + 1;
g[p - 1].push_back(i);
g2[i].push_back(p - 1);
}
{
queue<int> qe;
vector<bool> used(n, false);
for (int i = 0; i < n; i++) {
if (gsize(g2[i]) + gsize(g[i]) == 1) {
qe.push(i);
used[i] = true;
}
}
while (!qe.empty()) {
int v = qe.front();
qe.pop();
for (int u: g[v]) {
if (used[u]) continue;
f[u] = f[v] + 1;
qe.push(u);
used[u] = true;
}
for (int u: g2[v]) {
if (used[u]) continue;
f[u] = f[v] + 1;
qe.push(u);
used[u] = true;
}
}
}
cin >> q;
for (int i = 0; i < q; i++) {
int tx;
cin >> tx;
x.push_back(tx);
ans.push_back(0);
}
solve();
for (int tans: ans) cout << tans << ' ';
cout << nl;
}
Bonus: solve for $$$n, q \le 10^6$$$.
Don't forget to rate the problems!
PS: Solution codes probably will be added later.
UPD: explanations of the references:
Hope you liked them!
The quote is just a modification of the original title of KonoSuba.
This is just something I invented, no reference here. Hope you liked the short poem!
The quote is the meme "people die when they are killed" in the Fate series, changed to fit the programming context. And the title is just "Fate Zero" changed to "Sort Zero".
"Do you have a wish?" is a recurring phrase in the LN series HakoMari, referring to the wish granting "boxes". The title is just "Empty Box" changed to "Empty Graph". Also, "O_o" is a reference to one of the main characters, "O".
The quote can probably be traced to many origins, but I personally modified the quote "We are legion, for we are many" from the series 86. There were also a lot of 6's and 8's in the sample, which might have helped you guess.
I didn't really think about this one, just modified a quote from Vivy a bit.
UPD2: added solution codes (better late than never...)
Nice :)
;(
the gap between C and D is too large.
Maybe yes, but E1 was solvable, you could solve E1 instead of D
We expected C to be a bit harder and D to be a bit easier.
If you know Graph algorithms I think solving D would be easy however in general E1 is much easier
maybe. But for me, I solve D in 30 min but solve E1 in 1h30min(failed to find 6 and 15's apperance, not familiar with the method). And I think D doesn't really have much to do with Graph Algorithms.
Can you please explain the 6 and 15 case, cuz i dont understand it at all?
D isn't so hard, just D, like the ancient times, the C was easy let's gonna admit that
Problem D is one of the best problems I’ve ever seen. Overall, round is awesome. Hope to see more such great contests!
If the formal proof is quite hard, then why is it problem B ?
Do you expect people to solve without proof ?
That's a mistake on my part, I started wondering about the formal proof only on the day of the round, and it turned out to be unexpectedly difficult.
That's what people usually do...
Yes, but if you guess it wrong, you're gonna get huge amount of penalties and waste a lot of time.
Yea, but for most easy problems usually the gap between intuition and formal proof is not this large. Most of the time all you need to do is to formalize your intuition to turn it into a complete proof, which is definitely not the case for this problem.
Do you really think that those 13062 participants got Accepted on B all with proofs? :)
Of course not, it's bad that they solved it without proof. My point is, for easy questions, make the proofs easy too.
Or Make the problem with easy proofs. >_< (XD)
While I do think that it may not be appropriate for div2B when the proof is so difficult, I think it's quite intuitive that the optimal result arises when big numbers are paired with big numbers, leaving small numbers paired with small numbers (as opposed to pairing big numbers with small numbers). Even without that intuition, trying it on a few small examples should quickly demonstrate this trend. Add in the observation that adjacent numbers have a GCD of 1 (so none of the factors are "wasted" by the LCM) and we have a straightforward solution that's very easy to implement.
I didn't prove it during the contest myself, and that did cause me to hesitate before submission, but the intuition was strong enough that I decided to go ahead anyway.
But the intuition was strong enough that I got +2 :(
I think your intuition was correct, but you simply didn't test odd values. The 1 should have been at the beginning, not the end.
rearrangement inequality, and the fact: x and x+1 are coprimes, is a little bit hard to proof but the intuition helps. Sadly many Competitive programmers don't like maths
i just wrote brute force and saw the pattern for (1 to 9) ;) it was just swap(arr[i],arr[i-1]) from back....
Can E2 be solved using sqrt decomposition?
The editorial doesn’t show up in the bottom right of the problem pages
Now it should
I accepted f in the contest, my solution is $$$O(qn \log n)$$$. Maybe because of the large constant, it fst.
And I don't know why it get wa when I optimize it.
for vertices i,j the distance shall be $$$min(f[i]+f[j]+x,dis(i,j))$$$, f[i] means the distance between i and the nearest leaf. I sort f, and check whether the answer can be bigger than ans. Then for every node, it is a suffix of j that $$$f[i]+f[j]+x>=ans$$$ . I prework the suffix diameter and use O(1) lca to find the distance.
This was challenging and editorial is awesome. Thanks for these competitions
C can also be solved using binary search.
Find the minimum $$$k$$$ such that the array remains sorted if the first unique $$$k$$$ instances of the array are converted to $$$0$$$.
Solution: 168108392
Easiest Div2C ever?
Yes probably. Some newbies solved all A,B,C and still got negative delta.
Can someone please tell me why i get WA with this source for problem D (greedy solution)?
Take a look at Ticket 16031 from CF Stress for a counter example.
can you check for C ; https://codeforces.net/problemset/submission/1712/170962914
mircea_007 I did the same using priority queue but end up getting WA.
After the first two paragraphs of the editorial, another way is to proceed by induction. By Rearrangement Inequality, in any optional permutation, if $$$i<j$$$ and none of $$$p_i$$$ and $$$p_j$$$ are equal to $$$i$$$ or $$$j$$$, then we must have $$$p_i<p_j$$$. In particular, if $$$p_i=n$$$, then we must have $$$i=n-1$$$ or $$$i=n-2$$$. In addition, setting $$$j=n$$$ gives $$$p_n=n-1,n-2$$$. This implies that in an optimal permutation there are very limited possibilities for the last few elements of the permutation. We can use induction to prove that the best possibility is $$$p_n=n-1$$$ and $$$p_{n-1}=n$$$, so then we can construct an optimal permutation this way.
For subproof in problem D: When Imin is < u, can we take u to Imin and then Imin to v path instead of going till node 1? cc: Vladithur
Yes, going to node 1 just ensures that we always cross the min value.
I made video Solutions to the first 4 problems in case people are interested.
Great contest!
I don’t think we need that proof to solve B I just wrote the numbers down and i got my answers right away But btw nice editorial < 3
During the contest you may be right,but I don't think you can learn something from the problem after the contest.Actually,if it was D or E,probably you would hesitate to submit.
How do you arrive at the conclusion: a tuple $$$i \lt j \lt k$$$ is bad when $$$\mathrm{lcm}(i,j,k) = 2 \cdot k$$$ and $$$i+j \gt k$$$?
$$$lcm(i, j, k) = 2 \cdot k \implies 2 \cdot k < i + j + k \implies i + j > k$$$
I meant to ask how you get $$$\mathrm{lcm}(i,j,k) = 2 \cdot k$$$ in particular. Why $$$2 \cdot k$$$?
$$$lcm(i, j, k) < 3 \cdot k$$$, and $$$lcm$$$ must divide $$$i$$$, $$$j$$$, and $$$k$$$. This leaves us with only two possible values for it: $$$k$$$ and $$$2 \cdot k$$$.
Ohh okay. Thanks!
as a #008000 rated, I enjoyed solving C
Could someone point out my mistake in problem D? I don't understand what the tute is saying very well. https://codeforces.net/contest/1712/submission/168169178 I've also made this edit, but it still didn't work. https://codeforces.net/contest/1712/submission/168199569
Greedily changing the $$$k$$$ smallest values isn't always correct
Thanks, I hadn't thought of that.
I did this at first as well
Difficulties like ->Easy -> Easy -> Easy -> Hard -> Hard -> Hard -> Hard
E1 and E2 not hard
you didnt solve them tho?
Where does the "/6" and "/15" come from at problem E1? I ve seen this on many solutions.
I guess its an optimization for finding all triplets i,j,k with a fixed k, but i dont undersand it.
You will see, if(i==6 j==10 k==15), lcm(i,j,k)=30 < i+j+k, if(i==3 j==4 k==6), lcm(i,j,k)=12 < i+j+k, and you may enlarge i,j,k because if lcm(i,j,k) < i+j+k ,then X(lcm(i,j,k)) < X(i+j+k) ,since X is a positive integer.
And why is the cases (i==6 j==10 k==15) and (i==3 j==4 k==6) ? That's because 1/5+1/3+1/2 > 1 and 1/4+1/3+1/2 > 1
This is for finding $$$(i, j, k)$$$ where $$$lcm(i,j,k)=2k$$$ and $$$i+j>k$$$.
Note that because of the second condition, $$$\frac{k}{2} < j < k$$$. Since $$$j$$$ divides $$$2k$$$, $$$j$$$ must be $$$\frac{2k}{3}$$$.
Using the second condition again, $$$\frac{k}{3}<i<\frac{2k}{3}$$$, which implies $$$i$$$ is either $$$\frac{2k}{5}$$$ or $$$\frac{k}{2}$$$.
So the solution triplets in this case has the form $$$(\frac{2k}{5}, \frac{2k}{3}, k)$$$ (happens when $$$15$$$ divides $$$k$$$), or $$$(\frac{k}{2}, \frac{2k}{3}, k)$$$ (happens when $$$6$$$ divides $$$k$$$).
Nice. Thanks for the clear explanation.
Very easy greedy Solution of problem D
can you explain it as well plz?
Try this for an explanation.
As discussed elsewhere, when doing an update of an a_i you should always update to 1e9.
There are two possible strategies for doing the K updates of the a_i: 1. Update the smallest K of the a_i, or 2. Update the smallest (K-1) of the a_i, and then use the last update to update a neighbour of the largest of the updated a_i.
We can split the 2nd strategy into two: if K=1, then the final update is to the largest of the original a_i if K>1, then the final update will be to a neighbour of a_i already updated with value 1e9.
The final diameter of the graph is min( 2*smallest a_i, max of smaller neighbours a_i, a_i+1 )
So overall we have to calculate 4 numbers: v11 — strategy 1 — 2*smallest of updated a_i v12 — stratgey 1 — max of smaller of the a_i and a_i+1 v21 — strategy 2 — 2*smallest of updated a_i v22 — stratgey 2 — max of smaller of the updated a_i and a_i+1
diameter for strategy 1 is min(v11,v12) diameter for strategy 2 is min(v21,v22)
We pick the strategy for the larger diameter, so the final answer max( min(v11,v12), min(v21,v22) )
In the code: v22 is cnt — initially calculated for K=1, and overwritten with 1e9 if K>1 v21 is expressed as 2ll*vec[k-1].ff min(v21,v22) is put into "ans"
The smallest K values of a_i are identified by populating and sorting vector<pair<ll,ll>> vec. Then we explicitly update the a_i array.
v12 is calculated in maxi2 in a loop that looks at the neighbours min(a[i], a[i+1]); v11 is picked up from 2ll*vec[0].ff and we have maxi2 = min( maxi2, 2ll*vec[0].ff ) to calculate min(v11,v12).
The final output cout << max( ans, maxi2) selects the diameter from the better of the two strategies.
Video Editorial
thanks for the vid. much appreciated!
Finally became CM! Like B & D very much.
Why should i to find 6 and 15 in problem E1? How can I think that way. I see tourist's solution seemed iterate in 30 to find such special case. It shows that he is trying to find such special case.
For small numbers multiplication funtion has a low growing, so the idea is use BruteForce for those small values
Did anybody solve this clue by the author? I'm still stuck on it.
Edit: I saw the answer in editorial.
I really think this round is quite exquisite.
The bonus in every problem, the riddle in the beginning of each task(and ACGN lover here too!), and the nice problems.
Though I was one of those who was stuck in problem D just in lack of a binary search while using a bad greedy, I still have to upvote the problem preparation for all involvers！Thankyou！
B Bonus: try to prove the solution without the editorial!
Me: Use the data!
Could anyone please explain for what mysterious reasons this solution for D using ordered_set is not passing even small test cases like (n <= 1000)? (168216695)
Hand written Treap passed pretests, but FST'ed mostly likely due to bug in my implementation (168172650)
In my opinion, E1 and E2 are good problems. Thanks for the great contest!
In C iterating from last index, I checked if Ai < Ai-1. If yes then simply print number of distinct elements before Ai and break. If no then check whether count of Ai>1 && Ai!=Ai-1(i.e its duplicate lies in prefix of the array) then also print number of distinct elements before Ai and break.
https://codeforces.net/contest/1712/submission/168175298
Please help me where am I doing wrong
I am not confident that your idea would work in general, but here is an immediate error I found: when line 34 is executed, i.e., decreasing the hash value of an element if it appears multiple times in a row, the decrement --c at line 37 is still executed, even though the number of distinct values in the prefix has not decreased. Here is are three test cases I encourage you to check:
Each occurrence of 2 decreases the number of distinct characters, with the final test case even reporting a result of -1.
I think you intended to put a
continue;after line 34. I'm still a little skeptical of whether this would be enough to correct the program (since the logic feels like it still has loopholes), but that should be a good step forward.For prob D can someone explain this TC?
My solution is to replace the 2nd and 6th values with $$$10^9$$$. So the diameter here would be $$$d(2,6) = 4$$$ (going through one of the nodes with value 2), but the expected result is 2. Am I missing something here?
The length of the edge connecting node $$$2, 6$$$ is $$$2$$$, because $$$\min{a_2, a_3, a_4, a_5, a_6} = 2$$$.
Oh shit I misunderstood the problem for the whole night. Thank you!
In fact, we can prove in E1, $$$j = \frac{2}{3}k$$$ if $$$\operatorname{lcm}(i, j, k) = 2 \cdot k$$$ and $$$i + j > k$$$. So we can solve this problem in complexity $$$\mathcal O(n \sqrt n + \sum_{i = 1}^n \sigma_0(i))$$$ per testcase, which is faster than the editorial.
code 168224359
Yes, but I don't really like number theory, so I didn't want to include more of it in the solution.
C feels amazing, I can only think of BFS to solve it
Very good contest, one of the best I have seen yet :)
personally liked all the anime references.
But I can't understand D. /sad
Video Solution for D.
(Video Solution for C.)
thinks.
Here is my proof for B. The key claim is the following:
Claim: Let $$$a,b$$$ be positive integers such that $$$a\ne 1$$$ or $$$b\ne 1$$$ (or both), then $$$\operatorname{lcm}(a,b) \leq \frac{a^2+b^2-1}{2}$$$.
Proof: If $$$a\neq b$$$, then
Otherwise, $$$a=b$$$, then
Hence, the claim is proven $$$\blacksquare$$$
Now, we just use the claim (note the exception at $$$1$$$):
If $$$n$$$ is odd, then this is the desired bound. Otherwise, note that this quantity is not an integer, so we actually reduce the bound by $$$\tfrac 12$$$. One can check that this yields the equality.
this is very elegant!
Bonus Problem's Solution for A
We have to find $$$ans_1, ans_2, \ldots ans_n$$$
pfrom endxappears at indexiand $$$x<i$$$. This implies thatxis at wrong position, after its ideal positionx.ansarray will be computed inO(n)time.Implementation
About Bonus Problem of C
A-E2 video Editorial for Chinese ：
Bilibili
Comeback after almost a year ... I did pretty bad
for E2. "Turns out that for the current constraints, for every k, we can iterate over all pairs of 1≤i<j<k, where i and j are divisors of 2.k and check if the triplet is bad."
Why can we do this? I know that the sum of number of factors of all numbers from 1 to n is O(nlogn), but here we are iterating over each pair of factors for every number, what is the bound for that?
Judging by this answer, $$$\ll n \log^3 n$$$.
In editorial of problem D, what is the "another case to work"?
In problem E how can I find this $$$\frac{(r - l + 1) \cdot (r - l) \cdot (r - l - 1)}{6}$$$ .
It's just $$$C_{r - l + 1}^3$$$ — the number of ways too choose three elements from $$$r - l + 1$$$ elements (if we don't care about the order).
What is the solution for F Bonus?
Did you solve the original F problem first?
Can somebody help me out with Problem D : Link
My approach is ans = max( min(2 * mn , min(a[i],a[i+1]))) , where mn is the minimum value of a.
Any ideas about
Bonus Problem for D?I have solved problem D, but I can't think of any approach for doing it for all k.
Maybe Vladithur can create a separate blog post about the solutions of bonus problems.
I'm too lazy to do that(
As for the bonus of problem D, the basic idea is to extend the greedy solution.
If we try to extend the greedy solution of D. The complexity reaches
O(n^2).kwe need to doO(n)processing.k-1to solve fork, But there seems a lot of casework, and even after that complexity still seems to beO(n^2)in some worst cases.Can someone help me here ?
What I meant is that you process the values of $$$a$$$ in increasing order and then maintain a couple of sets / multisets to find the answer for every $$$k$$$. Try to find what information you need to know to get the answer for some $$$k$$$ (and is easy to maintain when you move on to $$$k + 1$$$).
I found C to be bit on a tougher side, compared to the no of solves during the contest :/
I am very confused about the use of i+= i & (-i) loop from many of the fastest solutions of problem E2. For example, this, this and this.
It seems to me that there are some hidden patterns that I am not aware of. Could anyone help me?
It's just one possible implementation of a fenwick tree.
.
Can anyone tell me why I got the wrong answer in https://codeforces.net/contest/1712/submission/168392728 of problem d?
Thanks for amazing round and tutorial but there is a small mistake Div2B Hint 3
lcm(x, x + 1) + lcm(x + 1, x) = x² + (x + 1)²-1
Fixed
why is this unrated now ?
We can probably prove problem B by induction? In addition, I don't understand this sentence in problem B:
Could anyone please explain it?
Sory for my poor English.
Для задачи E1. А как быстро найти такие $$$j$$$, что $$$1 \le i < j < k$$$? Перебирать все пары делителей для $$$2k$$$ довольно долго.
"If there is at least one operation left, there are two cases: k=1 and k>=2.If the first case, it is optimal to apply our operation near one of the maximums in the array to maximize min(a_i,a_{i+1})"
Would it not be more optimal to change a_i > = ans/2 to 1e9? Consider ans = 18, and the following array-
[0,1,4,5,6,9,10,11,15,17] and nodes [1,2,3,4,5,6,7,8,9,10]
Let k = 6. The binary search solution would turn this to-
[1e9, 1e9, 1e9, 1e9, 1e9, 9, 10, 11, 15, 17] and k = 1.
Now, if we change 15 to 1e9 (as mentioned in the solution), then d(9,10) = 2*9 = 18. But if we change 9 to 1e9, then d(6,8) = 2*10 = 20 > 18?
I meant $$$k = 1$$$ and $$$k \ge 2$$$ for the value initially given, not after we apply the operation some number of times.
I am not sure I follow, consider-
arr = [2,3,4,5,11,12], nodes = [1,2,3,4,5,6] and k = 1.
Let ans = 5. In the binary search, no a_i's would be turned to 1e9 as 5/2 = 2 !< 2. First if condition fails as operation wasn't applied. In the second if condition, k==1 and max(a_1,..,a_n) = 12 > 5, hence it will return true.
But, if we change 11 to 1e9, then d(5,6)= 2*min = 2*2 = 4 < 5?
The division is not floored, $$$\frac{5}{2} = 2.5$$$, not $$$2$$$.
Why tutorial for problem E1 is missing?
Because the main idea of the model solutions is the same in both versions, E2 just uses a data structure to speed up the last part of the solution, which you don't need to use in E1.
Thanks, it would be really helpful if you can mention this information at the start of the tutorial for E2. Or rather one should have made the editorial for E1 because I think people usually solve in order so they won't look up into E2 before they have solved E1 expecting that E2 must have a different solution.
Vladithur can u elaborate on the proof for greedy in div2 D. Why is it optimal to consider k-1 minimums in the first place?
This is somewhat similar to the binary search solution. In it you change the smallest values such that all $$$a_i \ge \frac{ans}{2}$$$, and then cleverly choose which remaining values to change. Turns out that there are only two cases: zero values left or $$$\ge$$$ one value left, so in our greedy solution, we also leave one value to do something with, hence we change the $$$k - 1$$$ smallest values.
For Problem E2 : It is mentioned as
Since i<j<k, a triplet is bad only when lcm(i,j,k)=k or (lcm(i,j,k)=2⋅k and i+j>k)but there is no explanation like why it is true or how do the author arrive to this condition. Vladithur Can you explain this ? Like its is easy to observe that if lcm(i, j, k) = k then it would be bad, but for 2k I am not sure. Also why not the values in between k and 2k.C :
wHERE IS MISTAKE , iT'S FALLING ON 2ND TEST CASE
https://codeforces.net/problemset/submission/1712/170962914
Casework O(nlogn) greedy solution for D:
Maintain array v and sorted array v2
One possible solution is to reduce all the k minimum elements to 1e9 then
ans=max(ans,min(v[i]+v[i+1],v2[k]*2)) ****
as elements 0......k-1 in v2 assigned 1e9 in v so the smallest element in v is v2[k]
Now reset v
Another possible solution is to assign the smallest k-1 elements to 1e9 then
ans=max(ans,min(max(v[i],v[i+1]),v2[k-1]*2)); ****
This is because we have an element v[i]. Next or previous element we can assign 1e9 so it becomes min(v[i],1e9) = max(v[i],v[i+1]) as we assign the minimum element to 1e9
Now if k>=2 Another possible solution is to assign the smallest k-2 elements to 1e9 then
ans=max(ans,min(1e9,v2[k-2]*2)) ****
as we have two consecutive elements of the form [1e9 1e9] in the array and the smallest element in the array is v2[k-2]
E2 is "small to large" technique ?