Idea: diskoteka
Tutorial
Tutorial is loading...
Solution
#include <bits/stdc++.h>
using namespace std;
#define int long long
int32_t main() {
int t;
cin >> t;
while (t--) {
int n, m, k;
cin >> n >> m >> k;
int x, y;
cin >> x >> y;
string ans = "YES\n";
for (int i = 0; i < k; ++i) {
int xx, yy;
cin >> xx >> yy;
if ((x + y) % 2 == (xx + yy) % 2) {
ans = "NO\n";
}
}
cout << ans;
}
return 0;
}
Idea: diskoteka
Tutorial
Tutorial is loading...
Solution
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int n, k;
cin >> n >> k;
vector<int> c(n);
for (int i = 0; i < n; ++i) {
cin >> c[i];
}
vector<int> last(k, -1);
vector<int> max_step(k), max2_step(k);
for (int i = 0; i < n; ++i) {
int step = i - last[c[i] - 1];
if (step > max_step[c[i] - 1]) {
max2_step[c[i] - 1] = max_step[c[i] - 1];
max_step[c[i] - 1] = step;
} else if (step > max2_step[c[i] - 1]) {
max2_step[c[i] - 1] = step;
}
last[c[i] - 1] = i;
}
for (int i = 0; i < k; ++i) {
int step = n - last[i];
if (step > max_step[i]) {
max2_step[i] = max_step[i];
max_step[i] = step;
} else if (step > max2_step[i]) {
max2_step[i] = step;
}
}
int ans = 1e9;
for (int i = 0; i < k; ++i) {
ans = min(ans, max((max_step[i] + 1) / 2, max2_step[i]));
}
cout << ans - 1 << "\n";
}
return 0;
}
Idea: pavlekn
Tutorial
Tutorial is loading...
Solution
#include <bits/stdc++.h>
using namespace std;
#define int long long
int gcd(int a, int b) {
if (a == 0) {
return 0;
}
if (b == 0) {
return 1;
}
if (a >= b) {
int r = a % b;
int k = a / b;
if (k % 2 == 1) {
return gcd(b, r) + k + k / 2;
} else {
return gcd(r, b) + k + k / 2;
}
}
return 1 + gcd(b, abs(a - b));
}
int calc(int a, int b) {
if (a == 0) {
return 0;
}
if (b == 0) {
return 1;
}
return 1 + calc(b, abs(a - b));
}
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int t;
cin >> t;
for (int _ = 0; _ < t; ++_) {
int n;
cin >> n;
vector<int> a(n);
vector<int> b(n);
for (int i = 0; i < n; ++i) {
cin >> a[i];
}
for (int i = 0; i < n; ++i) {
cin >> b[i];
}
set<int> cnt;
for (int i = 0; i < n; ++i) {
if (a[i] == 0 && b[i] == 0) {
continue;
}
cnt.insert(gcd(a[i], b[i]) % 3);
}
if (cnt.size() <= 1) {
cout << "YES" << endl;
} else {
cout << "NO" << endl;
}
}
return 0;
}
Idea: diskoteka
Tutorial
Tutorial is loading...
Solution
#include <bits/stdc++.h>
using namespace std;
#define int long long
int f(int s, int k) {
// (s + 20x) * (k - 4x)
// (-80)x^2 + (20k - 4s)x + (sk)
// -b/2a = (5k-s)/40
int x = (5 * k - s) / 40;
x = min(x, k / 4);
int res = s * k;
if (x > 0) {
res = max(res, (s + 20 * x) * (k - 4 * x));
}
x = min(x + 1, k / 4);
if (x > 0) {
res = max(res, (s + 20 * x) * (k - 4 * x));
}
return res;
}
int32_t main() {
int t;
cin >> t;
while (t--) {
int s, k;
cin >> s >> k;
int ans = s * k;
if (s % 10 == 5) {
ans = max(ans, (s + 5) * (k - 1));
} else if (s % 10) {
if (s % 2 == 1) {
s += s % 10;
--k;
}
for (int i = 0; i < 4; ++i) {
if (k > 0) {
ans = max(ans, f(s, k));
}
s += s % 10;
--k;
}
}
cout << ans << "\n";
}
return 0;
}
1848E - Vika and Stone Skipping
Idea: diskoteka
Tutorial
Tutorial is loading...
Solution
#include <bits/stdc++.h>
using namespace std;
const int K = 1e6;
#define int long long
int mod;
int dp[K + 1];
int cnt[K + 1];
const int MM = 3e6 + 7;
int inv[MM];
int cc = 0;
int mul(int a, int b) {
int bb = b;
while (bb % mod == 0) {
++cc;
bb /= mod;
}
return (a * (bb % mod)) % mod;
}
int dv(int a, int b) {
int bb = b;
while (bb % mod == 0) {
--cc;
bb /= mod;
}
return (a * inv[bb % mod]) % mod;
}
int32_t main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
int x, q;
cin >> x >> q >> mod;
inv[1] = 1;
for (int i = 2; i < MM && i < mod; i++) {
inv[i] = mod - inv[mod % i] * (mod / i) % mod;
}
int y = x;
for (int i = 3; i <= K; i += 2) {
if (dp[i]) {
continue;
}
for (int j = i; j <= K; j += 2 * i) {
dp[j] = i;
}
}
int ans = 1;
while (x % 2 == 0) {
x /= 2;
}
for (int i = 3; i <= K; i += 2) {
while (x % i == 0) {
ans = dv(ans, cnt[i] + 1);
++cnt[i];
ans = mul(ans, cnt[i] + 1);
x /= i;
}
}
int f = 1;
if (x > 1) {
ans = mul(ans, 2);
}
x = y;
for (int i = 0; i < q; ++i) {
int d;
cin >> d;
while (d % 2 == 0) {
d /= 2;
}
int k = d;
while (dp[k]) {
ans = dv(ans, cnt[dp[k]] + 1);
++cnt[dp[k]];
ans = mul(ans, cnt[dp[k]] + 1);
k /= dp[k];
}
if (cc) {
cout << 0 << '\n';
} else {
cout << ans << '\n';
}
}
return 0;
}
Idea: pavlekn
Tutorial
Tutorial is loading...
Solution
#include <bits/stdc++.h>
using namespace std;
const int MAXN = (1 << 20);
int a[MAXN];
int solve(int n) {
if (n == 1) {
if (a[0] == 0) {
return 0;
} else {
return 1;
}
}
int fl = true;
for (int i = 0; i < n / 2; ++i) {
if (a[i] != a[i + n / 2]) {
fl = false;
}
}
if (fl) {
return solve(n / 2);
}
for (int i = 0; i < n / 2; ++i) {
a[i] ^= a[i + n / 2];
}
return n / 2 + solve(n / 2);
}
int32_t main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; ++i) {
scanf("%d", &a[i]);
}
cout << solve(n) << endl;
return 0;
}
A,C,D and E are just math problems. Cf should do better.
Problem statements were too long.
look at As editorial :0
I agree with the general mood that A-C of the contest were too hard. D-F were okay, although in E $$$M$$$ should have been set larger than $$$O(Q \log x)$$$ to prevent the edge case where one of the prime powers reaches $$$M$$$.
Why should setters prevent corner cases instead of contestants preventing them in their code?
This is a Div 2 codeforces contest. Because this is Div 2 where the contestants aren't necessarily that strong, and because this is codeforces where the pace of the competition is very fast, it should be more about solving the problem than implementation and avoiding corner cases.
Also, it looks like the authors made the same mistake, hence the explanation that Test Case 15 (the first case with that corner case) was incorrect and the solutions had to be rejudged.
This is a Div 1.5 contest, so include purples in your thesis. "not strong div2 participants" wouldn't reach E anyway.
Btw my opinion on this problem is: why the fuck is it variable mod? Why not just use constant mod 1e9 + 7? It seems some coordinators or setters are addicted to using variable mod for no reason other than worsening the runtime of people's solutions.
I didn't know about test case 15, that slipping by testing is a big oof.
There are problems where variable mod is needed, in particular the problems where there are limited amount of parameters and it's possible to locally compute all the possible answers with a slower code then submit the result.
I agree that for this problem there is no such reason and a constant 1e9+7 is better.
Div2 contestant also should have rights to solve E with certain wisdom ,instead of getting cornered by this kind of modulo trick. In fact that’s almost irrelevant to the main idea of this problem, which i think is conducting the numbers of factors (forgive my terrible English)
well, stupid edge cases are no fun (the original problem when i tested had a constant mod > 1e9)
also, the setters themsevles got the case wrong....
I liked A and B, even though the latter was mostly implementation. C and D were too mathy and guessable though.
problem statement of A was pretty confusing for me.
same
I like your pfp, I am a veteran of SF3 as well!
same here :(
There exists an $$$O(n)$$$ solution for problem C, which is better than the editorial's $$$O(n\log a_i)$$$:
Submission
Can you explain your solution, i read but i can't understand
More readable code for same logic.
https://mirror.codeforces.com/contest/1848/submission/214137912
I can't prove it, but it's correct:)
The first few steps are the same as editorial. On finding out how to calculate $$$(a,b)$$$'s answer, write a brute force first. Then u can notice the patterns the same in my code:P
If someone can prove it I'll be very grateful.
Actually it the same as second solution of editorial: "Thus, we can determine the remainder cnt modulo 3 by looking at the pair (a/d, b/d) modulo 2".
Let a = 2**g * ra, b = 2**h * rb
ra, rb — odd numbers.
If g > h, then:
a / gcd(a, b) = 2**(g — h) * (ra / gcd(ra, rb)) — even number
b / gcd(a, b) = (rb / gcd(ra, rb)) — odd number
In case g < h, first will be odd, second even.
In case g == h, both are odd.
This is O(n log(n)) Only to calculate the number of trailing zeroes it takes log(n) times only.
Hi,
__builtin_ctz()is an inbuilt function of G++ that only have $$$O(1)$$$ complexity as far as I know. Correct me if not:)https://mirror.codeforces.com/blog/entry/59268
Though it does the bitwise calculation, it is too fast to get noticed, but the complexity is still O(log(n)). You can verify this by manually writing the code for it, and you will see that it takes the same time.
Click on compare the previous solution in the below submission.
https://mirror.codeforces.com/contest/1848/submission/214137912
I can't really agree. It seems on my local machine that
__builtin_ctz(x)is at least 30 times faster than the handwritten one. as the handwritten one has the complexity of $$$O(\log n)$$$, I would say that__builtin_ctz(x)has to be something like $$$O(1)$$$.Here's my code and results:
Test 1:
Test2:
change the
limin the first code to $$$10^8$$$, and it runs 0.2 s on average; change thelimin the second one to $$$10^8$$$, and it runs 6.1 s on average.What do u think about that?
upd: i did another test, i changed the
get(x)to:and $$$lim$$$ is still $$$10^8$$$. the average time is 0.24s! That means the inbuilt function of g++ seems to be even faster than adding two numbers together! What's ur opinion?
theoretically, it is still n log n but practically it is O(n)
idk how __builtin_ctz() is implemented,but i still think its real complexity should be real close to $$$O(n)$$$.
Godbolt says there's special "bsf" assembler instruction for it.
Stop posting misleading content you are discussing popcount but ctz only needs x&-x
and popcount isn't O(log n) too it's O(log w) like O(log log n).
how did you arrive at the intuition of using 2 highest power in solving it.
I'm stupid, I don't have intuition, I write brute force, I observe patterns, I draw conclusions, I implement, I pass
What was the logic in your brute force code that showed some pattern?
This contest problem statements are too long like a reading comprehension.
I think C is more difficult than D
May be I'm too foolish , but what is the tutorial of C's meaning???
Personally I don't think problem A is too hard for its position. The idea is fairly common in game theory puzzles like https://www.youtube.com/watch?v=dh4nEuhZBgg and just requires a bit of good intuition.
diskoteka [user:pavlovn]
My thoughts on this round:
The problems themselves were fine.
I do think A was not suitable to be a div2A, but that’s because I think div2A’s should be “submitbait”. The problem itself was fine, and may have done better as div2B.
Perhaps D was too math-y than ideal, but that’s fine.
Perhaps E could’ve had $$$M \geq 10^8$$$ or have been fixed, but it wasn’t much extra (and it is not too hard to generalize to all $$$M$$$). It is bad that the sequence is on OEIS though, so ideally this problem shouldn’t have been used.
Also, don’t listen to the haters, and please keep setting contests. Over time you’ll get better, and this contest probably provided good feedback!
ty very much
C was a beautiful problem!
Your tutorial for problem D should write (b+20x)(a-4x), not (b+20x)(a-x) in the 4th paragraph, as reflected in your code.
Easier for F is just doing binary lifting as soon as you know the power of 2 thing. It's basically the same solution but O(NlogN) and you don't have to think about reducing it to an array of size N/2.
For problem A, vika's friends can see her(vika) right? So they see her and try to modify their movement right? Can vika see her friends as well?
If we color the rooms like a chessboard, and at least one of her friends starts on the same color as her, that friend has the strategy, and can react to her movement(because she moves first) in such a way that no matter what she does, she eventually will be caught, this is explained in the editorial. So, whether Vika tries to avoid her friends or not, it doesn't matter.
please explain me this(Problem A):
1 2 1
1 1
1 2
I think when vita goes to right , friend will catch him . Perhaps, i misunderstood something.
both vika and her friend should perform a move .So they're just going to swap rooms forever and never meet each other.
a friend catches vika only if they find themselves in the same room after performing the move
During contest, I understood from statement so:
First vika goes, then her friend.
Thanks
This is the WORSR contest i've ever seen. Can you just stop writing such long description of problems?? If you want to show your ability of writing, you don't need codeforces. And problems DEF are so easy to think, the difficulty is too low to be the div2 DEF, it can be used in div3. The quality of the problems are also too low I think. And the awful test cases of problem E, can you just check your test cases before the contest? To my surprise, this contest has not been unrated, but I don't want to see such contest like this anymore. YOU DO NOT NEED TO CREATE THE CONTEST IF YOU DO NOT HAVE THE ABILITY OR YOU DO NOT WANT TO DO IT WELL!!!
when problem A has the longest tutorial, you know you did something wrong.
wthforces
I don't get why so much hate for this round. A is better than normal, I liked problem C quite a bit, and rest of problems were fine. (yet another round with no datastructure tho ;-;)
You could've used a segment tree in order to avoid the corner case in E lol
I wonder why they even used prime mod. Either used fixed or force this....
Problem — A was quite lengthy from a beginner perspective and difficult to prove.
Otherwise the rest of the contest (I mean A-D, I haven't read E and F) was fine.
(I struggled a little bit to decipher the language in A (and B for that matter) The story of friends catching "Vika" but having a greater affinity to make a move seems inconsistent and looks like a bad attempt to make the problem look real worldish when it isn't).
I agree A could've been worded better, but I don't think it's bad enough to greatly worsen round.
Sorry ,but I feel A lengthy and language is very tricky
In solution1 of problem C , it will be ai-4*bi in place of ai-4ai
can someone explain me the problem statement of B in easy way.. i have read it multiple times but i am not able to understand
thanks..
vika can paint any particular plank of her choice. you have to figure out which one so that she has to step over minimum number of plank
condition is that she will be jumping on only same color planks.
Problem A was way too unbalanced for its position and difficulty range. Its not a great idea for begineers
in task B, isnt it necessary to mention in the problem statement that k distinct colourst will definitely be present.
My solution for A :
So we need to check for every position (i,j) in n*m grid, if there is a position where both main person and anyone among the k people have the same distance to (i,j). So we iterate over every position and check whether the distance of main person is same as distance of any of k friends to that position. TC : O(n*m*k).
AC code : 214108505
I used the same implementation.
After reading editorial for problem A, do you still feel like this is DIV-2 A problem ?
diskoteka
Only in Vika-land, problem would need this much logical thinking and proof for problem A solution.
I think, by considering it as a problem that could be solved by brute force thus being easy to implement, it's a good Div 2A.
Because of problems like A, people start to hate math in cf-contests. The problem would be great if it were at the math contest, but not at cf.
Personally, I loved this round! I only solved ABCD, but this round had better problems than most recent Div2s in my opinion. I don't understand why the announcement blog post was downvoted. Keep making rounds!
ive done other coding contests but this was my first cf contest and i solved 0 questions, so skill issue for me ig
In problem C, I have a different way of finding the $$$cnt_i\% 3$$$ mentioned in the editorial.
Similar to solution 2, we will build the sequence from the end. It will be of the following form:
Now try to build a tree of all possible values before $$$x$$$. You will notice values in a particular level are either of the type $$$odd * x / 2$$$ or $$$k * x$$$ where $$$k$$$ is a positive integer. You will also notice that $$$k * x$$$ has a period of $$$3$$$, same as the period of $$$0$$$.
So the problem reduces to finding which of the $$$a_i, b_i, c_i$$$ is of the type $$$k * x$$$.
Here's my submission.
i am still not able to understand your approach, can you explain in more simple way. thanks.
Think of the first time $$$a_i$$$ becomes zero. It will be of the following form:
What would come before $$$k$$$? Let it be $$$y$$$, solving $$$|y - k| = k$$$ we would get $$$y = 2*k$$$. So the sequence now becomes
Similarly, try to find what would come before $$$2*k$$$. From here, multiple values might be possible for each position. For example, before $$$2*k$$$, both $$$3*k$$$ and $$$k$$$ are possible values.
Try to draw a tree of these values. $$$2*k$$$ will be the root and $$$3*k$$$ and $$$k$$$ are its children. It will be something like this: https://binary-tree-builder.netlify.app/?indextree=2*k,3*k,k,5*k,k,3*k,null,8*k,2*k,4*k,null,4*k,2*k.
In this tree, at each level, the values are either of the type $$$even*k$$$ or $$$odd*k$$$. You will also notice $$$even*k$$$ has a period of $$$3$$$, just like the period of $$$0$$$.
Now consider a situation where $$$a_1 = even*k$$$ and $$$a_2 = odd*k$$$. Then we know that both $$$a_1$$$ and $$$a_2$$$ won't become $$$0$$$ at the same time.
So we need all $$$a_i$$$ to be $$$even*k$$$ or all $$$b_i$$$ to be $$$even*k$$$ or all $$$c_i$$$ to be $$$even*k$$$. Only then we will get all zeroes at the same time.
The best explanation I found so far. But why (all ai or all bi or all ci) need to even*k. why not just check if all ci are even*k only? why we have to check ai and bi too?
We might have something like this:
As you can see, this will not result in all zeroes occurring at the same time.
Now consider the following:
This will result in all zeroes occurring at the same time. So its important that you check all $a_i$, $$$b_i$$$, $$$c_i$$$.
You're my savior
Thanks a lot !!
diskoteka, in the editorial of C, while writing Solution 1: I believe where you wrote
(ai, bi, ai — bi, ai — 2 * bi, ai — 3 * bi,
ai - 4 * ai)(and should have written:
ai - 4 * bi)Thank you diskoteka for great math problems, better than previous div(1,2) speedforces
E and F are well known math problems.
E: 2016 AMC 12B P16
F: Trivial special case of 2023 USA TSTST P9
It has been pointed out to me that the setup for F is actually identical to USA TSTST 2023/9 with $$$p^e=2$$$, although I’m not sure how much knowing the solution to the TSTST problem helps with the codeforces and vice versa. At the very least, one of the solutions to the TSTST problem proves that you will never print $$$-1$$$ (see the first claim in post #5 by a familiar-looking user).
Edit: sniped by the person who pointed this out lol
The first part of the TSTST problem tells you what the operation becomes after applying it $$$2^i$$$ times.
Oh true—the explicit computation of the $$$k$$$-th finite difference implies that for $$$k=2^i$$$, we have $$$\Delta^{2^i} f(n) \equiv f(n+2^i)+f(n) \pmod{2}$$$, since by Kummer's theorem/Legendre $$$\binom{2^i}{a}$$$ is even unless $$$a \in {0,2^i}$$$, which more or less reduces this problem to a straightforward binary search
Great round! Like these problems very much :3
We've found some alt ways to solve F, but they have higher time complexity.
A is too hard...
I think problem A is harder than problem B.
Different Approach For Problem C. Code
Can you explain how it works, like why does the highest power of 2 matters?
Then, if the cycle is "scroll" x times, the discount will be (b+20x)(a-x)
WHY NOT (b+20x)(a-4x)?
In C, I tried to find the number of operations to make ai zero manually in the hope of getting some pattern or some formula but all in vain. May be they should try not to give such mathematical things till C.
C is one of those problems which gives neither the pleasure when I solve them nor guilt when I could not.
Can someone break down C for me, I observed a few things during the contest.
Let a>b for all below expressions, if not we can just move to the next stage where b and |a-b| will fit the scenario(moving one parity ahead).
Every pair will eventually reach a loop of GCD(a,b), GCD(a,b), 0. So essentially we needed to check if every pair have the same parity (%3 values of the number of steps they reach 0). I also noted that we can go from a,b to a%b,b keeping the same parity, except for when (a/b ==1). And that is where it stopped for me.
I saw some people use a%2b to move a parity ahead, but where does that 2 come from?
I'll use pair notion meaning $$$(a, b)$$$ pair becomes $$$(b, |a - b|)$$$.
Consider some large $$$y$$$ and small $$$x$$$: $$$(x, y) \Rightarrow (y, y - x) \Rightarrow (y - x, x) \Rightarrow (x, y - 2x)$$$.
We made 3 operations meaning it wont change amount operations modulo 3, therefore you need to find largest $$$k$$$ i.e. amount of operations that: $$$0 <= y - k(2x) < 2x$$$ which is literally a definition of remainder.
can u prove doing this way + perform the said operation will not be more than logarithmic
I don't understand why do you need to prove this fact since it's literally gcd-like + you literally can implement it and test on big random numbers but here you are:
Since operation in pairs are not simmetrical you can't assume WLOG $$$x \leq y$$$ so I'll let you do $$$x > y$$$ by yourself ;)
So I'll be doing $$$x \leq y$$$ case:
Now consider two cases:
$$$4x \geq y$$$, then pair $$$(x, y)$$$ becomes $$$(y, y - x)$$$ where $$$y - x \leq \frac{3y}{4}$$$, $$$(y, y - x) \Rightarrow (y - x, x) \Rightarrow (x, y - 2x)$$$. Congrats! we have reduced second term by at least 3 quarters, since $$$y - x \leq \frac{3y}{4} \Rightarrow |y - 2x| \leq \frac{3y}{4}$$$
$$$4x \leq y$$$, then $$$(x, y)$$$ becomes $$$(x, y \ \% (2x))$$$. Congrats! we reduced second term by at least half
This yields next conclusion: rough estimate for upper bound of steps is $$$3*log_{4/3}(y)$$$ which is good enough. The rest of the proof is up to you
Typo in
deditorial:$$$(b+20x)*(a-4x)$$$
I found a very interesting pattern for problem C. For every i, we count number of trailing zeroes in binary representation of A[i] and B[i], now if A[i] has more, then this pair can be made 'dull' in 3k steps, if B[i] has more, then 3k+1 steps, and if they both have equal trailing zeroes, then 3k+2 steps. And we just have to ignore A[i]=B[i]=0 case.
Can someone provide me with clear insights on Problem C? Or provide resources to learn the topic. It is clear to me that the only thing that matters is the modulo 3 of the first occurrence of zero for each position of the array. But I find the tutorial very confusing on how to proceed further (maybe I am missing some number theory / modular arithmetic prerequisites)?
Consider several cases of
a[i]andb[i]. Such as:Both are odd. [O,O]
a[i] = 0occur from above observations?Similarly analyse cases of [O,E] and [E,O]
The case of [E,E] is interesting. You can observe that they will mimic the behave of the numbers we obtain by doing successive division by 2.
You know the story after this.
Did everyone lose rating? Half of the people must have gained acc to algo
I didn't understand the solution explained for problem C. Can someone explain it clearly.
me too
Who can explain the tasks C, D? I didn't understand the author's decision at all
problem C:
We can consider the i-th pair ($$$a_i$$$,$$$b_i$$$) ,but now we just focus on one pair. If we already have (0,x) , do some operations , when will 0 appear again at the same location ?
0 will alternate with a period of 3.
(0,x) → (x,x) → (x,0) → (0,x) → ...
If a>=2b , make 3 operations on (a,b) , what do we get?
Proof: (a,b) → (b,abs(a-b)) = (b,a-b) → (a-b,abs(b-(a-b))) = (a-b,a-2b) → (a-2b,abs((a-b)-(a-2b))) = (a-2b,b).
Similiarly , if b>=2a , make 3 operations on (a,b) , we will get (a,b-2a)
We know that if after $$$k$$$ operations (a,b) will change to (0,x) , 0 will stay in array a after $$$k+3l (l>0)$$$ operations. So we just calculate $$$r = k$$$ modulo $$$3$$$ for each pair. If two pairs' $$$r$$$ not equal, we can say "No". If all pairs have same value of $$$r$$$ , we can say "Yes".
Note that pair (0,0) always satisfy the situation , so ignore it .
how to prove complexity of urs code is log
It's similiar to the complexity of GCD and that's log.
But that we can prove .Here (a,b) where a>=2*b or b>=2*a remains in the same state for more times.Its not exactly same .from this state to say them same is not reasonable
Assume that a>=2*b.(if 2*b>a>=b , My code do the same thing as GCD)
GCD : (a,b) → (b,a%b) , 1 step makes the biggest num decrease from a to b.
Mycode :(a,b) → (a%(2*b),b).
I. 0< a%(2*b) < b. 1 step too , same as GCD.
II. b<= a%(2*b) <2*b . Then we go the second step: (a%(2*b),b) → (b,a%(2*b)-b). 2 steps makes the biggest num decrease from a to b.
so Mycode's complexity is no more than twice of GCD's . That's log.
(a,b)
three cases :
1) a>=2*b
2) 2*b>a>=b
3) a<b
fourth thing each state will be one of three
first case u proved in 1 or 2 steps will be same as (b,a%b)
second case u also proved in 1 step can get to (b,a%b)
third case can be reduced to (b,b-a) one of case 1 or 2(can take 3 steps)
a %b==a-b only if 2*b>a>=b
VERY WELL DONEhere is my implementation of the same approch (not even a single change ) https://mirror.codeforces.com/contest/1848/submission/214252557
if wanted its complexity can also be O(2*logn) or O(3*logn) -complexity at the heart
thank you very much!
Hey man, how were you specifically able to come up with the a >= 2 * b condition. Can you please explain your thought process behind it?
The tutorial is clear.
WTF you haven't said that sum of Ks over all test cases is at most 200000 in B!
It is unnessesary. k <= n is stated
Is the kind of reasoning used in problem E familiar to high-rated coders? Looking at the tutorial, I don't see any way I could have solved the problem on my own, so I want to know if people who solved it are well-versed with such kinds of reasoning or if it came to them naturally.
C probelm solution
Can someone please explain what does area to Vika mean? Is it something other that the number of blocks between Vika and one of her friends?
Simpler Solution for A:
Every move, the parity of R+C changes by 1 for both Vika and her friends. Therefore, a friend can catch up to Vika if and only if the sums of their coordinates have the same parity.
Because the friend moves after Vika, she can always decrease the distance between the two of them. By contrast, Vika can only increase distance by moving in the same direction as the gap between her and her friend (e.g. if Vika is three units to the right of her friend, she can only decrease horizontal distance by moving to the right). Because the map is finite, Vika will be forced to decrease the distance between her and her friend after a maximum of R+C moves, and so she will eventually be captured.
Nice .In both dimensions the distance cannot remain same .just follow vika steps .he will be caugth
Problem E Observation:
# of odd divisors = # of distinct initial fcan also be proved in the following manner:Consider all the pairs of [l,r] such that $$${\sum_{i = l}^r i = x}$$$
We are sure the the lengths
lenof all these segments have to distinctSo we need to find what all valid lengths are there. We have the following formula
lenis even. The case whenlenis odd is similar to prove.len = 2p and p>=1where did the formula (b+20x)(a-x) in D come from?
In any case, except for 5 and 0, the last digits will start following the pattern of 2,4,6,8 after a certain point. so we can make the formula as for not taking a discount 4 times we will have an increase in b by 20(2+4+6+8) and a decrease in the number of purchases left by 4.
if we do this process x times then the discount i can get is (b+20*x)*(a-4x).
and the formula in the editorial is partially wrong(missed 4*x).
I have created video editorial for problem A,B and C in english link. If you understand it, upvote.
Can anyone please give me a more intuitive idea of the induction of problem F? I mean how can we see it happen beside trying to prove the equation?
Try to solve problem Xor pyramid. This induction thing in F is just a special case $$$(n = 2^k)$$$ of Xor pyramid.
Lucas's theorem (for $$$p = 2$$$)
Maybe there is something wrong with problem A statement. What will happen when $$$n=1,m=1$$$?
In this case none of the girls can move to the neighbouring rooms, which violates your statement.
diskoteka Could you please, explain this case. Seems that the answer of MCS is
No.Finding $$$cnt_i\%3$$$ in problem C is not that hard I think.
$$$[a,b]$$$ will change like this $$$[odd,odd] \to [odd,even] \to [even,odd] \to [odd,odd] \to ... $$$
And $$$a$$$ can be $$$0$$$ in only $$$[even,odd]$$$.
(of course if both number is $$$0$$$ a is always $$$0$$$ and otherwise we can divide numbers by $$$gcd(a,b)$$$ so we can remove $$$[even,even]$$$ case)
Can anyone share there solution for D using ternary search. Mine is failing. My solution for D
Solution written in python for problem D, using ternary search.
In Problem D why are we only subtracting 1 from k after every iteration, shouldn't we take into account the fact that we are performing 4 * x number of operations and adjust k accordingly?
yes typo in editorial
cf should do better
A can be done through brute force too , just go to each cell , if distance from vika's starting cell == distance from any of her friend's cell , then vika can't escape
great contest, do consider making more contests please. [user:diskoteka][user:pavlekn]
I spent quite a long time upsolving C in coming up with the formulas to accelerate the procedure when $$$a>b$$$... I have no knowledge of gcd algorithms... how can it make the problem easier?
Alternative solution to F using SOS DP:
Notice that the value of $$$a_0$$$ after $$$k$$$ operations is the xor of all $$$a_i$$$ such that $$$\binom{k}{i}$$$ is odd.
Let's consider a binary mask $$$b$$$, where $$$a_0$$$ is the xor of all $$$a_i$$$ such that the $$$i$$$-th bit in $$$b$$$ is set. Initially $$$b=1$$$. An operation changes $$$b$$$ in the following way: $$$b \leftarrow b \oplus 2b$$$. This is equivalent to moving a row down in a Pascal's triangle that only considers parities.
Lucas's theorem states that $$$\binom{p}{q}$$$ is odd if and only if $$$q$$$ is a submask of $$$p$$$. Thus, the value of $$$a_0$$$ after $$$k$$$ operations is the xor of all $$$a_i$$$ such that $$$i$$$ is a submask of $$$k$$$. This can be computed in $$$O(n \log n)$$$ using SOS DP. We know that after $$$n$$$ operations all $$$a_i$$$ become zero, so the answer is the first moment that $$$a_0$$$ becomes zero and stays zero.
Submission
Yeah this was the solution I found (admittedly I didn't realize its truth was due to Lucas's Theorem). Let to a very beautiful implementation :)
F: Could the answer be -1? Why?
3 2 2 0
at this testcase isn't it -1?
I'm wondering how is one suppossed to come up with a proof such as one presented in tutorial E (or even think of proving the idea that number of odd divisors affects the answer)? Is that even Div2E level? The ideas used in the proof such as considering the parity of $$$cnt$$$ seem so random to me. I'd be grateful if any math gods would explain to me how to think about such problems or if you would show me more intuitive proof than the one in tutorial.
nvm