Consider all the possible pairs of (i, j) that may be different which would result in a +1 in the answer, minus the pairs such that a[i] = a[j] and you will get the pairs that a[i] != a[j], you can calculate the a[i] = a[j] part using a vector and binary search, and the total number is the same as (n k + 1) * (k / 2), Here is my code, maybe looking at it is a lot more simple R199691512, Nice round! Awesome work. x1 + y1 <= 5, x2 + y2 <= 20, so x1 + x2 + y1 + y2 <= 25. Then, while $$$dp[60-i] + dp[60] \leq m$$$, I add one copy of $$$i$$$ to the solution. (By the comment it seems like some runtime optimization trick but I am not sure). For example, you can try making $$$a_{i+1}$$$ bigger using a positive element. Thanks to all the authors for a cool contest! The answer is $$$x-1$$$. Wind_Eagle Why not make $$$N$$$ $$$K$$$ larger, like $$$300 000$$$ in $$$D$$$? But I request you not to spoil the problem or give hints to the problem in your blog titles. And when the number of negatives is 13, positives 7, the first strategy is better, taking 19+5+7=31 moves!. Codeforces Round #861 (Div. 2) - O(log N) solution for E + full video Block $$$x$$$ disappears when it reaches block $$$x+1$$$. D: Denote r=(k-1)/2. Sure, on practice you can optimize quadratic in 32 or 64 times, but it's still quadratic. $$$dp_{i,j}$$$ is similar to knapsack $$$dp$$$. Then, either $$$k\le 6$$$ or $$$(a, b, c, d - k) \in A_{k-1}$$$. I can guarantee that this kind of thing will not happen again, if it does, the administrator can shut down my account. because it's guaranteed that you can get a number with final two digits '90', ohh yeah now i undertood that, I was thinking something else the whole time. I and every other cper started like this. Checking a given base set takes $$$O(60^2)$$$ time. So does anyone have a explanation for this question now? 2), which will start on 29.07.2023 17:35 ( ).You will be given 6 problems and 2 hours and 30 minutes to solve them in both divisions.. One of the problems will be divided into two subtasks. How to do fractional cascading on an iterative segment tree? Let's calculate the expected time required for the block in row $$$x$$$ to "reach" the block in row $$$x+1$$$. 6 A. So, the final value is for each two values inside the array our value is: k1 i=0((ith bit of a and b equals to each other) i) i = 0 k 1 ( ( ith bit of a and b equals to each other) i). there exists some cubic solutions (like mine) which they wanted to pass as well ig, can anyone help me debug this. Why not Div. Thanks! Use $$$c_a$$$ times of $$$a$$$ and $$$c_b$$$ times of $$$b$$$, where $$$1\leq a x) swap(x, y); return x y; } void solve() { int n, m; cin >> n >> m; vector v(m); // given input: // 4 3 // 1 2 3 // 3 2 1 // 1 2 1 // 4 2 7. Solution. Assume by contradiction that such inequality is false. I'm a bit late, but I can compensate that by having a super-solution for problem E, which is faster than the authors' and also easier in my opinion. But correct answer is $$$2344$$$. 2), Invitation to SmallForces Monthly Contest #3, Codeforces Round 887 (Div 1, Div 2) Tutorial. There won't be much candidates for the answer so you can check all of them. Notice the unusual starting time. For C,I code for an hour but Failed at Pretest.Maybe D cost less time on coding than C. First I appreciate all the time and efforts for all the contributors to this round for the interesting problems. yea its a log of the int size, which is a constant, so its still a constant factor. If the answer is not a multiple of $$$x$$$, the answer is $$$< x$$$. So, in other words, knapsack $$$dp$$$ can be written as $$$dp_i |= dp_{i-a_j}$$$, where we iterate $$$j$$$ from $$$0$$$ to $$$n - 1$$$. If the maxEle < abs(minEle), then we increase the maxEle by doing the operation {maxIdx,maxIdx} while maxEle < abs(minEle). $$$w$$$ is the system word length (basically, the number of bits the processor stores in one unit; generally either 32 or 64). 2) +62 4 months ago SundryVictory +26 216392728 Can anyone explain the output of 7th test? 2) O(log N) solution for E + full video editorial. Agree! Is this solution worth rating 2800? Better to increase limitation for $$$N$$$ and $$$K$$$ to $$$300000$$$, because changing TL is linear function, but changing of limitations is quadratic function. I have a similar solution but with a segment tree instead of BS. First, there is something wrong in your if (n == 2). We'll do the same if mx is from a negative value, but the other way around. To count the number of lucky numbers whose first special digit is $$$a_n$$$, notice that if the first $$$n-1$$$ digits do not contain $$$a_n$$$, we can decrease each of the first $$$n-1$$$ digits by $$$a_n$$$, so that the first $$$n-1$$$ digits don't contain 0 and sum to $$$(2-n)a_n$$$ (all operations are mod $$$k$$$, of course). Maybe change the allowed runtime in problem D to 1 second? 2), Invitation to SmallForces Monthly Contest #3, Codeforces Round 887 (Div 1, Div 2) Tutorial. (with different ID's) .I created the second handle today(29/03/2023) and saved the handle on my PC.Then I registered for Codeforces Round 861(Div.2) and solved the A and B qns with id's "199686950" and "199683280" ;without realising that I was on the 2nd(KiritsuToKetsui) handle,and when I did recognise it I logged out and hastily submitted the same back in (kewang_zhili) handle.BUT I FORGOT THAT THERE WOULD BE A PLAG CHECK.Kindly excuse me for this time,as it was a genuine innocence,which shall not be repeated again @MikeMirzayanov. Suppose you find a valid interval $$$[l, r]$$$. (in string format). (fwiw, I think this and the last round are maybe a reason to have Div. if we choose digits_l[d] < digits[d] < digits_r[d], then obviously just greedily fill the rest with digits[d]. Plus, 10^5 is usually intended to be solved in at most $$$N\sqrt N$$$, not quadratic time. 2), Invitation to SmallForces Monthly Contest #3, Codeforces Round 887 (Div 1, Div 2) Tutorial. We know that $$$n$$$ is a multiple of each of $$$1, 2, \dots, x$$$, according to Fact 1 (and this is a sufficient condition for such an $$$n$$$ to exist with answer $$$x$$$). Preliminary score distribution: 750-1000-1500-2000-2500-3250. I can only come up with O(k+log(n)) solution. Can you win in $$$34$$$ moves? A Lucky Numbers. There are $$$\gcd(n-2,k)$$$ possible values of $$$a_n$$$ that fall into the first case, and others fall into the second case, so we can add them to $$$\frac{k^n-(k-1)^n}{k-(k-1)}-(k-1)^{n-1}$$$ to get the answer. 2), Invitation to SmallForces Monthly Contest #3, Codeforces Round 887 (Div 1, Div 2) Tutorial. The author is sslotin. If someone is interested, I've made a video, where I accessibly explain this formula (as well as all other problems): https://www.youtube.com/watch?v=i1i_7lqnGwA, https://codeforces.com/contest/1808/submission/199905893. In problem C, many solutions are giving wrong output on the test case 1 1000000000000000000 1000000000000000000 such corner test cases should have been there. Because $$$x>30$$$, we can't include $$$2$$$ latter type numbers into one set, so the total count will be trivial to calculate. So, it's enough to check intervals with $$$l = 1$$$, i.e., find the smallest $$$x$$$ that does not divide $$$n$$$. seems that there exists $$$O(k)$$$ and $$$O(\log n)$$$ solutions for E $$$O(k)$$$ solution: https://codeforces.com/contest/1808/submission/199658264, $$$O(\log n)$$$ solution: https://codeforces.com/contest/1808/submission/199673882, problem A the way luckiness is defined wouldn't the luckiness of 9 be 0, In problem D, I was trying an approach similar to https://atcoder.jp/contests/abc290/tasks/abc290_e with some modifications but I cannot understand what's going wrong. I'm a bit late, but I can compensate that by having a super-solution for problem E, which is faster than the authors' and also easier in my opinion. The only programming contests Web 2.0 platform, Competitive Programming Roadmap (target: [gray, blue]), 1569C - Jury Meeting: Not able to calculate n! TheScrasse Codeforces Round 889 (Div. I am not getting from Q statement & sample input/output. I can see my rating graph in the picture of picture :), I think my new dp predicts my future rating graph. Then, it's optimal to unlock $$$x \leq 2n$$$ cards, and use moves of type $$$1, \dots, x$$$, getting $$$a_1 + \dots + a_x - x + 1$$$ points. Let's solve task for $$$m=1$$$. 625 millions operations in predictable for GCC order it is not so much less than second. Official implementation: will be posted after the . In particular, this trick guarantees $$$|A_k|\le 5000$$$ for all $$$0\le k\le 62$$$. The choice of $$$a_1, \dots, a_{h}$$$ may seem arbitrary; let us try to justify it. curr = number formed finally after using different digits in recursion. my solution. Now you can simulate the process in both cases (positive and negative), and choose one that requires $$$\leq 31$$$ moves in total. With such $$$k$$$ this solution requires $$$\dfrac{n^2}{8} = \left(\dfrac{\left(n-k\right)\cdot k}{2}\right)$$$ comparisons. Here we need to take first a few closing brackets (x) and then a few opening (y). Submission: 216320776. 1 AC), Codeforces Round #854 by cybercats D1 + D2 PROCESS DP (video editorial), Educational Codeforces Round 115 Editorial, Educational Codeforces Round 116 Editorial. This observation explains why we start with approximately $$$\log_2(m)$$$ ones. Can you find other nodes faster? Upset when reading editorial :(. In this case we have one array and we need to calculate sum of absolute difference with every pair of elements. Please be considerate for the poor people who have this one at 5AM :), He is doing an exchange semester in Australia, For me who live in Korea, Both contests are good 09:05 UTC -> 18:05 GMT+9 14:35 UTC -> 23:35 GMT+9. We are with you. What's the answer if $$$n$$$ is not a multiple of $$$3$$$? By the way, if any author of a problem whose solution uses bitsets were retarded, then you are offending someone unexpected. Is the final stage of Belorussian National Olympiad so easy? So, now the numbers are of a equal lenght. Case 3 -> If there are some +ve and some -ve elements, We first try to convert all the -ve elements to +ve. 2) By dbsic211, 19 months ago, Thanks for participating, hope you enjoyed the problems! You cannot assume a bound for n when calculating time complexity. Why are the constraints on E3 so low? p.s. Given $$$(a, b, c, d)$$$, the instance of the problem is equivalent if we change the signs of the coordinates or we change the order of the coordinates. Number of indexes where a[i] == b[i] is the popcount from the bitmask of result of element-wise comparison. A: For any 100 consecutive numbers, there must be a number end with "09" and it will reach the maximum luckiness (9). No worries, thanks for the response and for preparing the contest! How do I get blue in codeforces in 1 month, Educational Codeforces Round 152 Editorial, UNIQUE VISION Programming Contest 2023 Summer(AtCoder Beginner Contest 312) Announcement, 2022-2023 Southern And Volga Russian Regional - Editorial, Teams going to ICPC WF 2022 (Egypt 2023) WIP List. https://codeforces.com/contest/1808/submission/199867234. Assume by contradiction that $$$|a|+|b|+|c|+|d-k|> 1 + 2 + \cdots + k-1$$$. i solved it after the contest end, i fixed every possible smallest and largest digit and then i took some digits from this range so that it fits in the range from L to R. I assume it is k^3 * logn, because limits seems to dictate that, yet my solution of k^3 logn had to be optimized several times before fitting in TL with a pragma, and finally getting uphacked :(. Bitsets have already been used in problems. Here is a similar atcoder problem, its editorial is amazing. One of the problems will be interactive, so please read the guide for . We're glad to invite you to take part in Codeforces Round 889 (Div. If $$$dp_{i,i} = 1$$$, you can update the answer with $$$a_1 + \dots + a_i - i + 1$$$, but you can't unlock any other card, so you have to set $$$dp_{i,i} = 0$$$ before transitioning to $$$i+1$$$. If you know the whole cycle (of length $$$x$$$), you can win in $$$n-x$$$ queries by asking for each node if it ends up in the cycle after $$$n$$$ moves. But, if we are, we may only add element from $$$[0, a]$$$, if our current anwser is the prefix of $$$r$$$, and $$$[b, 9]$$$ if it is of $$$l$$$. I think that it can be used only with lightweight operations under subsegments of array. Could anyone hack it or explain why it's correct? Wait somehow C is 1900 and D is 2100. The k even case is similar, only that now if $$$a_i$$$ is the first special digit, then $$$a_i+\frac{k}{2}$$$ also cannot appear before $$$a_i$$$. Then, we will enumerate all lucky numbers by the location of their first special digit. Longest interval always be under 44? How to update it? Else if digit is greater than or equal to maxi we will move with same digits (I=2 and a=6) Suppose l= 276814 So corresponding no will be 276888. That means that if you apply bitsets, you are guaranteed to at least divide the runtime by a $$$\log(n)$$$ factor. These components move equiprobably. With $$$n \leq 10^5$$$, we have $$$\frac{n^2}{32} \approx 3 \cdot 10^8$$$, suggesting that we're fine given the 3s time limit. This strategy requires $$$5 + 2(n-1) \leq 43$$$ moves. Codeforces Round 887 (Div 1, Div 2) Tutorial - Codeforces Can anyone prove it. Does the picture imply that this round will be the next big milestone in my downward trend in rating? Awesome work. If you either make everything positive or make everything negative, you can win in $$$n-1 \leq 19$$$ moves by making prefix sums or suffix sums, respectively. 2], I think I just did something crazy? 2) - O (log N) solution for E + full video editorial By SirRembocodina , history , 4 months ago , Hello, I've made a full video editorial for the latest CF round. : (div2 hard C). This round motivated me to solve more problems ;) How to do A? Problem 1855B could also be solved with Binary Search, I have no idea how it passed the pre-tests, but here is some code if anyone is interested. One can verify that to handle correctly all points with coordinates up to $$$1000$$$ it is necessary to compute $$$A_k$$$ for $$$0\le k \le 62$$$. Our main author solution on Polygon is exactly this O(log) solution with formula. Let $$$s$$$ be the number of sad students at the beginning. So we keep a track of maxi,mini of digits till i-1 and these digits will surely be in ans. (with different ID's) .I created the second handle today(29/03/2023) and saved the handle on my PC.Then I registered for Codeforces Round 861(Div.2) and solved the A and B qns with id's "199686950" and "199683280" ;without realising that I was on the 2nd(KiritsuToKetsui) handle,and when I did recognise it I logged out and hastily submitted the same back in (kewang_zhili) handle.BUT I FORGOT THAT THERE WOULD BE A PLAG CHECK.Kindly excuse me for this time,as it was a genuine innocence,which shall not be repeated again @MikeMirzayanov, Here is my code of problem B, but it got wrong in the pretest3. We expect to see many different solutions for this problem. WeaponizedAutist Perhaps you should give up . Will it be harder than Div 2 usually that we give? In most cases, you can make $$$2$$$ sad students happy in $$$1$$$ move. Feels like D is actually much easier. Codeforces Round #861 (Div. 2) - O(log N) solution for E + full video Fact 2: We can show that the answer never exceeds $$$42$$$ (for $$$n \le 10^{18}$$$).
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