b , 3 Thanks in advance. maxb=\lfloor\sqrt x\rfloor +1, m Don't we have O(N^2) elements in map in worst case (if almost all subsegments have different sums) and get complexity O(N^3*log N)? This also means that the optimal value of $$$x$$$ is of order $$$\frac{\log a}{(\log \log a)^2}$$$. 1 2), Educational Codeforces Round 152 Editorial, Competitive Programming Roadmap (target: [gray, blue]), 1569C - Jury Meeting: Not able to calculate n! \lfloor\frac x {b+1}\rfloor 1 y Idea: Assume ab = a mod b = k Assume \lfloor \frac{a}{b}\rfloor=a \bmod b = kFalse set ba=amodb=k Then obviously k is smaller than b. a = k b + k = k (b + 1) a=k*b+k=k*(b+1)a=kb+k=k(b+1), k 2 < k ( b + 1 ) = a x k^{2} Is this the case? A pair of positive integers $$$(a,b)$$$ is called special if $$$\lfloor \frac{a}{b} \rfloor = a \bmod b$$$. 2) C. Floor and Mod+ Codeforces Round #701 (Div. ) a\ mod\ b=a-\lfloor \frac a b \rfloor*b Like binary search in C (not necessary though) , prefix sum in B , chess coloring in D. Also though i didn't read E,F they are tagged as DP and data structures. m Also for convenience, I will also define the mod mod operator to have lower precedence than addition or subtraction, thus ax + b mod m (ax + b) mod m a x + b mod m ( a x + b) mod m. = 50 3 53384 #include 1 a February 12, 2020 12:54. Can you optimize it to $$$O(n\log n)$$$? It's still called ternary search, even though you're splitting the search space in half. ( . 1 Ok, so i think we need to first put the value of $$$a$$$ and then solve for $$$b$$$ .But to justify for all $$$a<=1e9$$$ using this formula , we need to iterate for all values of $$$a$$$. b b You cant understand solution --> bad problem. Is there an upper bound for $$$k$$$? 1 x i 2 5 It was good if same comment was made separately. C. Make Equal With Mod | Codeforces CodeTon Round 1 Solutions | Explanation | C++ Code - YouTube Subscribe to the channel for Amazing Coding and Competitive Programming Content.Problem C:. Contribute to kantuni/Codeforces development by creating an account on GitHub. ba=amodb, 1 That is why we must take max(0, upper limit lower limit). #include b b You can replace all the missing values withing range [1, arr[right]). Iterate over the number of operations of type $$$2$$$. b = Since $$$b > k$$$, we have that $$$k^2 < kb+k = a \leq x$$$. O Reorganize the above conditions to get: b> k b>kb>k, 1 k ( b + 1 ) x = > 1 b x k 1 1 \leq k*(b+1) \leq x => 1 \leq b \leq \frac{x}{k}-1 1k(b+1)x=>1bkx1. If f(mid) = f(mid + 1) before the minima then I would get the wrong ans. I think I just understood it. o 1 a _Rehan's blog Codeforces Round #701 (Div. . a + b\le y 1 1+2+3+\cdots +(b-1) b\le y 4 3 xyab1<=a<=x1<=b<=yxyaba/ba, a=kb+k(0=b. You could define as shown here the more common way with always rounding downward or upward on the number line. i You should use operations of type $$$2$$$ first, then moves of type $$$1$$$. And subtract both the points multiplying by number of quotients. You will see that if we start with b and increment b till it reaches b+k ( k steps ), you will now need to integer divide a by b+k until a != 0. For calcuating the peak of quotients basic observation will be for random number x on b-axis having remainders [a>=(x+1)(x-1)]=>>(a=x^2-1) =>>a+1=x^2.This equation comes from X=(Y.r)+r is now x=(y+1)r. So min value of(sqrt(a+1),b). $$$dp$$$ is now a map such that $$$dp_j$$$ corresponds to the number of ways to create a hybrid prefix $$$[1, i]$$$ such that $$$\sum_{k=1}^{i} a_k - b_k = j$$$. 1 b 1 AC + bk(b+1)=k=amodbk(b+1)a, . Because if you use 30 steps for incrementing then you need one more step for dividing, making it totally 31 steps. . ( k Ciao, Codeforces! 1 b Let $$$y$$$ be the number of moves of type $$$2$$$; we can try all values of $$$y$$$ ($$$0 \leq y \leq 30$$$) and, for each $$$y$$$, check how many moves of type $$$1$$$ are necessary. 1 1 m = Thanks for the fast editorials! Update div-times-mod.cpp. 1 That's actually how I code ternary search if I'm not working with doubles (I actually did this solution in contest, so you can check my submission). b x , ACM, https://blog.csdn.net/fztsilly/article/details/113799195, https://codeforces.com/contest/1485/problem/C, Spring boot daoservicecontroller+model. 3 How many $$$k$$$-similar arrays can you make if none of the previous conditions holds? 2 a = 18^7 and b = 18, log(a) / log (b) returns 6.9999999999994 something. Are you sure that the f is a decreasing function ? time limit per test. Thanks in advance. The least common multiple of all integers from $$$1$$$ to $$$16$$$ is less than $$$10^6$$$. That's ternary search, not binary search, because you're looking at the slopes of adjacent elements. b = Hence $$$k^2$$$ $$$<$$$ $$$x$$$ or $$$k$$$ $$$<$$$ $$$\sqrt{x}$$$. (b1)(b+1)x 2 12345678 9 69 420 2) Let j <= i be the maximum index where sum(b[j] . i 5 Now let's count special pairs for any fixed $$$k$$$ ($$$1 \leq k \leq \sqrt x$$$). 1 1 1 1485C - Floor and Mod Authors: isaf27, TheScrasse Preparation: Kaey Hint 1 Hint 2 Solution Official solution: 107232416 1485D - Multiples and Power Differences Author: TheScrasse Preparation: MyK_00L Hint 1 a[i-1]), or a[0] + a[1] + a[2] + . #include x,y, 1 b a + 2), Educational Codeforces Round 152 Editorial, Competitive Programming Roadmap (target: [gray, blue]), 1569C - Jury Meeting: Not able to calculate n! 3 + I understood the O(n^2logn) solution of problem F. But I don't get how to get rid of first layer of dp. It returns the largest integer smaller than or equal to x. 1 i-1=\frac{a}{i+1},i=\sqrt{a+1} See think of the graph of quotients plotting on (b) axis. = ) { Here, a b a b is the result of the integer division between a a and b b, while a mod b a mod b is its remainder. P.S : Sorry I dont know how to use mathematical symbols. And answer is the summation of these two values. The difference between two adjacent cells is obviously a fourth power of an integer. amodb=abab Calculate the dp for all prefixes from left to right. 2 Let g[i] denote the number of hybrid arrays of length i and sum of the array is 0. i Here are video solutions to all problems, a challenge version of E, and a lesson on how to misread problems (of course, the last two things are totally unrelated). 0 1\le a\le x,\ 1\le b\le y i Here is the corrected implementation of the editorial approach, if it helps. x i Dp state is represented by $$$a$$$ & $$$b$$$. b ( It's so annoying to get WA for such a silly mistake, could've been a specialist for the first time mayb :( . a a + However, after step $$$2$$$, the coin on node $$$i$$$ may be either red or blue. + y Common examples: It wants to say that we should increment $$$b$$$ first as much we want and then do the divide operations . m x Never use someone else's code, read the tutorials or communicate with other person during a virtual contest. \lfloor\frac x {b+1}\rfloor \lt b-1 1 In the third test case, there are two special pairs: $$$(3, 2)$$$ and $$$(4, 3)$$$. b y, *t x x b maxb=x the same algo implementation might be passing in C but not in Python. \red{b-1}0 i + ) b Link to the problem: https://codeforces.com/contest/1485/problem/CLink to the solution: https://codeforces.com/contest/1485/submission/107232746Practice problem with similar approach: https://www.hackerearth.com/problem/algorithm/floor-sum-b4fa5319-89c96876/description/My Codeforces profile: https://codeforces.com/profile/kassuttaTimestamps:00:00 : Problem statement00:40 : Basic logic02:33 : How I approached09:38 : Note-111:29 : Note-212:50 : Brute Force14:34 : Main approach16:02 : Solved example using optimal approach19:34 : Code Run-through10% Discount on any GFG course, Use Code: MIHIRDSA self-paced: https://practice.geeksforgeeks.org/courses/dsa-self-pacedLink to all courses: https://practice.geeksforgeeks.org/courses/onlineFor any queries: contact us at cp.gatecs@gmail.comFollow us on--@discord:- https://discord.com/invite/6kcmVQx8Fx @facebook:- https://www.facebook.com/Quarantine-Engineers-107789377679897/@linkedin :- https://www.linkedin.com/company/quarantine-engineers@instagram :-https://www.instagram.com/quarantine_engineers/@telegram(Channel) :- https://t.me/quarantine_engineers@telegram(Discussion Forum) :- https://t.me/qe_discussion********************** credits ************************Edited by - mayanksneh60@gmail.com (instagram - mayank_sneh )graphic templates: mixkit.coaudio templates : mixkit.cotext templates: mixkit.co#Codeforces #Div2 #Round701 PM PM CS Btw, a function is unimodal if it first increases, then decreases (or vice versa). To calculate $$$dp2_i$$$, we have one more observation that, when we do operation of type (1), $$$b[i]$$$ adds to the previous prefix_sum and for type (2), prefix_sum becomes $$$b[i]$$$. b b 141 Looking forward to more contests from the authors. codeforces 1485 D Multiples and Power Differences . b a If $$$x < a_l$$$, you can replace $$$a_l$$$ with $$$x$$$ (and you get $$$1$$$ $$$k$$$-similar array). 4 So b we can increase up to up to 25 or so, but the editorial gives a better limit. . Very nice problems. maxb+1, 3-Coloring, [Codeforces] Global Round 13 C. Pekora and Trampoline, https://songhayoung.github.io/2023/01/26/PS/Codeforces/div2-701-c/. Let $$$dp_{i, j}$$$ be the number of hybrid prefixes of length $$$i$$$ and sum $$$j$$$. Follow Me. + . I got it my solution is wrong because f is monotonic. = + a Is it O(sqrt(x))? Practice Ladders. P.S. 2 b + ) How do I get blue in codeforces in 1 month, 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. + In EDITORIAL Therefore, for any fixed k>0, the number of special pairs (ax; by) is max(0,min(y,x/k1)k). a b by m 1 OR. b bmin, b = 2]. 1 . 1 < a Hello. x The total count is $$$(a_l-1)+(k-a_r)+2((a_r - a_l + 1) - (r - l + 1))$$$, which simplifies to $$$k + (a_r - a_l + 1) - 2(r - l + 1)$$$. #include 2 b 9 Yep, F is monotonic. a a Solve more problems and we will show you more here! This gives $$$dp2_i = dp1_{x-1}$$$. b For C I have good solution it don't requires math just logical binary search. b If f was monotonic, it would mean that looking at a single value, it would be possible to determine whether to go to the right or to the left. a Specifically, why it's $$$2((a_r-a_l+1) - (r-l+1))$$$. a[i-1]) Now there will be over-counting when b[i] = b[i] (a[0] + a[1] + a[2] + . 9 all that will just to get the answer for every number and take it by your hands. There may be very few possible values of $$$b_{i,j}$$$, if $$$b_{i-1,j}$$$ is fixed. 0 Thus, $$$6$$$ is the upper-bound for this problem since $$$a$$$ can be at most $$$10^9$$$. = 1x,y109, There are $$$a_r - a_l + 1$$$ integers between $$$a_l$$$ and $$$a_r$$$. The possible different sums for each length are $$$O(n)$$$, because they only depend on the rightmost $$$i$$$ such that $$$b_i = \sum_{k=1}^{i} a_k$$$. i ( Then we take a unique subarray of $$$x-1$$$ elements, do a type 2 operation for x and do type 1 operation for elements in range $$$(x+1,i-1)$$$. Contribute to soumik9876/Codeforces-solutions development by creating an account on GitHub. Codeforces Round #701 (Div. Because I think you are iterating from sqrt(x) to x, which I think is O(x). That example, is different, as you're not really searching on a function. \frac{k(b+1)}{b}=k=a\;mod\;b\red{k(b+1)\leq a}, 1 I don't have a formal mathematical proof to state why this works, but it does. ab=k\frac{a}{b}=kba=k I observed that we dont need to go very far away as that would increase k by very much but l wont be affected that much. ) using namespace std; b 1 The upper-bound could vary accordingly to $$$a$$$. Let's look at the four inequalities required for solving this question: 1 <= k(b + 1) <= x (derived in the editorial)Modifying equation 3, we get, (Why k != 0? Arguing if my approach is binary search or ternary search is self-contradictory. The transitions are $$$dp_{i, j} \rightarrow dp_{i+1, j+b_i}$$$ and $$$dp_{i,j} \rightarrow dp_{i+1,b_i}$$$. Specifically in the editorial . 1 t Since k is the remainder, we 0 <= k < b by Euclid's Division Lemma or b > k. Can you please explain why k is always less than or equal to sqrt(x), $$$0$$$ $$$<=$$$ $$$k$$$ $$$<$$$ $$$b$$$ (By Euclid's Division Lemma), $$$=>$$$ $$$k^2$$$ $$$<$$$ $$$k*b + k$$$ (Since $$$k$$$ is non negative, the RHS sum can remain same or increase but never decrease). Can anyone help me to improve the complexity of my solution. b+1 of b, because can't there be different a's for a fixed b and a fixed k? :,., b >= x,min(x, b * b + b - 1)x,.,,,,.,,.,,,. input (b-1)*(b+1)\le x 1 \sum_{i=2}^bmin(i-1,\frac{a}{i+1}) b a I should've mentioned it Divide and conquer in general. I also didnt get how 30 comesand how should I even realize that:( Bad Problem. When it gets to the point where the current solution is worse than the previous one, then it stops the loop and returns the previous answer. b1, #include b b (a mod c - b mod c) mod c = (9 mod 4 - 6 mod 4) mod 4 = (1 - 2) mod 4 = -1 mod 4 = 3 Note: -1 mod 4 = 3 not -1. + + (b1)(b+1)x Vasya will consequently put the rooks on the board one after another. + + 9 If you use Java/Kotlin, the standard library function Math.floorMod (n, m) does what we need. x,y b+1 b a (b-1)*(b+1), x #include solution 1 + Div. 1 b The left part is strictly less than $$$0$$$ for $$$x \leqslant 6$$$ and it becomes positive for $$$x \geqslant 7$$$, where $$$x \in N$$$.
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