### chokudai's blog

By chokudai, history, 4 years ago,

We will hold AtCoder Beginner Contest 155.

We are looking forward to your participation!

• +60

| Write comment?
 » 4 years ago, # |   0 I hope this one can be a little easier.
 » 4 years ago, # |   -9 E is similar to this problem.
•  » » 4 years ago, # ^ |   -46 wish ,i had visited this comment before the contest ended ...
•  » » » 4 years ago, # ^ |   +39 Sakhiya07 You should wait for contest to end before pasting source of problem. Its just ABC and I expect repeated problems for educational purpose. wish ,i had visited this comment before the contest ended ... forget_it What would you get by copy pasting? I challenge you to copy paste one of that GYM's AC submission now and get AC. If you have access to that problem's submissions.
•  » » » » 4 years ago, # ^ |   0 What type of challenge you want me too complete? I didn't understand...
•  » » » » » 4 years ago, # ^ |   +1 Open one of AC submission of this problem (made before start of today's atcoder contest) and submit it on atcoder.
•  » » » 4 years ago, # ^ | ← Rev. 2 →   -35 Stop Disliking my comment
•  » » » » 4 years ago, # ^ | ← Rev. 2 →   0 no, because all you care about is your score and not what you learn
 » 4 years ago, # |   0 How to solve E and F ?
 » 4 years ago, # |   0 How to Solve D ?
•  » » 4 years ago, # ^ |   +15 Binary search (from -1e18 to 1e18). Then, once you set the guess for K, do another binary search to find out how many pairs are less than the guess.Final runtime: O(N * (logN)^2)
•  » » » 4 years ago, # ^ |   +10 You can do the checking process by two pointers, I think it is easier to code and faster.
•  » » » » 4 years ago, # ^ |   0 How can you solve it with two pointers?
•  » » » » » 4 years ago, # ^ |   0
 » 4 years ago, # |   +67
•  » » 4 years ago, # ^ |   +30 It must be: Difficulty
•  » » » 4 years ago, # ^ |   0 I agree. I expressed a bit simply. :)
 » 4 years ago, # |   +12 Contest is over!Difficulty on D was misplaced. Implementation wise, it was much harder than E.
 » 4 years ago, # |   +6 This round really made me autistic(cross out) self-close
 » 4 years ago, # | ← Rev. 4 →   +4 I guess this is one of the difficult ABC to me. (even though I got better rank than other contests just by solving A-B-C ).
 » 4 years ago, # | ← Rev. 2 →   0 Can anyone tell me the mistake I did in D, the first two samples are working and out of 72 TCs, almost half of them are working. I did a Binary Search from -10^18 to 10^18 and found out how many products will be less than the mid, and accordingly shift the high and low by comparing with k. https://atcoder.jp/contests/abc155/submissions/10161880 Also, do we require BigInt in E or is it doable without them? If yes, then can anyone tell me how?
•  » » 4 years ago, # ^ |   +1 We can do E without using BigInt too.
•  » » 4 years ago, # ^ |   +1 E doesn't need BigInt. In fact, I recommend you solve the problem by turning the String into an array of integers, then doing an O(N) sweep from right to left).
•  » » 4 years ago, # ^ |   +1 I didn't find the bug but I have some suggestions. You don't need long doubles. If a is a integer, a <= x is the same as a <= floor(x) and a >= x is the same as a >= ceil(x). You just have to be careful with negative values. I found it easier to compute 2*(the number of pairs) and then divide it by 2 instead of just compute it directly. I didn't understand why you count the number pairs with equal and smaller products in two different variables and use them differently. Shouldn't you always use their sum? I coded the same solution. https://atcoder.jp/contests/abc155/submissions/10145988
•  » » » 4 years ago, # ^ |   0 Thank you for these suggestions, and for having a look at my code. I was coding it without using doubles but I was having a difficulty as some point so I thought it would be easier using doubles. I thought of calculating both values because I thought that if kth product was not unique then I might miss it because if I calculate all the products lesser than mid, it would be less than k and all the products less than and equal to mid would be greater than k. I couldn't come up with a justification why their sum would be used, can you help me with that too?
•  » » » » 4 years ago, # ^ |   +1 Let f(x) = number of pairs with product <= x. The answer is the smallest X such that f(X) >= k.To have a intuition I think its enough to look at the first sample. $f(-13) = 0, f(-12) = 2$. So the answer for $k = 1$ is the same as $k = 2$ (-12).$f(-7) = 2, f(-6) = 4$, so the answer for $k = 3$ is the same as $k = 4$(-6).$f(7) = 4, f(8) = 5$, so the answer for $k = 5$ is 8.There is no need to compute < x and = x, just <= x.
•  » » » » » 4 years ago, # ^ |   0 Thank you so much for this clear explanation, I would incorporate these changes in my code, and will try to get an AC.
•  » » » 4 years ago, # ^ |   0 Hi Nson, could you please explain the div_floor and div_ceil functions as well as why do you need them?
•  » » » » 4 years ago, # ^ |   +1 div_floor($a$, $b$) = $\lfloor \frac{a}{b} \rfloor$ and div_ceil($a$, $b$) = $\lceil \frac{a}{b} \rceil$.I wanted to count the number of elements $x$ such that $a_i * x \leq lim$, this is: $x \leq \frac{lim}{a_i}$, if $a_i > 0$ $x \geq \frac{lim}{a_i}$, if $a_i < 0$ And then I used floor and ceil to only work with integers.
•  » » » » » 4 years ago, # ^ | ← Rev. 2 →   0 I see! Any particular reason why you coded these functions (floor, ceil) yourself and not used the ones from std?
•  » » » » » » 4 years ago, # ^ |   0 To use the floor or ceil I would have to compute $\frac{lim}{a_i}$ using doubles and that's what I was trying to avoid.I always don't use floating-point numbers when I can to avoid any precision issues.
•  » » » » 4 years ago, # ^ |   0 Also could you please explain where does the -1 come from in the following two lines: if(i < id) ans += id - 1; else ans += id; and if(i >= id) ans += n - id - 1; else ans += n - id;
•  » » » » » 4 years ago, # ^ |   +1 For $a_i > 0$ I want to count the number of $a_j$ such that $a_j \leq \frac{lim}{a_i}$.As the array $a$ is sorted, this is a prefix of it and with upper_bound I can find the size of this prefix. There is just one problem, if $i$ is in this prefix then I don't want to count the pair $(i, i)$ at the answer, but its enough to just remove 1 element(the $i$) from this prefix.For $a_i < 0$ it's basically the same thing, but the answer is a suffix of $a$.
 » 4 years ago, # | ← Rev. 2 →   0 Can E be solved using DP? If yes DP transitions, please :)
•  » » 4 years ago, # ^ |   0 The solution for E is greedy. Technically you could use DP, but there is no point. At every state there is only one move you should do.
•  » » 4 years ago, # ^ |   0 Hint:"Simulate the subtraction process"
•  » » » 4 years ago, # ^ |   0 And digit dp.
•  » » 4 years ago, # ^ |   0 Yes, a simple N*2 digit dp would do
•  » » » 4 years ago, # ^ |   0 Can you please elaborate ?
•  » » 4 years ago, # ^ | ← Rev. 2 →   +4 dp[idx][carry] is the minimum number of banknotes we (you and the cashier) used after processing all numbers starting from 0 to idx-1, such that we have carry = 1 or 0So, we start from the right, and decrease the idx as we go through the number N, the transition will look like this:dp[idx][carry] = min( dp[idx-1][0] + N[idx] , dp[idx-1][1] + (10 — N[idx]) )corner case: the first option can't be done when N[idx]==9 and carry==1in the base case we should return carry value
•  » » » 4 years ago, # ^ |   0 Hi,Can you/someone please elaborate solution a bit more? As in what does dp state denote (what does carry mean?). What is the intuition/logic behind it and what are we trying to do? Sorry, my peanut brain is unable to understand it. Any help would be appreciated.
•  » » » » 4 years ago, # ^ |   +3 You can check this code.Code
•  » » » » » 4 years ago, # ^ |   0 Can you please explain what does your dp state denote? And what does dp transitions denote?
•  » » » » » » 4 years ago, # ^ | ← Rev. 2 →   +3 dp(i,0)-no. of notes if the transaction is exact(it means that you don't have any carry or in simple words you paid the exact amount upto i)dp(i,1)-no. of notes if you gave one extra noteNow since at ith position banker has option to give you the remaining amount which was carried.So dp(i,0)=min(dp(i-1,0)+num(i),dp(i-1,1)+10-num(i))for other state consider giving more than required to the banker so we add one note.So dp(i,1)=min(dp(i-1,0)+1+num(i),dp(i-1,1)+10-num(i)-1).
•  » » » » » » » 4 years ago, # ^ | ← Rev. 7 →   0 Hi,Am sorry to bug you again. I thought I understood, but no. So for example, we are trying to solve for ...689 and let i be at '6'. dp[i,0] (689) = 6*100 (num[i]) + 89(dp[i-1, 0]) =. Can you write the second state...? dp[i,1] (689) = ...?
•  » » » » » » » 4 years ago, # ^ |   0 Does 'one note' here refer to a note with value=1. Or can the value be anything? If it's the latter, then in our transition state, why have we used dp[i][0]=min(dp[i-1][0]+num[i], dp[i-1][1]+10-num[i]);
 » 4 years ago, # |   0 Can someone tell me the exact solution of D?
•  » » 4 years ago, # ^ |   0 Binary search from -1e18 to +1e18. Then you take a guess of k and perform binary search for the number of numbers below that number. I would say this question was harder than E. ;_;
•  » » » 4 years ago, # ^ | ← Rev. 2 →   0 Are those, wordings of Agnimandur ?
•  » » » » 4 years ago, # ^ | ← Rev. 2 →   +7 For those Who still Didn't uunderstand D.Agnimandur Explained in CPCbasically split the input into the positive numbers and the negative numbers, and then sort them. call the positive array POS, and the negative array array NEG.then binary search from -1e18 to 1e18. let the "guess" of the binary search be G. if (G < 0), then iterate over all of the negative numbers. For each negative number, do a binary search over all the positive numbers, and find the largest positive number such that the product is <= G. If G > 0, then all of the negative * positives work. Then, iterate over NEG and at each index i do another binary search from i+1 to the end to find the number of pairs of negative numbers that have a product less than G. Do the same thing over POS.In total, after you iterate over the lists, if you've found X pairs less than G, and X is less than K, then G must be too small. Otherwise it works, and you transition the binary search accordinglyhttps://atcoder.jp/contests/abc155/submissions/10157539
•  » » » » » 4 years ago, # ^ | ← Rev. 2 →   0 thx, well explained!
•  » » » » » 4 years ago, # ^ |   0 Why for G -ve we have to check product<=G and for +ve we want product
•  » » » » » » 4 years ago, # ^ |   0 you wanna count all pair such that their product is below their G to validate whether G is the answer or not.
 » 4 years ago, # |   +3 Why doesn't ABC get updated on Atcoder calendar anymore? I rely on the calendar for contest participations and I've missed like all the ABCs recently because of it.
•  » » 4 years ago, # ^ |   0 Same here
•  » » 4 years ago, # ^ |   +4 use this insteadgoole canlendar link
•  » » 4 years ago, # ^ |   0 I suggest you use clist
 » 4 years ago, # |   0 what is wrong in My E?https://atcoder.jp/contests/abc155/submissions/10162814TIA
•  » » 4 years ago, # ^ |   0 Try 5.
 » 4 years ago, # |   +3 How to solve F?
•  » » 4 years ago, # ^ |   +34 Sort the A_i, so every operation concerns a subsegment of indices. Let D_i be the xor between B_i and B_{i-1}, where we consider that B_0 = 0. Our objective is to make D[] = 0. Notice that now every operation amounts to flipping the state of at most two elements of D.Build a graph on 1..n where indices i and j have an edge between them if an operation affects both. Notice that if we have a simple cycle x_1, x_2, ..., x_k then flipping the edge from x_1 to x_k has the same effect as flipping the other k-1 edges, so it's enough to take a spanning tree of each component. Now we can root each tree arbitrarily and greedily go from leaf to root, doing the correct operations to do everything except possibly the root equal to 0. If at the end the root is 0 we have to flip a single element in the tree in order to fix it. If no operation does that then it is impossible.
•  » » » 4 years ago, # ^ |   0 Thank you !
•  » » » 4 years ago, # ^ |   0 Thank you !
 » 4 years ago, # | ← Rev. 2 →   +2 Was this streamed yesterday ? They are solutions to today's contest . It shows to be streamed on 15th Feb 2020 https://www.youtube.com/watch?v=SG60Cp9pSog is this Scam ?
 » 4 years ago, # |   +9 Why cant they provide editorial in english when the majority is not from japan :/
•  » » 4 years ago, # ^ |   0 maybe the majority of the problem writers are japanese
 » 4 years ago, # | ← Rev. 2 →   0 https://atcoder.jp/contests/abc155/submissions/10169630Can anybody explain what's wrong with this submission for problem E?
•  » » 4 years ago, # ^ |   0 In the case 955 I believe the answer is 10(955, 956, 957, 958, 959, 960, 970, 980, 990, 1000, 0) but your code outputs 11.
•  » » » 4 years ago, # ^ |   0 Yeah, i got it thx man.
 » 4 years ago, # |   0 I'm much weaker than you all, could you tell me how to solve Problem C?
•  » » 4 years ago, # ^ | ← Rev. 2 →   0 Count the occurrence of each string then output all those strings which have highest number of occurrences. For counting I used map, and in order to output in lexicographical order I sort them using vector of string.Here is my solution
•  » » » 4 years ago, # ^ |   0 Why do you need to sort in lexicographical order..? Map sorts the strings and stores the count of each string right? You do not need any vector to sort again...
•  » » » » 4 years ago, # ^ |   0 YES. I know that, but in the moment I couldn't think of it. I was in hurry.
 » 4 years ago, # |   +3 when will the English Solution be published?`\
 » 4 years ago, # | ← Rev. 2 →   0 https://atcoder.jp/contests/abc155/submissions/10152047I appreciate if someone tells me what is wrong with this submission. It works fine with small tests, but gives wrong answer for the long one:For the test 314159265358979323846264338327950288419716939937551058209749445923078164062862089986280348253421170 it returns 249 instead of 243.the code is this: int main() { string s; cin >> s; int sum = 0; int last = 0; for(int i = s.length() - 1; i >= 0; i--) { int current = s[i] - '0'; if(last) current++; if(current > 5) { current = 10 - current; last = 1; } else last = 0; sum += current; } sum += last; cout << sum << endl; return 0; }
•  » » 4 years ago, # ^ |   +3 Your solution for test '555' gives answer 15 instead of 14 (1000 — 400 — 40 — 5)
•  » » 4 years ago, # ^ | ← Rev. 2 →   0 Greedy approach does not work for this problem, for example if N=555, your program outputs 15, but you can obtain 14 (1*1000 — 4*100 — 4*10 -5*1) wish comments would auto refresh, so this does not happen when I try to answer someone.
•  » » » 4 years ago, # ^ |   0 I think this https://atcoder.jp/contests/abc155/submissions/10180380 is greedy one. Isnt?
 » 4 years ago, # |   0 please write the solutions in English too and provide setters/editorial writers code along with it. giving contests without having access to proper editorials is kinda frustrating.
•  » » 4 years ago, # ^ |   +8 "For International Readers: English editorial will be published in a few days" They're doing a great work, give them some time. Also often you can understand the solution if you copy-paste the editorial into an automatic translator and read that.
 » 4 years ago, # |   0 Please Publish Editorials in english also ,so that non Japanese can understand....
•  » » 4 years ago, # ^ |   0 You can use google translate
 » 4 years ago, # |   +4 English editorial is out!