### chokudai's blog

By chokudai, history, 3 years ago,

We will hold AtCoder Beginner Contest 182.

The point values will be 100-200-300-400-500-600.

We are looking forward to your participation!

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 » 3 years ago, # |   +2 Six hours left =.=
•  » » 3 years ago, # ^ |   +1 20 minutes left =.=
 » 3 years ago, # |   +9 yay ! I am on a binge solving At Coders problems. They are really cut-the-shit and ad hoc awesome xD !!
 » 3 years ago, # |   +1 4 wa in problem C.Know why getting WA but still can't fix that.That's how every contest made my day
•  » » 3 years ago, # ^ |   0 I used lots of if-else :v :v
•  » » 3 years ago, # ^ |   0 You can see my solution it's easy to understand
•  » » » 3 years ago, # ^ |   0 Just tell that what is output for 55555555 for problem c? My accepted code give -1..but after getting accepted I feel.it is wrong for above test as for it ans is 2(555555).. Am I missing something?
•  » » » » 3 years ago, # ^ |   0 answer is 2..May be weak test case !!
•  » » » 3 years ago, # ^ | ← Rev. 2 →   0 Hey.Can u explain u r solution little bit?-emli-
•  » » » » 3 years ago, # ^ |   0 He is checking all possibilities of digits. For each i, he considered each bit equal to 0 as not chosen and each bit equal to 1 chosen.For each combination, he get the sum of digits and de number of used digits (# of one on bit representation). If the sum is divisible for 3, this combination is divisible too.Then he printed the biggest combination that sum is divisible for 3.
•  » » 3 years ago, # ^ |   0 You can make use of dynamic programming, everytime either leave a digit or add it to your total sum, if at the end the total sum is divisible by 3, then store the total no. of digits you erased from the given string of numbers. print the minimum of it. My solution : https://atcoder.jp/contests/abc182/submissions/17971476
•  » » » 3 years ago, # ^ | ← Rev. 2 →   +19 I used $2^{len(|N|)}$ brute force. I know there are a lot of more efficient solutions, but this is the fastest to implement.
•  » » » » 3 years ago, # ^ |   0 What is the brute force solution for this problem ?
•  » » » » » 3 years ago, # ^ |   +4 Statement says: we can erase some subset of digits. There are at most $18$ digits, so at most $2^18$ subsets which is quite small. Just run through all subsets, check if remaining number is divisible by $3$ and if so update the answer taking minimum of current answer and number of erased digits.
•  » » » » 3 years ago, # ^ | ← Rev. 2 →   +10 There is another brute force, That I think is easier to implement. At the first build array cnt[3] that cnt[i] is the number of digit x such that x % 3 = i. Now lets brute force on the number of digit 1 and the number of digit 2 we will remove. codeUPD: After I think a little bit more about it, It's unnecessary to remove digit 1 or 2 more than 2 times. This solution will be O(length of the number). code
 » 3 years ago, # |   +1 Tried to fix a runtime error on E for an hour QAQ Does anyone know any compilation flags that will be helpful in detecting out of bounds error?
•  » » 3 years ago, # ^ |   0 -fsanitize=address
•  » » » 3 years ago, # ^ |   0 Thanks!
 » 3 years ago, # |   +6 How to solve F ??
 » 3 years ago, # |   +7 How to solve F plss
•  » » 3 years ago, # ^ |   +42 hint: Transform the problem to finding the number of (y — X) instead of y.
•  » » » 3 years ago, # ^ |   +3 I think find the number of (y+X) is much easier!
•  » » » » 3 years ago, # ^ |   +33 How? There is no (y + X) is the problem?
•  » » » » » 3 years ago, # ^ |   +8 Y is the change. I think “add” is very convenient
•  » » » » » » 3 years ago, # ^ |   +33 Oh, I mean the same thing by (y-X). let z = y-X. find z and z + X.I thought your y is the y in the problem.
•  » » 3 years ago, # ^ |   +43 First of all notice that any number $X$ has unique minimal representation. This can be proved by greedy exchange (using the fact that each successive number is a multiple of the previous number).Now, notice the isomorphism (structure equivalence) between the number bases and the (minimal) representation of each number in the system of Yens. There are two properties which are maintained:1) If highest bit of $A$ is greater than $B$ in our Yen-system, then $A>B$.2) Each number has unique representation.So we can think of an equivalent problem: given a number $X$, how many numbers $b$ there are such that $X+b=a$ and $a$ and $b$ have no bit in common?This can be solved, after some thinking, with DP.Array g[] represents the representation of $X$ in Yen-systemWe introduce the concept of "span". span[j] represents the number of elements after the $j^{th}$ bits of X which have maxed out values of their bit. Then if g[j]>0, we can use this bit in the number $b$ such that DUE TO the process of carry-over, we don't have this bit in $a$. You can see the details of dp transition in my Beautiful AC Code
•  » » » 3 years ago, # ^ |   0 Can you explain the concept of Span a bit more. I couldn't get it. By the way your approach seems better than those of others to comprehend .
•  » » » » 3 years ago, # ^ |   +9 the "span" array is the count of consecutive elements whose bit is "maxed out" in $X$.For example, consider a representation where $A_{i}=3A_{i-1}$ FOR ALL $i$. This is exactly similar to base-3 representation, and every bit can have atmost $2$ as coefficient. When a bit in X has 2 as its coefficient, I say that it is "maxed out".So lets say the yen-array in given input is1 3 9 27 81 243 and $X=271$ so that it is $222001$ in our yen system. Now $dp[j]$ denotes the number of ways in which we can have $b$ (in the equation $X+b=a$) such that $b$ has no bits in common with $a$ and only bits from index $j$ to $n$ is possibly ON in $b$. That is every bit from $1$ to $j-1$ is definitely $0$.Now ,for example, if we want to have the $2^{nd}$ bit ON in $b$, and all before that OFF, we must have that bit equal to $1$, because BY THE PROCESS OF CARRY OVER, that bit will be OFF in $a$. And now, notice that carry over makes $3^{rd}$ bit $0$ in $a$. So the span of the $2^{nd}$ bit is 1, since now we can't have any bit as upto after span bits after $j$ as ON in $b$, because then there can't be a carry over.After that it is simple DP transition.
•  » » » » » 3 years ago, # ^ |   0 Are you considering bit numbering from left to right?
•  » » » » » » 3 years ago, # ^ |   0 Left to right, as in the example I have taken.
•  » » » 3 years ago, # ^ |   0 Nice explanation! Can you suggest some more problems like this?
•  » » 3 years ago, # ^ | ← Rev. 4 →   +1 I just came out this solution, not pretty sure that it is correct. However it does get me AC.The problem is same as "Find the number of integer $r$ with length $N$ (leading 0 is allowed) in base $A$ that shares no common digit with $X + r$, i.e., $\forall_{0 \le i \lt N}, r_i = 0 \lor (X + r)_i = 0$".Let $c$ be carry, $y = X + r$, we have ($N$ = 4): $\begin{array}{r} &c_3 &c_2 &c_1 &0 \\ &x_3 &x_2 &x_1 &x_0 \\ +\quad &r_3 &r_2 &r_1 &r_0 \\ \hline &y_3 &y_2 &y_1 &y_0 \end{array}$In particular, all digits $i$ satisfy ($m_i = A_{i + 1}/A_i$ is the limit of digit $i$): $y_i = (c_i + x_i + r_i) \pmod {m_i} \\ c_{i + 1} = \lfloor \frac{c_i + x_i + r_i}{m_i} \rfloor$Let $dp[i, c]$ = the number of valid ways to fill digits $[0, i)$ while $c_i = c$. This dp is computed from lower digit to higher digit due to carry-over.By iterating all possible values ($0 \le r_i \lt m_i)$ of $r_i$, we can get an correct state transition, but it will get TLE. Key observation is that we don't need to iterate all values of $r_i$. Since $r_i = 0 \lor y_i = 0$, there are only few valid conditions.We decompose $r_i = 0 \lor y_i = 0$ into 3 disjoint conditions: $r_i = 0 \land y_i = 0$ $r_i = 0 \land y_i \ne 0$ $r_i \ne 0 \land y_i = 0$ Let's consider when will these conditions be true? By substituting $r_i = 0, y_i = 0$ into the formula of digit $i$, we find that only when $(c_i + x_i) = 0 \pmod {m_i}$, condition 1 is true. We transfer to $dp[i + 1, c_{i+1}]$. Substituting $r_i = 0$ into formula, we find that only when $(c_i + x_i) \ne 0 \pmod {m_i}$ will $y_i \ne 0$ be true. We transfer to $dp[i + 1, c_{i+1}]$. Substituting $y_i = 0$ into formula, we find that only when $(c_i + x_i) \ne 0 \pmod {m_i}$ will $r_i \ne 0$ be true. Notice that in this condition, there will be a carryover. We transfer to $dp[i + 1, 1]$. That's all for the dp transition. Here is my submission. Chinese version
 » 3 years ago, # |   0 Is there a dp solution for C?
•  » » 3 years ago, # ^ |   0 Yes. link
•  » » » 3 years ago, # ^ | ← Rev. 2 →   0 Please elaborate what are you storing at dp[i][j]. I was thinking something like declaring dp[index][sum]---->Maximum digits required till index to make sum. Is that correct? thanks. snippet vector > dp(n + 1, vector(sum + 1, 0)); for (int i = 1; i <= n; i++) { for (int j = 1; j <= sum; j++) { dp[i][j] = dp[i - 1][j]; if (j - arr[i - 1] >= 0) dp[i][j] = max(dp[i][j], 1 + dp[i - 1][j - arr[i - 1]]); if (j % 3 == 0) { ans = min(ans, n - dp[i][j]); } } } 
•  » » » » 3 years ago, # ^ | ← Rev. 2 →   0 I'm storing the maximum number of digits that I can take,upto index i, such that sum%3==0. I think your code should work if you store the same. (PS. i'm very bad at iterative dp. xD)
•  » » 3 years ago, # ^ | ← Rev. 2 →   +1 Just try every subset with bitsets or recursion. my solution
•  » » » 3 years ago, # ^ |   +6 No need for that. Just count all digits modulo 3 and use this if(n<3 and c[0]==0 and (c[1]==0 or c[2]==0)){ cout<<"-1\n"; } else { cout<
•  » » » » 3 years ago, # ^ |   0 what do you mean by "count all digits modulo 3"? Please explain. Do you mean that if number is k digits long then for every digit calculate it's modulo 3?
•  » » » » » 3 years ago, # ^ |   0
•  » » 3 years ago, # ^ |   0 You can think of finding the longest subsequence which has sum divisible by 3. Then you can do number of digits — longest subsequence
•  » » 3 years ago, # ^ | ← Rev. 2 →   +1 O(logn) Solution using DP. My Code : Link
 » 3 years ago, # |   0 Does we have to use dfs in E?
•  » » 3 years ago, # ^ | ← Rev. 2 →   +2 No, just annoying implementation. I used prefix and suffix sums on the rows and the columns.
•  » » » 3 years ago, # ^ |   +2 Not sure if this is what you meant, but you can just iterate over each column twice, once up and once down, and then each row twice, once left and once right. If you go over a block, everything after it is dark, and if you go over a bulb, everything after it is lit. So you just copy paste the same for loop like 5 times, change the bounds and you get the answer. Not really annoying implementation
•  » » 3 years ago, # ^ | ← Rev. 2 →   +8 No, just optimize the brute force solution a little bit. The brute force solution is, for each bulb, and for each direction, start from the bulb, and go along with this direction until you meet a block, set each grid passed illuminated by the bulbs. And then, iterate all grids, count the number of grids illuminated by the bulbs. The time complexity is $O((H + W)N + HW)$.The brute force solution might get TLE, although I think it's time complexity is good enough to pass.While going along with one direction, you can stop not only when you meet a block, but also when you meet a bulb. If you do so, the time complexity will be $O(HW)$ cause you will pass a grid at most 4 times.
•  » » » 3 years ago, # ^ |   +5 Thanks! That brute force modification helps to not get TLE.
•  » » 3 years ago, # ^ |   0 Using this matrix struct with already clock_rotate, you only need to repeat 4 times rotating the matrix every rotation.Submission
 » 3 years ago, # |   0 please anyone explain solution for D
•  » » 3 years ago, # ^ |   0 Calculate prefix sums on the input. Also calculate the maximum sum you achieve on the input for every $0$ to $i$, where i is from 0 to N — 1 (0-based indexing). Then, you can find the maximum coordinate by iterating from 0 to n and adding the maximum sum in range(0, i) to your current coordinate. Codeauto solve () -> void { int n; std::cin >> n; std::vector a (n), pref (n + 1), best (n + 1); for (int i = 0; i < n; ++i) { std::cin >> a [i]; pref [i + 1] = pref [i] + a [i]; best [i + 1] = std::max({best [i], pref [i], pref [i] + a [i]}); } int64_t max = 0, x = 0; for (int i = 0; i < n; ++i) { max = std::max(max, x + best [i + 1]); x += pref [i + 1]; } std::cout << max << '\n'; } 
•  » » 3 years ago, # ^ |   +4 Let us define end[i] as coordinate after the end of ith round (let us assume 0-based index) then, end[i] = end[i-1] + prefix[i]Now, if your answer is after round i, and before round i+1ans = max(end[i], end[i]+prefix[0] , end[i]+prefix[1] , end[i]+prefix[2] ,..)ans = max(end[i], end[i]+max(prefix[0],prefix[1],..,prefix[i]))sudo code : prefix[0] = a[0] for i = 1 to n-1 prefix[i] = prefix[i-1]+a[i] end[0] = prefix[0] for i = 1 to n-1 end[i] = end[i-1]+prefix[i] max_so_far = prefix[0]; ans = max(0,prefix[0]); for i = 1 to n-1 max_so_far = max(max_so_far , prefix[i]); ans = max(ans, end[i-1] + max_so_far); 
•  » » » 3 years ago, # ^ |   0 Thanks man, really helped!
 » 3 years ago, # |   0 What's the idea for E? I spent 50 mins on it thinking it was easier than D but I couldn't get it to work on one of the sample tests; probably some implementation errors. D was easier than I thought though. Great contest, thanks!
•  » » 3 years ago, # ^ | ← Rev. 2 →   0 HintThe key idea for E is that you can remember if you've already visited a square when going in one of the 4 directions, so you can reduce the number of iterations. My solution: https://atcoder.jp/contests/abc182/submissions/17982339
•  » » 3 years ago, # ^ |   +15 There are a lot of ways. Shortest would be: assume light doesn't pass through a square with another light bulb, which doesn't affect the answer, and then for each light bulb iterate through all squares it illuminates and mark them (with the modified rules each square will be touched at most $4$ times, so it's $O(HW)$).
•  » » » 3 years ago, # ^ |   0 Thanks rchaves and tribute_to_Ukraine_2022, understood.
•  » » 3 years ago, # ^ | ← Rev. 3 →   +3 am I the only one who solved it with Binary Search ?
•  » » » 3 years ago, # ^ |   0 What's your idea with binary search?
•  » » » 3 years ago, # ^ |   0 Can you provide link of your solution?
•  » » » » 3 years ago, # ^ |   0 The cell will be illuminated if it has bulb closer than block in any of its four sides. Lets say cell is (r,c) and direction is right. Now do lower_bound on row 'r' for bulb position and block position. If bulb position is less than block position, that means there is no block between that cell and bulb position.So,cell will be illuminated. Similarly we can try in another directions with little modifications. MyCode:Link
 » 3 years ago, # |   0 How to solve E. My O((H+W)*N) solution gave TLE.
•  » » 3 years ago, # ^ |   0 solution with commentchar ss[1505][1505]; char ans[1505][1505]; /*-----*/ void solve(){ ll n,m,r,c; string s; cin>>r>>c>>n>>m; for(ll i=0;i>x>>y; --x,--y; ss[x][y]='.'; } for(ll i=0;i>x>>y; --x,--y; ss[x][y]='#'; } n=r,m=c; for(ll i=0;i=0;j--){ //row from down to up if(ss[i][j]=='.') ck=true; if(ss[i][j]=='#') ck=false; if(ck) ans[i][j]='1'; } } for(ll j=0;j=0;i--){ // column from right to left if(ss[i][j]=='.') ck=true; if(ss[i][j]=='#') ck=false; if(ck) ans[i][j]='1'; } } ll answer=0; for(ll i=0;i
•  » » » 3 years ago, # ^ | ← Rev. 2 →   +1 How many fake account do you have? RD_TheCoder ,Rudro25. BD_King. CF.
 » 3 years ago, # | ← Rev. 2 →   0 can someone explain sample test case 2, problem D, how the answer is 2?
•  » » 3 years ago, # ^ | ← Rev. 2 →   +1 The positions after each step are: Spoiler -2 -4 -3 -5 -4 -1 -3 -2 1 0 -2 -1 2 1 0 Therefore, the greatest coordinate occupied is $2$.
•  » » » 3 years ago, # ^ | ← Rev. 3 →   0 thank u very much. My bad, I wasn't go through all step edit:got AC now T_T
 » 3 years ago, # |   0 Problem C and D were easier to do than B for me. Can anyone explain how B should be done?
•  » » 3 years ago, # ^ |   0 You can just brute force for the possible value of the answer from 2 to 1000. Whichever value gives maximum count of divisions is the answer.
•  » » 3 years ago, # ^ |   0 Brute force numbers from 2 to 1000 and calculate how many numbers in A are divisible by them. The one with the maximum count of numbers divisible is the answer.
 » 3 years ago, # |   0 How to solve problem F?
•  » » 3 years ago, # ^ |   0
 » 3 years ago, # |   0 How to solve C? People are saying it can be solved using DP. I don't know DP. Can anyone please tell me which variation of DP should actually I learn(as DP is huge) to be able to solve this problem(C)??
•  » » 3 years ago, # ^ |   0 It's a standard knapsack problem. For every digit you either use it in the answer or you don't.
•  » » » 3 years ago, # ^ |   0 Pls share your solution.
•  » » » » 3 years ago, # ^ |   0 Code#include "bits/stdc++.h" using namespace std; // Hi string s; int dp[50][300]; int solve(int index,int sum){ // if you are at index with sum of digits equal to sum if(index >= (int)s.size()){ // if you finished all the numbers if(sum%3 == 0 && sum != 0)return 0; // valid solution return 1e9; // not valid } // if already calculated return the stored result. if(dp[index][sum]!=-1)return dp[index][sum]; // else solve int c = s[index]-'0'; // take digit and add it to sum, or don't take it and add 1. int res = min( solve(index+1,sum+c), solve(index+1,sum)+1); return dp[index][sum] = res; // store the current result to use later (maybe). } int main(){ ios::sync_with_stdio(0);cin.tie(0); memset(dp,-1,sizeof(dp)); long long n; cin>>n; s = to_string(n); int ans = solve(0,0); if(ans >= 1e9){ // no valid solution cout<<-1<<'\n'; }else cout<
•  » » » » » 3 years ago, # ^ |   +1 Just an optimization that will help you in solving a lot more problems, when you call the solve function, you pass $sum + c$ you should take module $3$, this way the sum will never be $\ge 3$. In the return statement, check if the sum is $0$ then the number is divisible by $3$. This way you will reduce the states and your code will be much more faster.
•  » » 3 years ago, # ^ |   0 u can use recursive, it will be O(2^digit)
•  » » » 3 years ago, # ^ |   0 By generating all possible subsets?
•  » » » » 3 years ago, # ^ |   0 yes, for easier implementation, I use string for storing the input
•  » » 3 years ago, # ^ | ← Rev. 3 →   +1 It can be solved using greedy approach as well. As for N to be divisible by 3, sum of its digit must be divisible by 3. And sum%3 can be either 0, 1 or 2. If its 0 than we have to do nothing. If it's 1 then we remove one digit (which is 1 mod 3 i.e. d%3=1) or 2 digits (which are 2 mod 3 .i.e, d%3=2). Similarity for the case where sum%3 = 2. If we aren't able to remove the digits in such a way then answer is -1.
•  » » » 3 years ago, # ^ |   0 Thanks. i feel so dumb for not seeing this. xD
•  » » 3 years ago, # ^ |   +4 I think the solution with bitmasks is pretty niceMy submission
•  » » » 3 years ago, # ^ |   0 Thanks! Can you suggest me any blog/resource to learn this technique?
•  » » » » 3 years ago, # ^ |   +3 Hi! I think a good resource for bitmask DP is this HackerEarth tutorial, I remember it helped me understand the logic behind it. After that, it's just solving problems, some problems that come to mind, from easiest to hardest, are 580D - Kefa and Dishes, this CSES task, 16E - Fish, 678E - Another Sith Tournament.
•  » » » » » 3 years ago, # ^ |   +6 Thanks a lot dude! You're awesome :'D
•  » » » » » 3 years ago, # ^ |   0 You can also try this
 » 3 years ago, # |   0 Can someone tell me what are the 21 solutions to the last sample of problem F, I don't get how there are so many.
 » 3 years ago, # | ← Rev. 3 →   0 Can anyone please tell me what's wrong with my code for E: https://pastebin.com/1WPa7pUD ? EDIT: I finally fixed it...I am so stupid!
 » 3 years ago, # |   0 is it possible to see the submission of other participants after the contest at Atcoder.??
•  » » 3 years ago, # ^ |   0 Yes, for example, see my submission in my comment here.
•  » » » 3 years ago, # ^ |   0 But I am not able to see the solution of particular participant in the standing section as we have in codeforces where we just double click on the submission to view the code.
•  » » » » 3 years ago, # ^ |   +3 Click the magnifying glass next to a user on the standings to see their submissions.
 » 3 years ago, # | ← Rev. 2 →   +11 I recorded a video of me solving the contest: https://youtu.be/VsUvBvgeb9Y. I got 64th, solving all problems in about an hour.
•  » » 3 years ago, # ^ |   +8 Your solution for solving F is fantastic.Thanks for sharing that.
 » 3 years ago, # |   +20 Here is my editorial for this contest.English editorial中文题解
•  » » 3 years ago, # ^ |   0 Can we use line sweep for E to solve for any H,W.
 » 3 years ago, # |   0 For problem C what am I missing , its WA https://atcoder.jp/contests/abc182/submissions/17965994Please help!!
•  » » 3 years ago, # ^ |   +4 https://atcoder.jp/contests/abc182/submissions/17972323 hope this helps!
•  » » » 3 years ago, # ^ |   +1 Yeah!!Got the better approach :-) But still dont know which case is missing in mineAnyway thnx :-))
•  » » 3 years ago, # ^ | ← Rev. 3 →   0 Spoiler985198Correct answer:1(remove 1)Your answer:2
•  » » » 3 years ago, # ^ |   0 Thnx a lot!!
 » 3 years ago, # |   +6 Here's a screencast of being destroyed by daylight saving time, also includes solutions to all problems