### chokudai's blog

By chokudai, history, 4 years ago,

We will hold AtCoder Beginner Contest 178.

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

We are looking forward to your participation!

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 » 4 years ago, # |   +8 Reminder: Contest starts in 10 mins!
 » 4 years ago, # |   0 Are there any other contest on atcoder for beginner? ABC are very good but aren't that frequeant and i tried AGC and i shouldn't have... so any other recommendation
•  » » 4 years ago, # ^ |   +6 The ARC's should be doable, they're gonna be frequent again soon.
 » 4 years ago, # | ← Rev. 2 →   -31 deleted
•  » » 4 years ago, # ^ |   +4 Inclusion Exclusion. Number of ways that we can pick any value from 0 to 9 = 10^N . Number of ways that 0 doesnt occur = number of ways to pick any value from 1 to 9 = 9^N. By same logic, number of ways that 9 doesn't occur anywhere = 9^N Number of ways that both 0 or 9 don't occur = 8^N Hence answer is 10^N — 9^N — 9^N + 8^N . Use mods wherever required
•  » » » 4 years ago, # ^ |   0 I tried (nC2)*2*10^(n-2). nC2 to choose any two places for 0 and 9. 2 because {0,9} has two permutations. 10^(n-2) to fill remaining places with 0-9. I am still not sure why this approach is wrong. Could someone point out?
•  » » » » 4 years ago, # ^ |   +9 You are counting duplicates for example,when 0,9 is fixed like this 0_9 and like this 09_ in both cases you are counting 099
•  » » 4 years ago, # ^ |   0 Why so many downvotes he is just asking the approach didn't say anything wrong or asked anything wrong U can just use Inclusion exclusion
•  » » » 4 years ago, # ^ |   +25 He asked during contest which is not allowed
•  » » » » 4 years ago, # ^ |   -11 I was just try to be first so i can get reply so i can get reply quickly without having intention to get solve yes i can give these immediate after contest but i have a online class during the contest for that i gave that during the contest to be honest i have no intention to get a solve from here during contest
•  » » » » » 4 years ago, # ^ |   +31 The issue is that even if you weren't going to cheat yourself, if someone answered you during contest, anyone else could have seen the answer too, which wouldn't have been fair.
•  » » 4 years ago, # ^ |   +1 In this question you need to find all sequences so: There is some i such that Ai = 0 (count of such i may be > 1) There is some i such that Ai = 9 (count of such i may be > 1) (both conditions must be met)Let's iterate k — number of positions i such that Ai = 0/9Number of ways to create "good sequences" with 0/9 is 2^k — 2 (on every position you have 2 ways to put 0/9) (-2 because you don't need to create sequences 00..000 and 99..999)Let's count C(n, k) — the number of ways to select a set of i positions from n positions, where we want to put 0/9Finally we want to fill the remaining positions with number 1..8. It is equal to 8^(n — i)And final number of ways for some k equal to C(n, k) * (2^k — 2) * (8^(n — i))P.S. more details about Binomial Coefficients you can read here Binomial Coefficients
 » 4 years ago, # |   0 Any ideas on how to do F? If the frequency of any number in first array is greater than n — (frequency of that element in the second array) than answer is "No" else answer exists.I made a logic that the maximum frequent number in the first array will get the least frequent number from the second array. If there are multiple numbers with the same frequency then the smallest number with the smallest frequency goes to the largest number with the largest frequency. I was not quite able to implement it.
•  » » 4 years ago, # ^ |   +15 I understand, I thought there's just 1 minute left which won't really make any difference. One cannot implement it that quick. Sorry though I should have been a bit more patient
 » 4 years ago, # |   0 Solved E but coudnt solve C and D :(.
•  » » 4 years ago, # ^ |   0 approach for e
•  » » » 4 years ago, # ^ |   -8 There are only two cases to consider, if we only consider results such that Xi <= Xj. If Yi <= Yj, then the distance is (Xj + Yj) — (Xi + Yi) Otherwise, the distance is (Xj — Yj) — (Xi — Yi) By breaking it down into these cases, I have gotten rid of the absolute value function making it much easier to reason about the distances.So, we simply pick points with minimum and maximum x+y, and compute the distance. Then pick points with minimum and maximum x-y, and compute the distance. One of those two distances is your maximum.This can be done in O(n).My submission
•  » » » » 4 years ago, # ^ | ← Rev. 2 →   +25 Copy-pasted from: stackoverflow Why can't you provide the link directly instead of copy pasting the text XD
•  » » » » » 4 years ago, # ^ |   +4 its simple as you prob. dont want to jump from one site to another for a simple reason ,so posting a simple basic explanantion is not that big of issue
•  » » » 4 years ago, # ^ |   0 The manhattan distance of two points is the distance of the parallel diagonals of these points.The points at a distance x from a given point p form a diamond.
 » 4 years ago, # |   0 How to do D?
•  » » 4 years ago, # ^ | ← Rev. 2 →   +4 Simple DP. very similar to thisJust coins here are {3, 4, 5 .. N}
•  » » » 4 years ago, # ^ |   0 But using coin change dp method does not give us the different permutations right? Like for sum = 7. the answer = 3 ( {3,4} , {4,3} , {7} ). But dp gives us only answer = 2 ( {3,4} , {7} ). Please correct me if I'm wrong
•  » » » » 4 years ago, # ^ | ← Rev. 2 →   0 Imagine filling DP array for n = 7 by this code: Spoiler // dp[i] = how many ways to get this sum dp[0] = 1; // there is one way to get sum 0: take nothing for(sum=3; sum<=n; sum++){ // what sum do we want to get for(coin=3; coin<=sum; coin++){ // going through the coins, note that coins bigger than sum will not be considered because we can't get that sum using it. for example it's impossible to get sum of 5 using atleast once coin 7 dp[sum] += dp[sum-coin]; dp[sum] %= mod; // taking mod } } cout << dp[n]; dp array at the end will be: dp[0] = 1; // {0} dp[1] = 0; // {} dp[2] = 0; // {} dp[3] = 1; // {3} dp[4] = 1; // {4} dp[5] = 1; // {5} dp[6] = 2; // {3+3, 6} dp[7] = 3; // {4+3, 3+4, 7} why 4+3 gets considered? answer: imagine sum = 7, coin = 3 the number of ways to get sum 7-3=4 will be added to dp[7]Hope you understood
•  » » » » » 4 years ago, # ^ |   0 can you please describe how can we implement it by top-down dp approach?I have implemented this but it isn't correct.
•  » » » » » » 4 years ago, # ^ | ← Rev. 3 →   0 Code#define int long long int dp[N]; int sol(int n){ if(dp[n] != -1)return dp[n]; dp[n]=0; for(int i=3;i<=n;i++){ dp[n] += sol(n-i); }return dp[n]%=mod; } void solve(){ int n; cin >> n; memset(dp, -1, sizeof(dp)); dp[0] = 1; cout << sol(n) << '\n'; return; } 
•  » » 4 years ago, # ^ |   +10 I did it using Combinatorics, Stars and bars. Divide the number into x sums, (1 to s/3) and for each, find all possible combinations, by keeping each of it exactly 3 initially.
•  » » » 4 years ago, # ^ |   +2 I also did the same,this approach is bit intuitive as this revises the P&C concepts :)
•  » » » » 4 years ago, # ^ |   0 Yeah, but dp would be a lot easier.
•  » » » 4 years ago, # ^ |   0 Could you elaborate more on this approach?
•  » » » » 4 years ago, # ^ | ← Rev. 2 →   +11 Sure. Firstly let's notice that we can only have sequences of size from 1 to s/3. (because each one should be at least 3) Then, let's call these "number of groups" as x, example: For s=9, we have, x=1, {3} ; (with 6 still remaining) x=2, {3,3} (with 3 still remaining) x=3, {3,3,3} (with 0 remaining) Now, we can distribute the remaining sum, for each x, among the x groups, in (n+r-1)C(r-1) ways, This is, distributing n identical objects among r people (stars and bars). So iterating over each x from 1 to s/3 by calculating the combination for each iteration, it's easy to find the answer. Submission
•  » » » » » 4 years ago, # ^ |   0 thank u a lot
 » 4 years ago, # |   0 How to solve E? I tried to find smallest and largest points on x and y axis(total 8 points) and found maximum among them.
•  » » 4 years ago, # ^ |   +3
•  » » » 4 years ago, # ^ |   +4 Thanks! Completely forgot to break mod, went in the wrong direction. :(
•  » » 4 years ago, # ^ | ← Rev. 3 →   +1 I removed the mod (absolute value) by considering 4 cases.case 1 : xi > xj and yi > yjwe need to find now max(xi-xj + yi-yj) = max((xi+yi)-(xj+yj)) = max(xi+yi)-min(xj+yj)for other cases I multiplied coords by -1 and solved by case 1 againcase 2 : multiply all x by -1case 3 : multiply all y by -1case 4 : multiply all x and y by -1
•  » » 4 years ago, # ^ | ← Rev. 3 →   +11 Manhattan distance between two points is: |$x$a — $x$b| + |$y$a — $y$b|. Now if can be evaluated in 4 ways: $dist$ = ($x$a — $x$b) + ($y$a — $y$b) $dist$ = ($x$a — $x$b) — ($y$a — $y$b). $dist$ = -($x$a — $x$b) + ($y$a — $y$b). $dist$ = -($x$a — $x$b) — ($y$a — $y$b).Now if you simplify them a little then the distance can be one of the two below: 1) $dist$ = ($x$a + $y$a) — ($x$b + $y$b) 2) $dist$ = (-$x$a + $y$a) + ($x$b — $y$b)Now the distance will be maximum from above expressions. submission link
•  » » » 4 years ago, # ^ |   0 Thanks. Here is shorter implementation I did with your help.
•  » » » » 4 years ago, # ^ |   0 Oh, i see. It's really concise. 80% of time i overkill (: problems.
•  » » 4 years ago, # ^ |   0
•  » » 4 years ago, # ^ |   +5 the approach to solve was same as this question.you were just given distance function instead of f(i, j) codevoid solve() { int n, x, y; cin >> n; vector sum(n), diff(n); for (int i = 0; i < n; i++) { cin >> x >> y; sum[i] = x + y; diff[i] = x - y; } int maxone = INT_MIN, maxtwo = INT_MIN; int minone = INT_MAX, mintwo = INT_MAX; int ans = INT_MIN; for (int i = 0; i < n; i++) { maxone = max(maxone, sum[i]); maxtwo = max(maxtwo, diff[i]); minone = min(minone, sum[i]); mintwo = min(mintwo, diff[i]); ans = max({ans, maxone - minone, maxtwo - mintwo}); } cout << ans; } 
•  » » 4 years ago, # ^ |   +3 Alternatively, you can calculate the convex hull of the given points and brute force the vertices of the polygon.
•  » » » 4 years ago, # ^ |   +6 This will TLE if to much points are on the convex hull, it could be all.
•  » » » » 4 years ago, # ^ | ← Rev. 2 →   0 Yes, you're right. However, I got an Accepted verdict. I guess the test cases are not strong enough.
 » 4 years ago, # |   +10 Pure Mathematics Contest!!
•  » » 4 years ago, # ^ |   +4 A and B was no math ;)
•  » » 4 years ago, # ^ |   0 and D was also of dp ..No math
•  » » » 4 years ago, # ^ |   0 You could do it with stars and bars
 » 4 years ago, # |   0 Maybe E is just a too well known problem, but F solution was more obvious to me than E solution.
•  » » 4 years ago, # ^ |   0 how to solve F ? It seems easy at first.
•  » » 4 years ago, # ^ |   0 how to solve F ? Think about 70 mins and more but still fail to figure
•  » » 4 years ago, # ^ |   0 What was the approach for F?
•  » » 4 years ago, # ^ |   0 Can you please explain your solution to E, I don't know which one test case was failing, I got 17 AC and 1 WA
•  » » » 4 years ago, # ^ |   0 This might help https://atcoder.jp/contests/abc178/submissions/16705345 I sorted the points considering points to be equal if they lie in the same circle centered at the origin
•  » » » 4 years ago, # ^ |   0 Put three comparators. one comapring x1-y1 other with x1+y1 one with y1-x1 Take the max of difference of the end terms and max of three would be the answer.
•  » » » 4 years ago, # ^ |   0 Here is my take on E,ans=(x2-x1)+abs(y2-y1) ,x2>x1Now we have two options for answerans1=(x2+y2)-(x1+y1) ==== max(x+y) -min(x+y) for optimal answerans2=(x2-y2)-(x1-y1) ==== max(x-y) -min(x-y) for optimal answer.Final answer would be the max of two.
•  » » » » 4 years ago, # ^ |   0 Hey can you please suggest which one test case is failing for this solution
•  » » » » 4 years ago, # ^ |   0 I have a doubt. The two cases are coming from assumptions, like case 1 comes from y2 >= y1. So how is it that when we calcuate the answer, we don't check for these restrictions and still get the correct answer. Can't it be that for i and j which gives max(xi + yi) — min(xj + yj), yi is actually less than yj which makes that pair invalid?
•  » » » » » 4 years ago, # ^ |   0 Actually I have this same doubt and I have used the same assumptions that invalid pairs wont affect the best answer for the problem and got a AC, I was really trying to find a intuitive or mathematical proof for the same but havent got one, I would really appreciate if someone comes up with a proof for the same.
•  » » » » » 4 years ago, # ^ |   0 Here we are only concerned about the largest valueLets get our definitions clear firstoption1 = (xj-xi) + (yj-yi) for yj>=yioption2 = (xj-xi) + (yi-yj) for yi>=yjNow for yj>=yi we can never have option2>option1 and vice versa.So though we are considering some invalid pairs our answer is never going to be affected
•  » » » » » » 4 years ago, # ^ |   +1 Nice. Thanks!
•  » » 4 years ago, # ^ |   0 Can you check why I am getting WA on F. here is submission
•  » » 4 years ago, # ^ |   +19 More than being well known, E is pretty easily searchable I think.
 » 4 years ago, # |   +7 MathCoder literally :V
 » 4 years ago, # |   0 What are the solutions for problems E and F?
•  » » 4 years ago, # ^ |   +3 e is well-known, u can google it
•  » » » 4 years ago, # ^ |   0 I googled couldn't find the same problem, can you please provide a link where solution of this problem is explained. like gfg or something?
•  » » » » 4 years ago, # ^ |   0 hereyou're welcome :))
 » 4 years ago, # |   0 Hi, this is my code for F problem I am failing in 3 test cases and I couldn't figure out those TCs,you can find my code in this link please help mehttps://ideone.com/7PaZ9kMy submission link https://atcoder.jp/contests/abc178/submissions/16710181
•  » » 4 years ago, # ^ |   0 can I get any test case for which my code is failing?
•  » » » 4 years ago, # ^ |   0 there is dropbox link to all atcoder test cases in one of the rng_58's blog
 » 4 years ago, # |   +118 F solution: Reverse array B, then A will be in ascending order and B will be in descending order. If we now look at positions where A[i] == B[i], they will form a segment [l, r], and will all have equal element A[i] == B[i] == C. Then you just need to swap all positions i from this segment with such positions j that A[j] != C and B[j] != C. And if there are not enough such positions j then I think it's not hard to prove that reordering is not possible at all.
•  » » 4 years ago, # ^ |   0 I think reordering is not possible only in the case where lets say x occurs m times in A and k times in B and both k>=(n+1)/2 and m>=(n+1)/2. By pigeonhole principle!
•  » » » 4 years ago, # ^ |   +8 No consider A = [1 1 1 1 2] B = [1 1 2 3 4], maybe we can say that frequency of some element must obey $f_A + f_B <= n$ ...
•  » » » » 4 years ago, # ^ |   +8 Ya I missed it thanks , Amazing F!
•  » » » » 4 years ago, # ^ |   0 How to prove that answer is always Yes when every f_A + f_B <= n
•  » » » » » 4 years ago, # ^ |   0 That is shown from answer of madn. When A is ascending and B is descending, and lets say $f_A + f_B = N$ for boundary condition, maximum intersection is possible if both A and B have near equal frequencies, intersection will be of length $\lfloor N/2 \rfloor$. So we have adequate numbers ($\lceil N/2\rceil$) to put in this positions of overlap, so answer is always possible
•  » » » » » » 4 years ago, # ^ |   0 oh,actually its very obvious...what am i thinking...Thanks!
•  » » 4 years ago, # ^ |   +25
•  » » » 4 years ago, # ^ |   0 great idea man
•  » » 4 years ago, # ^ |   +15 This is the best solution I have ever read!!
•  » » » 4 years ago, # ^ |   0 Yeah I find it very simple and easy to see correctness. In my contest I was just wondering why they gave A, B in sorted order, maybe it was hint toward this solution
•  » » » 4 years ago, # ^ |   +3 Yes, key insight is that only a[i] = b[i] = c will be the only value, and form an interval
•  » » » » 4 years ago, # ^ |   0 Can you prove it ?
•  » » » » » 4 years ago, # ^ | ← Rev. 4 →   +5 if there are two different value that a[i] = b[i], let a[i] = b[i] = c, a[i+1] = b[i+1] = d, and c != da[i] < a[i+1], so d > cb[i] > b[i+1], so d < ccontradiction
•  » » 4 years ago, # ^ | ← Rev. 2 →   +3 I am not claiming following is correct but I also cannot see why it is wrong. It must be wrong because it gives runtime error. The process was: Find net frequency of each element in the two arrays. If it is greater than n, then it is not possible to rearrange. Otherwise we use a max-heap to store pair of {"frequency in B", "number"}. Now we loop A from 1 to N and simply pop from heap and check it with A[i]. If not equal then we place it in our final array. Otherwise we pop another value and place it in final array. We push back the changed frequency. I think it runs into dead end where it no longer finds a suitable pair, but why it happens? Thanks!
•  » » 4 years ago, # ^ |   0 i think thats the easiest code to read and understand.
 » 4 years ago, # |   0 Man problem C ruined the round for me!:) Can anyone please explain the approach for it?
•  » » 4 years ago, # ^ |   +3 10^n-2*9^n+8^n Inclusion exclusion
•  » » » 4 years ago, # ^ |   0 Wow, thanks! Very interesting!
•  » » » 4 years ago, # ^ |   +1 Can you please help me to understand what wrong with this approach.Take two numbers and assign 0 and 9 to them , ways=n*(n-1)remaining (n-2) numbers can be arranged in 10^(n-2) ways making ans=n*(n-1)*10^(n-2)Thanks.
•  » » » » 4 years ago, # ^ |   0 The 10 ^(n-2) replicates in many cases when n>=3 see lets say you place 0,9 at 1 2 so possible values are 0 9 1, .....0 9 9 SImilarly suppose in another permutation you fixed 0 9 at position 1 3 and other 10 values accordingly at remaining position .Observe both the vectors carefully.0 1 9........0 9 9.You can clearly see 0 9 9 appearing common in both thus WA!
•  » » » » 4 years ago, # ^ |   0 There will be repeation, like in your solution you will count (2 1 9 2) twice.
•  » » » 4 years ago, # ^ | ← Rev. 2 →   +5 Intuition behind this:There are $x$ ways of choosing which of $x$ numbers to place in a spot, each choice is independent of each other, so for $n$ places there are $x^{n}$ ways.Number of ways of placing numbers in the range $[0, 9]$ -> $10^{n}$.Number of ways of placing numbers in the range $[0, 8]$ or $[1, 9]$, that is the number of ways of generating an array that is guaranteed to be missing either a $0$ or an $8$ -> $9^{n}$ each.But note that we have double counted the number of ways to place $[1, 8]$ (both missing) while removing the bad array count, so add back the number of ways of doing that which is $8^{n}$.
•  » » 4 years ago, # ^ | ← Rev. 2 →   0 You can use DP for accounting for double counting
•  » » » 4 years ago, # ^ |   +1 Thanks! I tried to do a dp approach but i didn't manage to solve it during the contest! In my opinion, D was easier than C.
•  » » » » 4 years ago, # ^ | ← Rev. 2 →   0 Exactly! I was stumped by C as well. Here's my DP solution to C Code#include using namespace std; #define int long long const int M = 1e9+7; vector>dp; void solve(){ int n; cin>>n; dp.assign(n+2,vector(4,0)); dp[0][0] = 1; for(int i=1;i<=n;++i){ dp[i][3] = ((dp[i-1][1]+dp[i-1][2])%M+10*dp[i-1][3]%M)%M; dp[i][2] = (dp[i-1][0] + 9*dp[i-1][2]%M)%M; dp[i][1] = (dp[i-1][0] + 9*dp[i-1][1]%M)%M; dp[i][0] = 8*dp[i-1][0]%M; } cout << dp[n][3] << endl; return; } int32_t main(){ int t=1; while(t--){ solve(); } return 0; } 
•  » » » » » 4 years ago, # ^ |   0 Nice solution!
•  » » 4 years ago, # ^ | ← Rev. 2 →   +3 Answer = (total sequence with length n and all element <= 9) — (total sequences such that it contains neither 0 nor 9)first term above is simply pow(10,n) for the second term I used inclusion exclusion:A = Set of sequences that doesn't contain 0B = Set of sequences that doesn't contain 9n(A union B) = n(A) + n(B) - n(A intersection B)n(A) = n(B) = pow(9,n)n(A intersection B) = pow(8,n)
•  » » » 4 years ago, # ^ |   0 Nice explanation, thank you!
 » 4 years ago, # | ← Rev. 3 →   +4 what's wrong in my solution of F? getting wrong answers on 6/50 test cases?https://atcoder.jp/contests/abc178/submissions/16721049UPD: I did some changes in my above code but still it's giving wrong answer on 3 test-cases :( https://atcoder.jp/contests/abc178/submissions/16724818
 » 4 years ago, # |   0 Anyways to do C or D without DP ?
•  » » 4 years ago, # ^ |   0 C is simple inclusion — exclusion. My one line code : cout << sub(add(power(10, n), power(8, n)), add(power(9, n), power(9, n)));
•  » » 4 years ago, # ^ |   0 D could be solve by brute force the number of elements in the sequence and the do a star-and-bar-like counting trick.
•  » » 4 years ago, # ^ |   +5 My approach of D (Without dp ) : First of all fix the length of the sequence . Suppose length is 5 ; so a1 + a2 + a3 + a4 + a5 = S ; now for all i , a[i]>=3 ; using stars and bars technique , no of solution of that equation is ncr(s-3*length +length -1 , length-1);here length = 5 ; final answer = sum of solutions for each fixed length .  Codeint ans=0; for(int len=1;len<=s;len++) { if(s<(3*len))continue; int sum=s-3*len; int yo=ncr(s-(3*len)+len-1,len-1); ans=(ans+yo)%mod; } cout<
•  » » 4 years ago, # ^ | ← Rev. 2 →   0 I did D using combinatorics. It is the modified version of distributing n balls into r urns such that each urn contains at least 1 ball. Solution of standard problem: (n-1)C(r-1).Here problem can be reformulated as if you can use any number of urns (because the length of sequence can vary) and each urn should contain at least 3 balls. Solution: summation of (n-1-2*i)C(i-1) for all i, where i is the number of urns and n is the sum we want. The term -2*i because we first put 2 balls in every urn so that problem is now transformed into a standard one containing at least 1 balls, for each i.https://atcoder.jp/contests/abc178/submissions/16705277
 » 4 years ago, # |   0 For problem C: I'm thinking like this, place 0 and 9 in any two places (nC2) then other n-2 places will have 10 possibilities hence (10^(n-2)). So overall answer is nC2 *(10^(n-2)).Can anyone say what fact am i assuming wrong/missing. Thanks :)
•  » » 4 years ago, # ^ |   +1 nC2 * 10^(n-2) will duplicate
•  » » » 4 years ago, # ^ |   0 can you give correct combination formula for this problem ?
•  » » » » 4 years ago, # ^ |   +1 I used dp https://atcoder.jp/contests/abc178/submissions/16709001 Here, dp[n][0][0] means sequence of length n and 0 zeros and 0 nines, dp[n][0][1] means sequence of length n and 0 zeros and atleast 1 nine, dp[n][1][0] means sequence of length n and atleast 1 zero and 0 nines, dp[n][1][1] means sequence of length n and atleast 1 zero and atleast 1 nines
•  » » » 4 years ago, # ^ |   0 so the 0's and 9's are getting duplicated?
•  » » » » 4 years ago, # ^ |   0 In cases when you put 9, 0 like this: ab90 09cd sequences may coincide, when a = 0, b = 9, c = 9, d = 0: 0990 0990
•  » » 4 years ago, # ^ |   0 There are repetitions in your formula. For n = 4 the answer is 974. Check with your submission
 » 4 years ago, # |   +21 atdynamicprogrammer
 » 4 years ago, # |   +6 lose again because of my speed
 » 4 years ago, # |   0 Was trying to do D , using DP but gave TLE... https://atcoder.jp/contests/abc178/submissions/16721075 Can someone pls help what is wrong
•  » » 4 years ago, # ^ |   +3 Index is not necessary in the state.
 » 4 years ago, # | ← Rev. 3 →   +43 Editorial: Acout<2*n,the answer is "NO". Otherwise the answer is "YES".You can construct it as follow.We can find all the positions that a[i]=b[i],and store them in one "vector".We can let count[i] = the number of times that i appear in the vector.If for all i that count[i]*2<=vector.size(),we can construct them like this probelm:https://codeforces.com/contest/1381/problem/COtherwis we can swap the most appearing elements with other positions that are not in the vector,to satisfy the condition above.
•  » » 4 years ago, # ^ | ← Rev. 2 →   +3 In $C$ constraints are smaller, so if someone has weak combinatorics(like me) then dp is more intuitive imo. here is my sol using dp: Spoiler#include using namespace std; constexpr int mod = 1e9 + 7; struct M { unsigned v; M(long long a = 0) : v((a %= mod) < 0 ? a + mod : a) {} M& operator+=(M r) { if ((v += r.v) >= mod) v -= mod; return *this; } M& operator-=(M r) { if ((v += mod - r.v) >= mod) v -= mod; return *this; } M& operator*=(M r) { v = (uint64_t)v * r.v % mod; return *this; } }; uint32_t add(M a, M b) { return (a += b).v; } uint32_t mul(M a, M b) { return (a *= b).v; } uint32_t sub(M a, M b) { return (a -= b).v; } int n; const int maxn = 1e6; int dp[maxn][2][2]; int go(int id, bool zero, bool nine) { if (id == n) { return zero & nine; } int &ref = dp[id][zero][nine]; if (~ref) { return ref; } int ans = 0; // 8 ways to place a numbers from [1, 8] ans = add(ans, mul(8, go(id + 1, zero, nine))); // place a 0 here ans = add(ans, go(id + 1, 1, nine)); // place a 9 here ans = add(ans, go(id + 1, zero, 1)); return ref = ans; } void solve() { memset(dp, -1, sizeof dp); cin >> n; cout << go(0, 0, 0) << '\n'; } signed main() { ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0); solve(); return 0; } 
•  » » » 4 years ago, # ^ |   0 Yeah, dp can also work!
•  » » 4 years ago, # ^ |   +8 Hey, in problem D, I think you also need to add 1 to dp[i].
•  » » » 4 years ago, # ^ |   0 Fixed!
 » 4 years ago, # |   0 Is there a case to be handled for mod of negative numbers in C ?
•  » » 4 years ago, # ^ |   0 if a<0 then use mod-abs(a) will work.
•  » » 4 years ago, # ^ |   0 Don't forget to add mod when you are subtracting two numbers under mod.
•  » » » 4 years ago, # ^ |   +1 I just have learnt a very valueable lesson.
•  » » 4 years ago, # ^ |   0 we can do this : ans = (a-b+MOD)%MOD;
•  » » » 4 years ago, # ^ |   0 How does it work?Can you please explain or give me any reference?
•  » » » » 4 years ago, # ^ |   0 I'll leave that for you to figure out :)
 » 4 years ago, # |   +11 And.. AtCoder becomes MathCoder.
 » 4 years ago, # |   0 c explanation ?
•  » » 4 years ago, # ^ |   +9 Total Number: 10^nWithout digit '9': 9^nWithout digit '0': 9^nWithout both: 8^nFinal Result: 10^n — 9^n — 9^n + 8^n
•  » » » 4 years ago, # ^ |   0 :O i feel dissapointed about myself
•  » » 4 years ago, # ^ |   0 Total numbers you can make using all digits (10^n) — numbers with all digits but 9 (9^n) — all but 0 digits (9^n) + all but 0 and 9 (8^n) (because they were deleted twice);
 » 4 years ago, # |   +54 Maybe this is an off-topic.I think Atcoder should show counts of solve for each problem in dashboard/tasklist(like CF). I know it is in the standings page, but viewing standings itself some number of times is well-disgusting.
 » 4 years ago, # |   0 I have tried to solve F using bipartite graph matching but it failed just in test 16 :)))
•  » » 4 years ago, # ^ |   +8 WA or TLE?
•  » » » 4 years ago, # ^ |   0 TLE, actually I was expecting it! I was checking a random edge to find a match each time.
•  » » 4 years ago, # ^ |   +16 I don't think it can work , the nodes and edges are too many !
•  » » » 4 years ago, # ^ | ← Rev. 2 →   +6 It worked :)) submmision #16726993I guess test-cases are not that much strong.
•  » » 4 years ago, # ^ |   +8 of course it will :))))
•  » » 4 years ago, # ^ |   -8 maybe you can try using a segment tree to build the graph.
•  » » » 4 years ago, # ^ |   +8 But the nodes are too many :)
•  » » » » 4 years ago, # ^ |   -8 yeah, it is.
 » 4 years ago, # |   +9 I found my solution for F is realy complex(Though it is right)!!Can you share your idea for F?
•  » » 4 years ago, # ^ | ← Rev. 2 →   +6 Resulting ordering of B array is some cyclic shift of itself. (If no shifting is possible, no solution).
•  » » » 4 years ago, # ^ |   +9 Thank you for your reply:)But can you prove it?And how to achieve it?
•  » » » » 4 years ago, # ^ | ← Rev. 2 →   0 I did not scratch a full proof yet.For each element in B, we find what shifts are not possible. This can be done using set/segment tree.
•  » » » » » 4 years ago, # ^ |   +9 Can you share your code.
•  » » » » » » 4 years ago, # ^ |   0
•  » » » » 4 years ago, # ^ | ← Rev. 2 →   +6 my F solutionWhenever there exists index i where A[i] is equal to B[i], shift array B one time. Loop until there are no such indices in one interation. I use two pointers to maintain the shift
•  » » » » » 4 years ago, # ^ |   0 Can you prove it?
•  » » » » » » 4 years ago, # ^ |   +6 Sorry, I didn't prove it during contest. I generate all possible cases for small N and it seemed work.
•  » » » » » » » 4 years ago, # ^ |   0 alvinvaja , I really like the quality of your profile picture. May I know which camera was used?
•  » » » » » » » » 4 years ago, # ^ |   0 It wasn't mine. Got it from pinterest
•  » » » » » » 4 years ago, # ^ |   +6 I think mine is similar where I take a number, x, in both a and b and if the start of b's interval of x isn't over the end of a's interval of x, I calculate how many shifts it would take to make b's start over a's end of x's interval (if b's start isn't already over a's end) and take the maximum shift for all numbers that are both in a and b. Maybe there's a counterexample but I couldn't find one. After the greatest shift needed is done, for that number x in b that needed the largest shift, all the numbers < x in b should be assigned to a number bigger than it in a and all the numbers >= x in b are shifted up to match numbers bigger than it in a until n, and then they match to lower numbers that are <= x because the shift circles it to the beginning of a. Unless the array is impossible I don't see how any numbers can conflict yet.
•  » » » » 4 years ago, # ^ |   +12 I used the same approach, but I also can’t prove it. But I know how to achieve it. Each position of B can be expressed as "not shifting in range [l, r]" which is equivalent to ~[x >= l and x <= r] = (~(x >= l) or (x >= r + 1)).Now the requirement is just AND product of all boolean clauses which is nothing, but 2-SAT. I thought that they put some problems to practice Atcoder library, turned out that I overcomplicated things up. lol (They didn’t add library yet, so I got compile error.) code
•  » » 4 years ago, # ^ |   +5 you can see this solution I just tried to solve F using this and got AC. Logic is simple as well as code. Here is my code for this approach
•  » » » 4 years ago, # ^ |   0 Thanks:)
 » 4 years ago, # |   0 Can someone check why I'm getting WA on C . here is my submission
•  » » 4 years ago, # ^ | ← Rev. 2 →   0 Result of a1 — a2 + b1 may be negative, so just add MOD to it before taking remainder
•  » » » 4 years ago, # ^ |   +1 Thanks, it worked!
•  » » 4 years ago, # ^ |   0 "ll ans = (a1 — a2 + b1)%M" should be ll ans = (a1 — a2 + b1 + M)%M;
•  » » 4 years ago, # ^ |   0 ll ans = (a1 - a2 + b1 + MOD)%M;
 » 4 years ago, # |   +1 Just 2 min short to get AC in problem F. Nice F anyway
•  » » 4 years ago, # ^ |   0 What was your idea behind F? Was it greedy?
•  » » » 4 years ago, # ^ |   0 The idea is tough to explain for me. I try. 1.A is in ascending order and B is in ascending order. So, just try to pick up elements like two pointer from start to end. Then, pick the unpicked elements too. Now, check the answer. If its OK then print.2. Otherwise, reverse B once again and go like 1. Check the answer. If OK then print it, else there is no answer.
 » 4 years ago, # | ← Rev. 2 →   +7 A Combinatorics solution for D — RedistributionRecall that, by the famous Stars and Bars Theorem, the number of non-negative integral solutions of $x_1 + x_2 + \dots + x_r = n$is ${n + r - 1} \choose {r - 1}$Let us denote this value as $stars(n, r)$. Clearly, the value is 0 if $n$ is negative.Moreover, if we add a constraint that $x_i \geq t$, then, we can replace $x_i$ by $y_i$, where $x_i = y_i + t$This transforms the problem to finding the non-negative integral solutions of $y_1 + y_2 + \dots + y_r = n - r \cdot t$Clearly, the number of valid solutions is $stars(n - r \cdot t, \ r)$So, for each $r$ from 1 to $n$, we sum up the values of $stars(n - 3 \cdot r, \ r)$.
 » 4 years ago, # |   0 DP Solution for Problem C — Ubiquity Code#include using namespace std; #define int long long const int M = 1e9+7; vector>dp; void solve(){ int n; cin>>n; dp.assign(n+2,vector(4,0)); dp[0][0] = 1; for(int i=1;i<=n;++i){ dp[i][3] = ((dp[i-1][1]+dp[i-1][2])%M+10*dp[i-1][3]%M)%M; dp[i][2] = (dp[i-1][0] + 9*dp[i-1][2]%M)%M; dp[i][1] = (dp[i-1][0] + 9*dp[i-1][1]%M)%M; dp[i][0] = 8*dp[i-1][0]%M; } cout << dp[n][3] << endl; return; } int32_t main(){ int t=1; while(t--){ solve(); } return 0; } 
•  » » 4 years ago, # ^ |   0 can you Please explain the states and transition.
 » 4 years ago, # | ← Rev. 2 →   +7 A great contest.Unfortunately,I nealy AK(AC 5 problems),but I'm in trouble with the sixth problem.All of them are Math problems(.(xiao xue ao shu in Chinese)And there's Chinese in English task:配点。Hope the abc will be better!：）
 » 4 years ago, # |   +22 So I did a randomized solution for F. I try three ways: Reverse $B$. Random shuffle $B$ for $T$ times. Rotate $B$ randomly for $T$ times. $T = 20$ is sufficient to get AC, and it runs in 61 ms. Of course, I have no proof (well not even an intuition) for this. :DSo, I'll be glad if someone can hack this solution or give some proof/intuition for it.
•  » » 4 years ago, # ^ |   0 Assume odd n, and n/2 '1' in a[], and n-n/2 '1' in b. Then we need to rotate exactly n/2 positions.
•  » » » 4 years ago, # ^ |   +8 Reversing B works in this case.
•  » » » » 4 years ago, # ^ |   +11 Ok, let the sympobls be '2', not '1'. And a '1' in first position of both arrays. Then the reverse of b[] does not work, but still we need to rotate exactly n/2 positions.
•  » » » » » 4 years ago, # ^ | ← Rev. 2 →   +8 Yeah it prints No for this case. gen// py n = 200000 print(n) a = [1] * (n // 2) + [2] * (n // 2) b = [1] + [2] * (n // 2) + [3] * (n // 2 - 1) print(*a) print(*b) 
 » 4 years ago, # |   +17 How to prove that on F, if a solution exists, one of the cyclic shifts of b is also a solution?I wrote my solution based on it and it passed.
•  » » 4 years ago, # ^ |   -12 Isnt there a lot of cyclic shifts possible?
 » 4 years ago, # | ← Rev. 2 →   +21 My Unofficial tutorial (A-F) for this contest. And the Chinese version.
•  » » 4 years ago, # ^ |   0 What is the codeforces problem similar to F?
•  » » » 4 years ago, # ^ |   +6
 » 4 years ago, # |   0 Just a thought: How to solve F if the elements are unsorted in both the arrays?
•  » » 4 years ago, # ^ |   +5 You can look at my solution, which does not depend on whether the arrays are sorted or not.
•  » » 4 years ago, # ^ |   +8 You can always sort them and use your solution for the sorted input...
 » 4 years ago, # |   0 Can somebody please help me find the only test case that is failing for my solution of problem E?
•  » » 4 years ago, # ^ |   +6 Changing the initial value of your variable ans1 to LLONG_MIN should solve your problem, because it's not necessary that x+y or x-y is always greater than -1.
•  » » » 4 years ago, # ^ |   0 Thanks dude, Another stupid mistake
 » 4 years ago, # |   0 What's wrong with my solution of D?I used dp, $dp_{ij}$ means the sequence‘s length is $i$ and its sum is $j$. Spoiler#include using namespace std; const int MOD = 1e9 + 7; int dp[3000][30000]; int main() { ios::sync_with_stdio(false); cin.tie(nullptr); int s; cin >> s; for (int i = 3; i <= 9; i++) dp[1][i] = 1; for (int i = 1; true; i++) { if (i > s / 3) break; for (int j = 3 * i; j <= 9 * i; j++) { for (int k = 3; k <= 9; k++) { dp[i + 1][j + k] += dp[i][j]; dp[i + 1][j + k] %= MOD; } } } int ans = 0; for (int i = (s + 8) / 9; i <= s / 3; i++) { ans += dp[i][s]; ans %= MOD; } cout << ans << "\n"; return 0; } 
•  » » 4 years ago, # ^ |   +1 The elements of the sequence can be arbitrary positive integers rather than digits.
•  » » » 4 years ago, # ^ |   0 Yes, I get it! Thank you so much!
•  » » 4 years ago, # ^ |   +1 Can be simplified using a 1D dp array. Starting from i=4, dp[i]=dp[i-1]+dp[i-3] and then take the mod.
•  » » » 4 years ago, # ^ |   0 Nice solution:D
 » 4 years ago, # | ← Rev. 2 →   0 Why do you have the problem whose solution is available in CPH? The most popular book on CP, talking about problem E.
 » 4 years ago, # |   0 Sry for a newbie question, how to we see the test cases at atcoder???
•  » » 4 years ago, # ^ |   +3
•  » » » 4 years ago, # ^ |   0 This page might not updated for a while... There are no test cases for last or last-to-last ABC
 » 4 years ago, # |   0 How to solve E??
 » 4 years ago, # | ← Rev. 4 →   0 Can someone plz tell me why my code for task C is failing for 5 test cases? #include using namespace std; typedef long long ll; #define mod 1000000007 ll help(ll a,ll b){ if(b==0)return 1; if(b==1)return a; ll ans = 1; if(b&1){ ans = help(a,b-1)*a; } else { ans = help(a,b/2); ans = ans * ans; } return ans%mod; } int main(){ ll n; cin>>n; ll x = help(10,n); ll y = help(9,n); ll z = help(8,n); ll res = x-2*y+z; // error in this line cout<
•  » » 4 years ago, # ^ |   0 I guess you forgot to take mod from the res at the end
•  » » » 4 years ago, # ^ |   0 I have changed it after you said, but still those same 5 test cases are failing. I'll update the latest code in my comment. Can you plz look into it again. Thanks!
•  » » » » 4 years ago, # ^ |   0 Since it is c++, mod can also give negative result, you need to handle this
 » 4 years ago, # | ← Rev. 2 →   +25 F can be solved by maintaining an invariant as follows: Hint1Let $count(x)$ denote the number of occurrences of $x$ in $B$ + that in $A$.If there's an $x$ where $count(x) > n$, then there is no solution. Else, there is.We need to construct a solution iteratively while maintaining that invariant, that there's no $x$ where $count(x)$ exceeds $n$. Note that $n$ is decremented each iteration, and also $count(x)$ represents the count of $x$ in the remaining elements in $A$ and $B$. Hint2Let us call $x$ critical, if $count(x) = n$.There can be at most 2 critical numbers. Let us assume there's only one, and it'll be easy to see how the same idea solves the other case.After an iteration, the invariant will be broken iff there's $x$ s.t. $count(x)$ exceeds $n$, hence in the previous iteration $count(x)$ was $n$ (i.e. $x$ was a critical number) and its $count$ didn't change. Hint3Hence, if the current iteration is $i$ and there's a critical number, then assign it to $b_i$, so that its count decrements in the next iteration and the invariant is maintained. But what if that critical number is equal to $A_i$? Then actually its count will be decremented too.So either way, you can maintain the invariant. If there's no critical element, just put any number and the invariant will be maintained anyway.Thanks to Yousef_Salama for showing me this idea, you can check his submission.
 » 4 years ago, # |   +13 I have written unofficial English editorial, you can find it at: editorial
•  » » 4 years ago, # ^ |   +1 great!!
 » 4 years ago, # |   0 For F, I am reversing B and then checking if B[i]=A[i]. If so at index=p, I am reversing the array from B[0] to B[p] and from B[p+1] to B[n-1]. This works as for the reversed B, all numbers left of B[p] will be >= B[p] and to right of B[p] will be <= B[p]. Whereas in A, all numbers left of A[p] will be <= A[p] and to right of A[p] will be >= A[p]. Theoretically it should work but I am getting wrong answer in 3 of the 60 test cases and I cant figure out why. Can anyone point out the fault in my algo? Submission: https://atcoder.jp/contests/abc178/submissions/16733148
•  » » 4 years ago, # ^ | ← Rev. 2 →   +8 Check this test case: 4 2 3 3 4 1 2 3 3 Correct answer: Yes Your answer: No 
•  » » » 4 years ago, # ^ |   +1 Thank you!
 » 4 years ago, # |   +43 Problem F is EXACTLY the same with Perm Matrix (just with smaller constraints).I've seen this problem,so I got an unexpected high rank.My atcoder account: sjcsjcsjc
•  » » 4 years ago, # ^ |   0 It's nothing new at atcoder, many time I got the exact questions which I had previously solved or seen somewhere. Problem E was also copied in the same contest.
•  » » » 4 years ago, # ^ |   +3 Problem E is just a standard problem of changing Manhattan distance to Chebyshev distance.Problems using this technique are everywhere.
 » 4 years ago, # |   0 Hi can someone please tell why my submission for problem C is working in atcoder c++ compiler but not in my local compiler.I have tried recursive dp solution for this problem and it is working well . But in some large input example for n = 869121 it gives segmentation fault in local compiler as well as in online cpp compiler but works well upon submitting can some please tell why this happening ? Can this be due to memory limit or recursion depth ? please help