### AhmedEzzatG's blog

By AhmedEzzatG, 4 months ago,

Hello, Codeforces!

Hope you enjoyed the round! Editorials for all problems are out now. I hope this editorial be informative and helpful.

Hint 1
Solution

Hint 1
Hint 2
Solution

Hint 1
Hint 2
Hint 3
Solution

Hint 1
Hint 2
Solution

## 1557E - Assiut Chess

Hint 1
Hint 2
Solution

• -34

 » 4 months ago, # |   0 Auto comment: topic has been updated by AhmedEzzatG (previous revision, new revision, compare).
 » 4 months ago, # |   +17 even though the round has many problems, but no one can disagree that the problems were great and beautiful.
 » 4 months ago, # |   +124 So there is no explanation on how this wrong solution passed the system tests?
•  » » 4 months ago, # ^ |   -237 ok. It's maybe wrong. but it's just hard to make an Interactor counter all such solutions I guess. We tried to kill all the wrong solutions of the testers. actually, there were only 2 testers who could solve the problem.
•  » » » 4 months ago, # ^ |   +107 Good idea,but weak interactor :(
 » 4 months ago, # |   +10 I think using math formula in C more beautiful and easier than editorialMy solution using some binomial theorem and flt 125416610It faster than editorial solution
•  » » 4 months ago, # ^ |   +3 can you explain what did you do?
•  » » » 4 months ago, # ^ | ← Rev. 3 →   +31 $\binom{n}{0}+\binom{n}{2}+ \cdots = 2^{n-1}$You can calculate this by using Binomial Theorem, which you subtract $(1+1)^n$ with $(1-1)^n$
•  » » » » 4 months ago, # ^ | ← Rev. 2 →   +4 However, there's an exception. When $n=0$ , the left part is 1. That's because $(1-1)^n$ will become undefined. Don't ignore that during exam :)
•  » » 4 months ago, # ^ |   0 explain please, I solved it pure math too, you can look at my 125416186 to see what things I already understood so that you could skip them if you want to.
•  » » 4 months ago, # ^ | ← Rev. 2 →   0 flt is Fermat's little theorem?
•  » » » 4 months ago, # ^ |   0 Ah, I got it. To calculate mod divide you have to find the inverse of the denominator where Fermat's little theorem is used.
•  » » » » 5 weeks ago, # ^ |   0 thanks
•  » » 4 months ago, # ^ |   0 thx!
•  » » 3 months ago, # ^ |   0 Dude, your logic for when n is even was really funny. Good one!
 » 4 months ago, # |   0 can somebody tell me abt my WA https://codeforces.com/contest/1557/submission/125416469
•  » » 4 months ago, # ^ |   +7 m[max_ele] = 0and 0 can be a valid element in your input array. so thats reducing your cnt1 by 1 in test 3. and hence a WA.
•  » » » 4 months ago, # ^ |   0 thnks broo
•  » » 4 months ago, # ^ |   0 If input array have duplicates your m[a[i]] will be over written and hence some data is lost this will produce an error suppose that array is 2 2 2 2 then always a[i]=2 and m[2] will be over written 3 times and eventually will store the last value and previous will be lost
 » 4 months ago, # |   +12 Back to back segtree problems for the past 3 rounds.
 » 4 months ago, # |   +23 Can you share the interactor as well? Would be interesting to know how does it decides where to move the king.
•  » » 4 months ago, # ^ |   +21 We try a lot of strategies for each cell, some greedy (for example, we try to make the king stay in the mid cells, Or try to go Up\downs if possible), and others random. maybe I will share the code later.
 » 4 months ago, # |   +83 Randomized approach for E: First, I put my queen at one of ${(4,4),(4,5),(5,4),(5,5)}$(random choice) For each move, I calculate $S$, which is the set of square which have a possibility of king's square. The size of $S$ is going small with the progress of the game. Next, I make some random horizontal or vertical move. As a reminder, through this progress, $S$ is going small. The important thing is, when we know where's the king, we can checkmate in $8$ turns. For example, if we have only $5$ candidates of king's square, we can finish the game in additional $40$ turns. When $S$ becomes small enough, try all candidates. It seems that this solution uses far less than $130$ queries, or can you kill my solution?my code
•  » » 4 months ago, # ^ |   +71 We can actually checkmate the king in 5 (4 if the queen stays out of the edges) moves by checkmating on the edge!I think that can get rid of the randomization part as if we place the queen at (4, 4), there are at most 31 squares the king can initially occupy. Thus, we have an upper bound of 4 * 31 = 124 queries.However, in practice, the solution generally uses <10 queries.Code: 125423506
•  » » » 4 months ago, # ^ |   -8 Whoa this is genius O_o
•  » » » 3 months ago, # ^ | ← Rev. 2 →   +5 Just in case this helps anyone, I tried to implement a cleaner and commented version of the code above. The zig-zag idea is really nice!127869222
 » 4 months ago, # |   +1 Can someone elaborate what are the leaves of seg tree used in D? If rows would have up to 10^5 columns the seg tree would be trivial, but with 10^9 I wasn't able to come up with seg tree structure and editorial didn't help neither.Even the "dpi,j = maximum number of rows (from row 1 to row i) that make a beautiful grid, and has 1 in column j at the last row." doesn't make much sense to me, as there can be up to 10^9 j indexes.
•  » » 4 months ago, # ^ | ← Rev. 3 →   +3 yes they didn't wrote it explicit, but in the end they wrote you can use coordinate compression.that would reduce columns from $10^9$ to $2m$. Exampleyou have segments: $[1,10],[2,20],[3,30]$you can process them as $[1,4],[2,5],[3,6]$i will still yeild the same answer for the problem.and you only $\text{dp}[i-1]$ to compute $\text{dp}[i]$ . so you can use one single row of $2m$ size and use segment tree to maintain it. all m updates in $\log{n}$ time.i did got the idea in the end and also implemented but i am getting WA at test 4.if anyone can help be debug.
•  » » » 4 months ago, # ^ |   +3 I used the same approach and got AC. I think you made a mistake in the update function. While updating the segment tree we need to check whether a higher value is updated to lower value. 125463800
•  » » » » 4 months ago, # ^ |   0 thanks, yes i did that check with ppgt() function which will propagate any changes that were made along the path we need for query/update, i missed a part in ppgt() to reset current root's lazy value to null after upddting its children. silly mistake, took me hours to find out.
 » 4 months ago, # |   -26 Solution for first three questions. https://www.youtube.com/watch?v=vBqSLVoXYsI
 » 4 months ago, # |   +3 Love this round. Hoping to have more lovely rounds like this in the future. (◍•ᴗ•◍)❤
 » 4 months ago, # |   +4 My heart starts bitting faster every time I see 5 problems in a contest. I was at the top of my abilities to solve C today
 » 4 months ago, # |   -11 SegmentTree for D is such an overkill.
•  » » 4 months ago, # ^ |   +11 Can you please brief your approach without segment tree for D.
•  » » 4 months ago, # ^ |   0 Do you mean you can do it without segment tree?
•  » » » 4 months ago, # ^ |   +32 We can get every set of rows $i$ that for each column $j$, $grid_{i,j}=1$. Then construct a DAG by connecting the edges of each row in the sets from the first to the last, and find the longest path on the graph. The comlexity is $O(n+m)$.
•  » » » » 4 months ago, # ^ |   0 How did you do it without a log factor?
•  » » » » » 4 months ago, # ^ |   0 Yes you are right. It's difficult to get rid of a log factor on the work before DP. I mean you can find the longest path on the DAG in $O(n+m)$.
•  » » » » » » 3 months ago, # ^ |   0 how does this work tho? doesnt longest path in dag have complexity O(V + E), how do you make sure there aren't too many edges cause then i would assume it becomes O(N^2). I looked at your code and it looks like there are only N edges but I don't get how something like that would work unless you are compressing the dag?
•  » » » » 4 months ago, # ^ |   +3 Can you explain how to generate the graph in O(n+m)?
 » 4 months ago, # | ← Rev. 2 →   0 Hello , I didn't get the logic behind 2nd problem . https://ide.geeksforgeeks.org/8684Y9kbwQ this is my idea . But got wrong output on pretest 2. Can anybody help me in this ?
•  » » 4 months ago, # ^ |   0 I'd suggest you clean out your code a little before asking for help!
•  » » » 4 months ago, # ^ |   0 Sorry . You can check this one .https://ide.geeksforgeeks.org/8684Y9kbwQI have cleaned unnecessary things.
•  » » » » 4 months ago, # ^ |   0 What is your thought process exactly for this question?
•  » » » » » 4 months ago, # ^ | ← Rev. 6 →   0 First I determined the number of increasing subarrays and stored each and every segments' first and last number in a vector (vv). Then , checked if the number is greater than given number (m) or not . If yes , then output (NO) [the number i determined , that is the minimum number of breaks]. else { sort the vector (vv); then checked vv[i].first > vv[i-1].second (i>0) , because only then we can merge the whole thing and get sorted array (final and).If at any point , this condition gets violated , simply output(NO). else at the end print(YES).}AhmedEzzatG, Can you check why it is showing wrong answer ?
•  » » 4 months ago, # ^ |   0 Try for this testcase 5 3 2 5 4 7 8Output:No
•  » » » 4 months ago, # ^ |   0 Yeah , it's giving NO
 » 4 months ago, # | ← Rev. 3 →   0 Why do I get WA on pretest 2? https://codeforces.com/contest/1557/submission/125391145 Brief explanation: If $n$ is odd, the answer is just $(2^{n-1}+1)^k$. Otherwise, if $A_n(j)$ is the number of ways for the bits from 0 to $j$ such that $And\geq Xor$, I get the recursion $A_n(j+1)=2^{nj}+2^{n-1}A_n(j)$, which solves to $A_n(k)=2^{nk-k}+2^{kn-n+1}-2^{nk-n-k+2}$. Where is the flaw?EDIT: found it, wrong recursion, it should be $A_n(j+1)=2^{nj}+\left(2^{n-1}-1\right)A_n(j)$
 » 4 months ago, # |   0 AhmedEzzatG Do you mind looking into why this solution fails on D? I used the approach described in the editorial during the contest, but really have no clue why the 4th pretest is failing. It's giving me headache. Thanks.125401562
•  » » 4 months ago, # ^ |   +41 Hi, sorry for delay. you have a small bug in segment tree code in line 68. return tree[v] = max(query(2 * v, l, m, ql, qr), query(2 * v + 1, m + 1, r, ql, qr));it must be return max(query(2 * v, l, m, ql, qr), query(2 * v + 1, m + 1, r, ql, qr));
•  » » » 4 months ago, # ^ |   +13 Thank you so much, I appreciate it. A dumb mistake that yet again cost me -200 delta.Regardless, I liked D very much (the only problem I attempted, so I can't say for others) and I had a great time solving it, although it gave me some trouble. Thanks for the round!
 » 4 months ago, # | ← Rev. 3 →   +121 really quick pure combinatorics sol for C:We'll look at each bit independently. The bit for the LHS will be 1 iff all the bits are 1 and 0 otherwise. For the RHS, the bit will be 1 iff the number of bits that are ones is odd, and 0 otherwise. In order to handle both at the same time, you need to split into cases based on the parity of $n$.Suppose $n$ is odd. Then, you can never have the case that the LHS is strictly greater than the RHS, because if the AND is equal to 1, then the number of bits that are 1 is odd, meaning the XOR is 1. Thus, you must have equality. If both the LHS and RHS is 1, then it must be all ones so that is 1 way. Then if both sides are equal to 0, we must have exactly an even number of ones, or $\binom{n}{0}+\binom{n}{2}+\cdots + \binom{n}{n-1}$which is equal to $2^{n-1}$. Thus, in the odd case, the answer is $\left(2^{n-1}+1\right)^k$.Now suppose $n$ is even. This time, it is possible to have the strict inequality by taking all ones. You then must consider two cases: when we have equality for both sides and when we have the strict inequality.We tackle the first case. Both bits can not both be equal to one, as that would require all the bits to be one, for an even total, and an odd number of ones, contradiction. Thus we need both to be 0, meaning not all bits are equal to 1, and we must have exactly an even number of ones. This corresponds to $\binom{n}{0}+\binom{n}{2}+\cdots+\binom{n}{n-2}$Again, this can be seen to be equal to $2^{n-1}-1$ through whatever method you want. Thus, there's a total of $\left(2^{n-1}-1\right)^k$ arrays in this case.Now suppose we have a strict inequaltiy. What we'll do is fix the bit where we have a 1 on the LHS and 0 on the RHS, and iterate over the positions for that bit: suppose the $i$'th bit will satisfy this. For the $i-1$ bits before it, we must have equality, which gives $\left(2^{n-1}-1\right)^{i-1}$ from the analysis of the previous case. Then, we have exactly 1 way to get the $i$'th bit, which is to take all ones. Then, the remaining $k-i$ bits don't matter, so there are $\left(2^n\right)^{k-i}$ choices here. Putting both cases together and summing over all $i$, we have $\left(2^{n-1}-1\right)^k+\sum_{i=1}^k \left(2^{n-1}-1\right)^{i-1}\cdot \left(2^n\right)^{k-i}$different arrays in the case that $n$ is even. Compute this now however you want.If you wanted a closed form for this, you can see that the summation is actually equal to $\frac{\left(2^n\right)^k-\left(2^{n-1}-1\right)^k}{2^{n-1}+1}$where you use the factorization $a^k-b^k=(a-b)\cdot\left(a^{k-1} b^0 + a^{k-2}b^1 + \cdots + a^0b^{k-1}\right)$EDIT: sorry for the typoes oops as you can tell i am terrible at proofreading
•  » » 4 months ago, # ^ | ← Rev. 2 →   0 hey, I implied your method: can you tell me, why I'm getting the wrong answer my submission: 125428860could you please amend it as well?
•  » » » 4 months ago, # ^ |   +26 if n is even,it should be $\left(2^{n-1}-1\right)^k+\sum_{i=1}^k \left(2^{n-1}-1\right)^{i-1}\cdot {(2^n)}^{k-i}$
•  » » » » 4 months ago, # ^ |   -114 Yes, of course... thanx This dickhead has wasted my 3 hours posted the wrong solution. And I was thinking something is wrong with my modular template.
•  » » » » » 3 months ago, # ^ |   +3 something is wrong with your mindset.
•  » » 4 months ago, # ^ | ← Rev. 2 →   +4 In the 4th paragraph, "we must have exactly an even number of zeroes" => "we must have exactly an even number of ones"
•  » » 4 months ago, # ^ |   +12 Nice comment, however be careful with the case: $2^{n - 1} + 1 \equiv 0 \mod M$. In this problem it won't occur, but it could happen if $M = 998244353$, for example: $2^{249561088} + 1 \equiv 0 \mod 998244353$and the answer is $2^{n k} + k \cdot 2^{n(k - 1)}$ for this special case.
•  » » » 4 months ago, # ^ |   -8 How did you conclude that for M = 1e9 + 7, ((2^(n-1) + 1) mod M) won't ever be 0. And why is the answer 2^{n k} + k * 2^{n(k — 1)} for this case?Also, I'd be glad if anyone could recommend learning resources (preferably text) for the same.
•  » » » » 4 months ago, # ^ |   0 compute all $2^n, n = 0, \cdots M - 1$ or some primary number theory.You may need to learn markdown, mathjax or tex.
•  » » » » 4 months ago, # ^ | ← Rev. 2 →   +13 Use contradiction.Assume that there exists an $n$ such that $2^n+1\equiv 0\bmod 10^9+7$Assume that $n$ is the smallest root of that equation.Assume that $m$ is the smallest root of the equation $2^m-1\equiv 0\bmod 10^9+7$It's easy to see that $m|10^9+6$ and $m\mid 2n$ but $m\nmid n$.So we can know that $2|m$ and $\frac{m}{2}\mid n$ and $\frac{2n}{m}$ is an odd number.It's easy to see that $2^\frac{m}{2}\equiv \pm 1\bmod 10^9+7$However $2^n=(2^\frac{m}{2})^\frac{2n}{m}$So we can know that $2^\frac{m}{2}\equiv -1\bmod 10^9+7$.According to $m|10^9+6$ we can know that $\frac{10^9+6}{m}$ is an odd number.So we can know that $2^{5\times 10^8+3}\equiv -1\bmod 10^9+7$In other words $(\frac{2}{10^9+7})=-1$However $(\frac{2}{p})=(-1)^{\frac{p^2-1}{8}}=1$Contradiction.Sorry for my bad English :(
•  » » 4 months ago, # ^ |   +3 hey, can u tell me that in your even n case, in sub-case of inequality, while setting i bit as most significant, for k-i bits answer should be (2^n)^(k-i) because we do not have to set any specific bit in all n numbers as same (in those k-i bits)..and in your case of 2^(k-i), we imply that ith bit is same in all n numbers(in those k-i bits), which is not necessary..
•  » » 4 months ago, # ^ |   0 Could you explain how exactly did you arrive at the closed form equation?
•  » » » 4 months ago, # ^ |   0 It's is a geometric series so one can get it by applying the formula for summing a geometric series and then simplifying; however I actually got it by working through all the steps: $f=\sum_{i=1}^{k}\left(2^{n-1}-1\right)^{i-1}\left(2^{n}\right)^{k-1}$Divide by the ratio of consecutive terms (one would normally multiply, but dividing seems simpler here): $\frac{2^{n}}{2^{n-1}-1}f=\sum_{i=0}^{k-1}\left(2^{n-1}-1\right)^{i-1}\left(2^{n}\right)^{k-1}$Take the difference between the sum from 1 to k and the sum from 0 to k-1: $\left(\frac{2^{n}}{2^{n-1}-1}-1\right)f=\frac{\left(2^{n}\right)^{k}}{2^{n-1}-1}-\left(2^{n-1}-1\right)^{k-1}$Simplify: $\left(2^{n}-\left(2^{n-1}-1\right)\right)f=\left(2^{n}\right)^{k}-\left(2^{n-1}-1\right)^{k}$ $\left(2^{n-1}+1\right)f=\left(2^{n}\right)^{k}-\left(2^{n-1}-1\right)^{k}$ $f=\frac{\left(2^{n}\right)^{k}-\left(2^{n-1}-1\right)^{k}}{\left(2^{n-1}+1\right)}$
 » 4 months ago, # |   -6 could some one tell me why this is wrong https://codeforces.com/contest/1557/submission/125421372
•  » » 4 months ago, # ^ |   +3 You could have a contiguous subarray but you might have a value that splits that contiguous array into parts E.g- 1 2 3 7 4 and k=2 your solution would give yes but the answer is no because 1.2.3.7 is contiguous but 4 needs to be put in between 3,7 so it needs 3 subarrays
 » 4 months ago, # |   0 Could someone explain why time limit is exceeded on this, https://codeforces.com/contest/1557/submission/125366350
•  » » 4 months ago, # ^ |   0 Don't use files. Use doubles instead of floats. Don't split the input string in a loop. Fixed solution: 125475544
•  » » » 4 months ago, # ^ |   0 Okay that makes sense, thank you
•  » » » » 4 months ago, # ^ |   0 Np, have a nice day!
 » 4 months ago, # |   0 PLEASE help with this I was unable to find my mistake WA on testcase 3https://codeforces.com/contest/1557/submission/125378949
•  » » 4 months ago, # ^ |   0 try case 1 4 7 3 with K =2 , i think your solution would give yes but the answer is no
•  » » » 4 months ago, # ^ |   0 thats not the case .... anyhow I got my mistake ->it is map is initialized with 0 for the last element in sorted it should be initialized with a element we cannot expect in the array I now changed it with INT_MIN _>it worked THANKS FOR REPLLYING
 » 4 months ago, # |   +8 I think it's meaningless to set the space of problem D so tight…… It's hard for $O(m\log10^9)$ to get pass, since it's easily getting MLE.But also great problems, thanks for the contest!
•  » » 4 months ago, # ^ | ← Rev. 2 →   -28 Discretization please! Codeforces Round NEVER set space limits tightly!
•  » » » 4 months ago, # ^ |   +13 But a time complicity of $\mathcal{O}(n + m\log{}10^9)$ was given by the problem writer as a accepted one.
•  » » » » 4 months ago, # ^ |   +5 I think #define int long long is the culprit here. Maybe try removing that line and see it fits inside the memory limit or not.
 » 4 months ago, # | ← Rev. 2 →   +24 I have a different solution for E.In the beginning the possible positions for KING is the whole chessboard. And if the king do a move , some positions will become impossible. So you can do $30$ random walk , to make the possible positions become very few(may be greater than one).And for each possible position , you try to checkmate it . You can do it for a known position in at most 8 steps : move the queen to that row . move the king. if the position is in your left , move left and move to the row where the king is ,vice versa. And remember to maintain the possible positions after every move .It seems that it use only about total $40$ moves for a test case.code
•  » » 4 months ago, # ^ |   0 we cannot rely on random walks what if the king is oscillating back and fro in two squares and doing random walks yr queen doesn't give check I know that probability of this is less but is not 0 so this approach can fail once in a million
•  » » » 4 months ago, # ^ |   0 In fact if you try to checkmate a position , you can also find some positions impossible .So I think it is impossible to be hacked .
•  » » » » 4 months ago, # ^ |   0 if the king only oscillates in two squares then the possible positions will be 64-16=48 and if u are unlucky enough u can exhaust 130 moves
•  » » » » » 4 months ago, # ^ | ← Rev. 2 →   0 I mean if you try to checkmate the positions one by one , you will continue finding some impossible positions .
•  » » » » » » 4 months ago, # ^ |   0 yaa makes sense
 » 4 months ago, # |   +10 I find the interactor for E a little "dumb": in all testcases, I made no more than 20 moves to kill the king lol (if the output of the testcase is the number of moves)
 » 4 months ago, # |   0 Thanks a lot for a detailed editorial.
 » 4 months ago, # |   0 problem E was super cute and I recommend everyone to just take a look at it :)
 » 4 months ago, # |   -7 Well, D is a good problem. I solved many problems that can directly construct schemes through dp arrays before. But problem D ISN'T such problem, so I got Wrong answer on pretest 4. This is a good reminder to me that the record transfer is used to construct the scheme, rather than backward pushing according to the dp array.But I'm still disappointed that I didn't passed D in the round. I am so close to the correct solution!
 » 4 months ago, # | ← Rev. 2 →   -14 why I'm getting the wrong answer my submission: 125428860could you please amend it as well?
 » 4 months ago, # |   +1 thank you for the nice editorial
 » 4 months ago, # |   0 What would be the rating of Div2C question?
•  » » 4 months ago, # ^ |   0 1700 maybe
•  » » » 4 months ago, # ^ |   0 Yeah. Thanks!
 » 4 months ago, # |   +13 AhmedEzzatG Thank you, Sir, The problems were really Good, We can learn a lot from them.
 » 4 months ago, # |   +3 i did this in D problem: if the first and second row have something in common(which I am precomputing) then we can just move on to next row and set previous column as current index and if they don't have anything common, then we have to find the minimum of the two cases: removing the first row and removing the second row. So I call for two functions, with previous values as current index for one and the previous index for the other.Recursive Code: int fn(int prev, int cind, set> adj[]) { if(cind >= n)return 0; if(dp[cind] != -1)return dp[cind]; if(isMerging(cind,prev)) return dp[cind] = fn(cind, cind + 1, adj); return dp[cind] = min(fn(cind, cind + 1, adj),fn(prev, cind + 1, adj)) + 1; } But I am doing isMerging step in O(n), can we do it in less than that? Also, I was not able to print the rows deleted.
 » 4 months ago, # |   0 on what concept problem c is based upon.?
•  » » 4 months ago, # ^ |   0 Dynamic Programming and Combinatorics
 » 4 months ago, # |   0 Can Anyone please figure out mistake in my C- problem code.... void q3(){ int n,k;cin>>n>>k; int ans = 1; int y = 1; y =power(2,n-1,mod); if(n&1){ int f = power(1+y,k,mod); pl(f);// just for printing.... }else{ y--; for(int x = 0;x<=k-1;x++){ ans = (ans*y)%mod; ans = (ans+mod+power(2,x,mod))%mod; } pl(ans);// just for printing.... } } 
 » 4 months ago, # |   0 hello everyone In Problem B is showing runtime error in testcase 2 but the code seems to be fine have a look guys. https://codeforces.com/contest/1557/submission/125441413 kindly help if you can. Thanks
•  » » 4 months ago, # ^ | ← Rev. 2 →   0 See you are not taking the vector as input when k is equal to n, therefore it results in a runtime error but even though your approach is wrong.15 41 2 4 3 6for the above input, your count is 2 but it should be 4 to make it sorted.
•  » » 4 months ago, # ^ |   0 even if n==k u have to take whole input :-)
 » 4 months ago, # | ← Rev. 2 →   +3 Problem C Video EditorialI solved it in contest and I show my thought process so it might help some of you
 » 4 months ago, # |   0 can someone help me explain why I am getting TLE on test 78 when there are 64*60 operations at max Link to submission : https://codeforces.com/contest/1557/submission/125445885
 » 4 months ago, # |   0 In the solution for question C, how did he/she make the base case: dp(-1, 0) = 1 and dp(-1, 1) = 1?
•  » » 4 months ago, # ^ |   0 not sure, I am using dp(-1,0)= 0, and dp(-1,1)= 1 (Java solution)
 » 4 months ago, # |   0 Can someone explain my WA? All my tests is AC, but here WA on pretest 2. https://codeforces.com/contest/1557/submission/125395144
•  » » 4 months ago, # ^ |   0 try for 14 32 4 3 5
•  » » » 4 months ago, # ^ |   0 it should print "NO"
•  » » » » 4 months ago, # ^ |   0 I understand, thank you.
 » 4 months ago, # |   0 How my approach is wrong for the submission of 1557B — Moamen and k-subarrays problem ? Please help 125447732
•  » » 4 months ago, # ^ |   0 try for14 32 4 3 5it should print 'NO'
•  » » 4 months ago, # ^ | ← Rev. 2 →   0 Got it now thanks spidermonkey
 » 4 months ago, # |   +42 Hey, In problem A we proved that taking the maximum number in one subsequence is better than taking the two greatest numbers (instead of only one) in one subsequence.My question is how we can generalize this argument for the whole problem as there might be the other cases except for taking the two greatest number.AhmedEzzatG
•  » » 4 months ago, # ^ |   +40 I proved it a bit differently:Suppose the array is split into 2 sub-sequences: $A$ with $k$ elements and sum $S_A$, and $B$ with $n-k$ elements and sum $S_B$, and further assume that $k \leq n-k$.Step 1. We can show that if $x \in A$ and $y \in B$, then $x \geq y$.Proof by contradiction. Assume that $y > x$, then we can swap $x$ and $y$ to get a larger result: $\frac{S_A - x + y}{k} + \frac{S_B - y + x}{n-k} = \frac{S_A}{k} + \frac{S_B}{n-k} + (y-x)(\frac{1}{k} - \frac{1}{n-k}) > \frac{S_A}{k} + \frac{S_B}{n-k}.$Step 2. If $A$ has more than 1 element, we can move the smallest element of $A$ to $B$ and increase the result. Let $x$ be the smallest element of $A$. By the previous step, $x$ is larger than any element of $B$, so after moving $x$ from $A$ to $B$ both the average of $A$ and average of $B$ will increase.
•  » » » 4 months ago, # ^ |   0 I had the same problem with the editorial proof, but I easily understood yours. Truly elegant. Thanks
•  » » » » 4 months ago, # ^ |   0 Editorial "proof" should be in quotes. :)
•  » » » 4 months ago, # ^ |   +13 One interesting observation — if both subsequences have the same length ($k = n - k$) then the result does not depend on how we do the split and is equal to twice the total average.
•  » » » 4 months ago, # ^ |   0 Thanks for the help.You are amazing!!
•  » » » 4 months ago, # ^ | ← Rev. 3 →   0 The proof is not so trivial to be arrived at during the contest for div2 problem A. So, I think, most of the participants just made a guess and hoped for the best.
•  » » » » 4 months ago, # ^ |   0 On the other hand, it's still problem A of div2, so if you get a hunch and go over some examples to "convince" yourself that it's the right observation, it's probably OK to trust your intuition and submit fast, and only try to prove it formally later.
•  » » » » » 4 months ago, # ^ |   0 For most problems of this level the challenge is to figure out the answer, and once you've done it the proof is trivial.
•  » » » 4 months ago, # ^ | ← Rev. 3 →   +5 A correction — the statement in the first step is true only if $k$ is strictly less than $n - k$ (otherwise $\frac 1 k - \frac 1 {n-k} = 0$).If $k = n - k$ the result is the same no matter how we split, and if $x \geq y$ for any $x \in A$, $y \in B$ it is as good as any other split.
•  » » » 4 months ago, # ^ |   0 Thank you, bro. Very helpful!
 » 4 months ago, # |   0 I do three problems only just a little.~~~QAQ
 » 4 months ago, # |   +1 Problem C，forgot to add enough mod,and failed twice.100 points T T
 » 4 months ago, # | ← Rev. 2 →   0 In problem D,why adding $dp[i][j] = max_{k\in C_i} {dp[i - 1][k]}$ step as call modify(1,1,N - 1,mx) is useless and wrong? In my opinion,tutorial's definition points out that $grid[i][j] = 1$ so the transfrom $dp[i][j] = max_{k\in C_i} {dp[i - 1][k]}$ is not exist.And that's why add modify(1,1,N - 1,mx) will casue wrong?:/
 » 4 months ago, # |   +31 In Problem E, this is a false promise: If at any point you make an invalid query or try to make more than 130 queries for each test case, the game will terminate immediately and you will receive a Wrong Answer verdict. after 30+ submissions of getting RTE and TLE I realized that my code making a move of the type : $(6, 1)$ -> $(6, 1)$. I should have got WA verdict but didn't.
•  » » 4 months ago, # ^ |   0 But I got a "Wa" verdict. 125371395
•  » » » 4 months ago, # ^ |   0 It is a bit random. I got many TLEs. There is no reason I should get TLE.
•  » » 4 months ago, # ^ |   +17 I also wasted a lot of my time trying to debug why my code was getting TLE, when it should be getting WA. Honestly, this E should have been tested more, I don't know how it got published with such a weak and inconsistent interactor
 » 4 months ago, # |   0 Are the n and k the opposite way round in solution C than they are in the question?
 » 4 months ago, # |   0 Hello and Hi!, I am new to DP and tried to write a solution for C after contest. My (wrong) approach is as follows: If $k$ is $0$ print $1$ otherwise define a $dp$ such that $dp[i]$ is the number of ways to arrange bits till the $i$ th position and win. Thus, $dp[0] = \begin{cases} 2^{n-1} & \text{if$n$is even} \\ 2^{n-1} +1 & \text{otherwise} \end{cases}$Now I have $3$ possible cases when Moamen wins at $i$ th bit. When the bitwise-and and xor of $i$ th bits are both $1$, when they are both $0$ and when they are $1$ and $0$ respectively. I deduced a formula using these $3$ cases $dp[i] = \begin{cases} 2^{ni-n} + dp[i-1] \cdot (2^{n-1}-1) & \text{if$n$is even} \\ dp[i-1] \cdot (2^{n-1} +1) & \text{otherwise} \end{cases}$I have read the edi and also saw the combinatorial method in the comments but cannot figure out the falsity of my method. Please do tell.
•  » » 4 months ago, # ^ |   0 Did you figure it out?
•  » » » 4 months ago, # ^ |   0 No :(
 » 4 months ago, # |   0 I can't understand "You can make i-th bit = 1 in an even number of indices, then Andi will be 0 and Xori will be 0 too." only one position is 1, then the xor could 1 why is 0?
•  » » 4 months ago, # ^ |   0 maybe it means you choose even number of 1
•  » » » 4 months ago, # ^ |   0 i know it means by the author's code,but thank you
 » 4 months ago, # |   +3 For Problem C, I think the editorial is wrong because $dp[i][0]$ is just a product of $2^n$ which corresponds to all possible cases.Let's define two vectors of size $k$ : allcases[i] : number of ways to buid an array of size n from bit $0$ to bit $i$ (included) dp[i] : number of ways to buid an array of size n from bit 0 to bit i s.t. (a[1]&...&a[n])[i...0]>=(a[1]&...&a[n])[i...0] with (x)[i...0] = $2^i x[i] + ... + 2^0 x[0]$ Then for allcases: allcases[0] = $2^n$ allcases[i] = $2^n$ allcases[i-1] And for dp: n is odd. two cases for bit i: all 1 : AND=1, XOR=1 even number of 1 : AND = 0, XOR = 0. number of cases : C(n,0) + C(n,2) + ... + C(n,n-1) dp[i] = (1 + number of cases) * dp[i-1] n is even. two cases for bit i: all 1 : AND=1, XOR=0 then for the bit 0...i-1 we can use all possible cases i.e. allcases[i-1] even number of 1 (except n) : AND = 0, XOR = 0. number of cases : C(n,0) + C(n,2) + ... + C(n,n-2) dp[i] = allcases[i-1] + (number of cases) * dp[i-1]
•  » » 4 months ago, # ^ |   0 @ustaritz in both the odd and even cases, why not also consider all 0. if all 0 , XOR = 0 and AND = 0. That is also a valid case, right?
•  » » » 4 months ago, # ^ |   0 The all 0 case is considered. It corresponds to an even number of 1 equal to 0.
 » 4 months ago, # |   0 I got WA on test 43 125484134 , test 43 seems random...., can somebody tell me why
•  » » 4 months ago, # ^ |   0 I just change min() to max() and AC,why
 » 4 months ago, # |   +2 I dont know what someone expected for when trying to copy code to on top of the leaderboard of virtual participation. Lmao
 » 4 months ago, # |   0 I hope that each example of the problems can be more valuable.
 » 4 months ago, # |   -8 To be honest，the author's solution expression is not so clear,it give me some troubles. I don't know what some nouns mean, no clear explanation ，as a newboy i really sad。
•  » » 4 months ago, # ^ |   -8 Build a segment tree of pairs (value, index) initially with { 0 , −1 }.i want to know what’s the value's mean,index's mean and what they can do.having many unclear expressions,it's not only me can't understand.I can give some suggestions? I hope the author can explain clearly what those nouns can do in solutions. gaussing the means by reading code is not a good thing.
•  » » » 4 months ago, # ^ |   -8 And the author's code have many details,it really wast time
 » 4 months ago, # |   0 In the bit manipulation question for the odd number case , I am having 2 approaches : dp[k] = (y+1) *dp[k-1]; -(1) instead of dp[k] = (y*dp[k-1]) + dp[k-1] ; — (2)where 'y' = nc0 + nc2 + nc4 + ....+nc(n-1),basically number of ways we can have even 1s.The 1st one is giving wrong answer, the 2nd equation is giving right answer, aren't they both the same,or is it because of some modulo properties,it is changing?The test case where it is failing :5 200000 200000 24567 23423 1 200000 200000 1 200000 0My Code : https://ideone.com/RZAe22
 » 4 months ago, # |   0 Could somebody take a glance at this: 125611384Do you have any idea, why this getting TLE?
 » 4 months ago, # |   0 AhmedEzzatG For your dp solution of problem C, when n is even, why dp[i][1] adds dp[i-1][0] in the end? I thought the definition of dp[i][1] is the number of ways from 0-th bit to i-th bit to build And = Xor, then if dp[i][1] adds dp[i-1][0], how's the latter cntEven * dp[i-1][1] gonna work? really confused.
•  » » 3 months ago, # ^ | ← Rev. 2 →   0 The editorial for problem C seems to be incorrect and definition of dp[i][] was also quite vague. Lost so much time due to this. Anyway, dp[i][0] is for AND > XOR for the ith bit only i.e AND[i] > XOR[i], where AND[i] means ith bit of AND and vice-versa for XOR. Similarly for dp[i][1] AND = XOR for ith bit only. Also one correction that dp[i][0] = 2*n * dp[i−1][0] should actually be dp[i][0] = 2^(n*i).
 » 4 months ago, # |   +3 In the solution of D it said: $dp_{i,j}$ — maximum number of rows (from row $1$ to row $i$) that make a beautiful grid, and has $1$ in column $j$ at the last row. What does it mean here "last row"? Is it the row number $i$ or it is the last row of the biggest beautiful grid?If it is the row number $i$ then the following doesn't make sense. It should be just 0. Otherwise, $dp_{i,j} = max_{k\in C_i}{dp_{i-1, k}}$. Otherwise $dp_{i, j}$ should somehow depend on $j$, but the right hand side doesn't contain $j$ in any way
•  » » 4 months ago, # ^ |   +3 The tutorial of problem D truly has confused me a lot. Now I've figured it out: Is it the row number i or it is the last row of the biggest beautiful grid? The last row is not necessarily $i$ or $i-1$, it means the last row of the biggest beautiful and in that row at index $j$ it stores $1$.The Otherwise part is also not correct. It should be $dp_{i,j}=dp_{i-1,j}$
 » 4 months ago, # |   +22 Why is this blog getting so many negative votes, Codeforces community doesn't appreciate the authors' effort!
•  » » 4 months ago, # ^ |   -21 That's just because the terrible interactor of problem E.I don't know any other reasonable causes.
•  » » » 4 months ago, # ^ |   +19 I'm really sorry that this happened but we did everything we could to make all the wrong solutions fail. But it is difficult to find a general way for all solutions. But now I've worked on it and it's a bit better. We apologize once again for that.
•  » » 4 months ago, # ^ |   +2 Because the tutorial is badly written... and some of the problems' pretest is tremendously STRONG.
 » 4 months ago, # |   +5 In the tutorial of problem D, the following statement is not correct: Otherwise, $dp_{i,j}=\max{..}$ Should be: If $grid_{i,j}\not=1$, $dp_{i,j}=dp_{i-1,j}$.
•  » » 4 months ago, # ^ |   0 Yes, I am confused about this, thanks.
•  » » 4 months ago, # ^ |   +5 Sorry for this mistake, I have modified it now.
 » 4 months ago, # |   0 Is the solution for problem A wrong? I think we can not sum things up like that
•  » » 4 months ago, # ^ |   0 What do you mean?
•  » » » 4 months ago, # ^ |   0 If we devide sequence to the biggest number and the others. He proved that moving second biggest num to the group of biggest number wont have better solution. But this does not mean its true for other cases, I know there is another solution, you can read comments above
•  » » » » 4 months ago, # ^ |   0 Yes, this has been pointed out above.
•  » » » » 4 months ago, # ^ |   +1 You can generalize the proof to any size of the sequence, not just 2. However, I think the proof in this comment is better than the editorial proof.
•  » » » » » 4 months ago, # ^ |   0 Maybe you can, but you did not. :)
 » 4 months ago, # |   0 The first edi that has -100 downvotes :/
•  » » 4 months ago, # ^ |   0 This is not the worst, the worst is the announcement of a contest with several hundred downvotes :/
 » 4 months ago, # |   0 Can somebody please help me figure out, why my code, which quite literally implements the solution described in the editorial (except that I am scanning from left to right instead of top to bottom), not work? It gives runtime error because my queen somehow enters the 8th column, whereas it should be able to checkmate while it is in the 7th column. Any help in figuring out the error (either in the solution or the implementation) would be appreciated. Code#include using namespace std; #ifdef LOCAL #include "debugger.h" #define dbg(...) cerr << "LINE(" << __LINE__ << ") -> [" << #__VA_ARGS__ << "]: [", debug(__VA_ARGS__) #else #define dbg(...) 5 #endif int main() { int test; cin>>test; while(test--){ int curfile=1,currank=8; // curfile=current column, currank=current row cout<>king; if(king=="Done"){ break; } else if(king=="Right" || king=="Up-Right" || king=="Down-Right"){ if(firstmove){ currank--; } else{ curfile++; cur=1; } } else if(king=="Left" || king=="Up-Left" || king=="Down-Left"){ cur=1; currank--; if(currank==0){ currank=8; } } else{ cur++; if(cur==9){ curfile++; cur=1; } else{ currank--; if(currank==0){ currank=8; } } } firstmove=false; assert(curfile<8); cout<
 » 4 months ago, # |   +1 Why so much hate around this contest? It wasn't bad at all, C and D were interesting enough and the other tasks were good, maybe only E wasn't finally prepared but it's not the point to hate the whole thing :\ what a heck
 » 4 months ago, # |   0 Can somebody pls explain what's wrong with if statements in B solution 125803557? 3 tests passed succesfully, but on the 4th test i got an execution error. i have already implemented my own solution 125803196, but it also got an execution error on 4th test case, so after many unsuccesful tries to figure out what's wrong (i'm pretty sure that everything is correct) i deсided to check author's solution and just rewrite it (first attachment) but i left if statements (which i think are correct), sent the code and again got execution error and only when i removed if statements all tests passed correctly. how and why? i'll be grateful for any explanations.
 » 4 months ago, # |   0 AhmedEzzatG I don't understand the following part: If equal is false at any moment, then you can choose any subset of indices to contain 1 in the i-th bit.You could select the odd number of indices to contain the i-th bit and then And would become less than Xor.
•  » » 4 months ago, # ^ | ← Rev. 2 →   +1 If $equal$ $=$ false that mean $And$ > $Xor$ in the previous bits. Therefore, whatever the value of $And$ or $Xor$ in i-bit, $And$ will remain greater than $Xor$.
 » 4 months ago, # |   0 I have a solution to problem E that only needs 16 moves in worst case, verified by DFS.https://codeforces.com/contest/1557/submission/125862124I believe it can be made better, but that means slower runtime and will be hard to verify.
 » 3 months ago, # |   +3 A simple random solution for E: 126166286. Just do random moves and try to keep a list of valid starting positions for the king each time. When just one starting cell remained, simply trap the king.
 » 3 months ago, # |   0 Why this entry has so many downvotes?
 » 3 months ago, # | ← Rev. 2 →   0 About B,I use map nxt,directly link the next value,but I failed Test3 in 58th case,I don't know why,anybody can find the mistakes in may code?It will help me a lot,thanks so much! My code here... #include using namespace std; #define rep(i,a,n) for (int i=a;i>___T; for(int cs=1;cs<=___T;cs++) #define get(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin()) typedef long long ll; typedef pair PII; typedef vector VI; const ll mod=1e9+7; const int Max = 1e6+10; ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} void solve(){ int n,k; cin >> n >> k; VI a(n),b(n); rep(i,0,n){ cin >> a[i]; b[i] = a[i]; } sort(all(b)); map nxt; rep(i,0,n-1)nxt[b[i]] = b[i+1]; int i = 0,piece = 0; while(i < n){ while(i + 1 < n && a[i+1] == nxt[a[i]]){ i++; } piece++; i++; } if(piece > k){ cout<<"NO"<
•  » » 2 months ago, # ^ |   0 same for me if u can help me i my code in a post i failed in 3rd one why ?
 » 2 months ago, # | ← Rev. 6 →   0 guys for problem number 2, my solution it seems right but don't get AC, it is like i got vector of pair then first part i put {number, index} then i sort them for every number index i go right and left in array and check if it is the next number in sorted array, plz help me to see my mistake, thanks , sry for my English. int t; cin >>t; while(t--) { int n, k; cin >>n>>k; vii v(n); lp(i, n) { cin >>v[i].f; v[i].s = i; } if(k == 1 || k == n) { cout <<"Yes\n"; continue; } sort(all(v)); int cnt = 0; lp(i, n) { cnt++; int l, r; l = r = v[i].s; r++; l--; i++; while(i= n) break; i--; } cout<<(cnt <= k ? "Yes\n" : "No\n"); }
 » 2 months ago, # | ← Rev. 2 →   0 Can someone please help me in B, I can't figure out what's wrong in this approach My Solution.Edit: I figured it out.