### the_nightmare's blog

By the_nightmare, history, 20 months ago,

1527A - And Then There Were K

Author: loud_mouth
Idea: Bignubie

Editorial
Solution (Loud_mouth)
Solution (the_nightmare)

1527B1 - Palindrome Game (easy version)

Author: DenOMINATOR
Idea: shikhar7s

Editorial
Solution (DenOMINATOR)
Solution (shikhar7s)

1527B2 - Palindrome Game (hard version)

Author: DenOMINATOR
Idea:DenOMINATOR

Editorial
Solution(Greedy) (DenOMINATOR)
Solution(DP) (DenOMINATOR)

1527C - Sequence Pair Weight

Author: sharabhagrawal25
Idea: rivalq

Editorial
Solution (sharabhagrawal25)
Solution (mallick630)

1527D - MEX Tree

Author: mallick630
Idea: CoderAnshu

Editorial
Solution (shikhar7s)
Solution (the_nightmare)

1527E - Partition Game

Author: rivalq
Idea: rivalq

Editorial
Solution (rivalq)
Solution (the_nightmare)

• +166

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 » 20 months ago, # |   +118 contest was the_nightmare
•  » » 20 months ago, # ^ |   +5 haha, nice one XD
 » 20 months ago, # |   -47 nightmare round
 » 20 months ago, # |   +23 How to solve E using divide and conquer DP? (and especially how to maintain the cost around?)
•  » » 20 months ago, # ^ | ← Rev. 2 →   +13 link here is my code for divide and conquer techniquei took the idea from the code given below and understood it linkbasically what we are doing here is we are maintaining a persistent segment tree on every ith index which will provide us with the information that if we consider a segment of [i,j] then what will be its cost. The basic idea here is to use segment tree with range updates and point query.You could see from my code how to update ranges its pretty straightforward.now that we can find out the cost of any segment in log(n)complexity all we have to do is calculate the dp which can be calculated with the help of divide and conquer the only hard part of this method was the persistent segment tree part which was difficult to understand and actually think by yourself(atleast for me it was very new idea)
 » 20 months ago, # |   0 In problem B1, when all the elements of the string is 1, then how Bob wins?
•  » » 20 months ago, # ^ |   +6 It is given in the input section that string $s$ contains at least one $0$
•  » » » 20 months ago, # ^ |   0 But for this, why Draw is not the correct answer?
•  » » » » 20 months ago, # ^ |   +1 Yes, technically it should be DRAW but to avoid confusion we omitted that case
•  » » » » » 19 months ago, # ^ | ← Rev. 2 →   0 i have implemented dp for B2, but it's giving me incorrect output, pls help me find the bug const int N = 1e3; ll dp[N/2 + 1][N/2 + 1][2][2]; void solve() { int n; cin >> n; string s; cin >> s; int cnt00 {}, cnt01 {}, mid {}, rev {}; for(int i = 0; i < n - 1 - i; i++) { if (s[i] == s[n - 1 - i] && s[i] == '0') { cnt00++; } if (s[i] != s[n - 1 - i]) { cnt01++; } } if (n % 2 && s[n/2] == '0') { mid = 1; } if (dp[cnt01][cnt00][mid][rev] < 0) { cout << "ALICE" << '\n'; } else if (dp[cnt01][cnt00][mid][rev] > 0) { cout << "BOB" << '\n'; } else { cout << "DRAW" << '\n'; } } int main() { fastio(); for(int i = 0; i <= N/2; i++) { for(int j = 0; j <= N/2; j++) { for(int mid = 0; mid < 2; mid++) { for(int rev = 0; rev < 2; rev++) { dp[i][j][mid][rev] = INF; } } } } dp[0][0][0][0] = 0; for(int i = 0; i <= N/2; i++) { for(int j = 0; j <= N/2; j++) { for(int mid = 0; mid < 2; mid++) { for(int rev = 0; rev < 2; rev++) { // i -> cnt of symmetric 01 pairs // j -> cnt of symmetric 00 pairs if (i > 0) { dp[i][j][mid][rev] = min(dp[i][j][mid][rev], 1-dp[i-1][j][mid][0]); } if (j > 0) { dp[i][j][mid][rev] = min(dp[i][j][mid][rev], 1-dp[i+1][j-1][mid][0]); } if (mid > 0) { dp[i][j][mid][rev] = min(dp[i][j][mid][rev], 1-dp[i][j][0][0]); } if (rev == 0 && i > 0) { dp[i][j][mid][rev] = min(dp[i][j][mid][rev], -dp[i][j][mid][1]); } } } } } int tc = 1; cin >> tc; while(tc--) { solve(); } } 
 » 20 months ago, # |   +4 orz
 » 20 months ago, # |   +6 Does someone know of any problem similar to C?
 » 20 months ago, # | ← Rev. 2 →   0 what is the time complexity in B2 dp approach. 1.is it O(n^2) for one test case as it depends on no of 00 pairs and no of 01 pairs? 2.also if n^2 per test case how it passes the judge in 1 sec as n^2*t=1e9 ?
•  » » 20 months ago, # ^ |   +4 We precompute the dp and use it to answer all test cases
•  » » 20 months ago, # ^ |   +3 Dp is pre-computed not run for each test case
 » 20 months ago, # |   -13 Alternative solution to A: SpoilerKeep doing $n=n$ & $(n-1)$ while $n$ & $(n-1)>0$. Print $n-1$.
•  » » 20 months ago, # ^ |   0 I think this will TLE in python
•  » » » 20 months ago, # ^ |   +2 It does not. PyPy 3 submission: 116877296
•  » » 20 months ago, # ^ |   0 Yes, you can do that too, but it would take too much time and result in TLE
•  » » » 20 months ago, # ^ | ← Rev. 2 →   +2 The complexity is logn for this so it won't TLE. At each iteration, you're setting at least 1 set bit to 0.
•  » » 20 months ago, # ^ | ← Rev. 2 →   0 Hey! I am not able to understand why I am getting TLE for this solution :( Solutionll ans=n; while(n--){ ans=ans&n; if(ans==0){cout<
 » 20 months ago, # | ← Rev. 3 →   0 MY ISSUE PLEASE HELP ( PROBLM B1) !!ok when the number of zeros are even for example example 0 0 0 0 0 0 A pay1 1 0 0 0 0 0 B reverse 0 0 0 0 0 1 A pay1 1 0 0 0 0 1 B pay1 1 1 0 0 0 1 A reverse 1 0 0 0 1 1 B pay1 1 1 0 0 0 1 A reverse 1 0 0 0 1 1 B pay1 1 1 0 0 1 1 A pay1 1 1 1 0 1 1 B reverse 1 1 0 1 1 1 A pays1 1 1 1 1 1 1 isn't this optimal here every 4 changes means DRAW and extra 1,2,3 means BOB winsAnd when zeros are odd: this part of editorial its not clear to me :( Alice will change s[n/2] from '0' to '1' and play with the same strategy as Bob did in the above case. This way Bob will spend 1 dollar more than Alice resulting in Alice's win. example 0 1 0 1 0 1 0 1 0 A pay1 0 1 0 1 1 1 0 1 0 B pay1 0 1 1 1 1 1 0 1 0 A reverse 0 1 0 1 1 1 1 1 0 B pay1 0 1 1 1 1 1 1 1 0 A pay1 1 1 1 1 1 1 1 1 0 B reverse 1 1 1 1 1 1 1 1 1 in this case every 2 means DRAW and its repeating pattern of pay1 as AB BA AB BA then if cnt_of_0 is odd and cnt_of_0/2 is odd we will have 1 zero extra which will be paid by B means ALICE wins.but if cnt_of_0/2 is even we will have 1 zero extra which will be paid by A means BOB wins.PLEASEEEEE HELPPPPPPP!!UPD: Understood!!How to solve this. Thanks everyone
•  » » 20 months ago, # ^ |   0 Actually, for example $000000$ game goes-$A$ pay $1$ $100000$$B pay 1 100001$$A$ pay $1$ $100101$$B pay 1 101101$$A$ pay $1$ $111101$$B reverse 101111$$A$ pay $1$ $111111$$B$ winsNow, analyse $010101010$
 » 20 months ago, # |   0 After spending about 20-25 minutes I understood the logic of problem C ( yes I'm still a noob), but I was wondering how does one come up with that logic during contests ( I know practice, practice practice) but I suck at dp and I've been trying to improve it, so if anyone has dp sheets that can build my foundation it'll be of great help thanks :), I've been doing classic dp problems like knapsack, longest common subsequence type questions and even started with matrix chain multiplication recently.
•  » » 20 months ago, # ^ |   0 The states are easier to come up during contests if you really try to, most probably you just take what the problem asks and derive subtasks as a prefix, eg: Kadane-ish (or multiple prefixes across multiple sequences, eg: LCS), suffix, subarray of the original sequence, eg: matrix chain multiplication. I'm sure after a lot of $practice$, things would become somewhat more intuitive and reflexive.
 » 20 months ago, # |   +29 Alternative solution to E:First steps are also coming up with the $dp$ and writing the brute-force transition formula. Then, by considering $last(a_{r + 1})$, we can prove the following property: $c(l - 1, r) + c(l, r + 1) \leq c(l - 1, r + 1) + c(l, r)$ Therefore, $c$ satisfies Quadrilateral Inequality, where a divide-and-conquer solution works in $O(nk\log n)$ time.Note that calculating $c$ needs a two pointers trick similar to 868F - Yet Another Minimization Problem.
•  » » 20 months ago, # ^ |   0 I am using divide and conquer dp optimization for problem E. can you help me why i am getting TLE code
•  » » » 20 months ago, # ^ |   +1 for(int k = mid; k >= optl; k--) { $k$ should be enumerated from $optr$ to $optl$, not $mid$ to $optl$. Otherwise, the parameter optr is unused.
•  » » 20 months ago, # ^ |   0 How to prove complexity of two pointers trick?
 » 20 months ago, # |   0 Anyone please, help me to understand..For problem B1 help me to figure out the answer for this test case 00100
•  » » 20 months ago, # ^ |   0 ALICE — 10100 BOB — 10101 ALICE- 10111 BOB — 11101 (REVERSE) ALICE — 11111ALICE --> 3 BOB -->1 BOB WINS
•  » » » 20 months ago, # ^ |   0 Why it is not possible?ALICE — 10100 BOB — 00101(REVERSE) ALICE- 10101 BOB — 11101 ALICE — 10111(REVERSE) BOB — 11111ALICE --> 2 BOB -->2 DRAW
•  » » » » 20 months ago, # ^ |   +1 it's not that you can't do that . you can .But you know what , the word optimal is mentioned in the question, means if i got a chance to play then i tried my best to win , so if bob put a 1 in the string instead of reversing he will land in the winning position , instead of a draw. You can't just brute force and say bob or alice win or its a draw. its not mandiatory that if i have a chance to reverse the string then i have to reverse it , so that i will be relived from that 1 dollar penalty, you can't do that .
 » 20 months ago, # |   0 rivalq the_nightmare I am confused in the editorial for E. Aren't the k mentioned in the dp transitions and the k mentioned in the big oh notation different?
•  » » 20 months ago, # ^ |   +3 Yes they are different. The one in dp transitions you can regard as a temp variable.
 » 20 months ago, # | ← Rev. 2 →   0 Problem B2 — Palindrome Hard Please explain this case for string: 1000 A reverses -> 0001 (A=0 B=0) B pays -> 1001 (A=0 B=1) A pays -> 1101 (A=1 B=1) B reverses -> 1011 (A=1 B=1) A pays -> 1111 (A=2 B=1) So BOB should win. But by the above code, it's making Alice the winner. Please guide me where I am doing a mistake in the implementation.
•  » » 20 months ago, # ^ |   0 Alice can win this way:A pays -> 1001B pays -> 1101A reverses -> 1011B pays -> 1111B = 2 A = 1, so Alice wins. Bob has no other moves.
•  » » » 20 months ago, # ^ |   0 According to you Alice wins..But according to Jyotirmaya Bob wins...So what is the exact ans...Both of you correct.
•  » » » » 20 months ago, # ^ |   +2 Alice wants to win right? So she would do exactly what I stated. Bob has no other move than just to lose. It is not logical to make a move that will allow your oponent to win.
 » 20 months ago, # | ← Rev. 2 →   0 In fact , some users of the Chinese online judge : Luogu said that the difficulty of these problems is not monotonically increasing and they suggested that you should have changed the order of problem B and C. the_nightmare
•  » » 20 months ago, # ^ | ← Rev. 2 →   +4 There is only one additional case to be dealt in B2 if you look at the editorial. That would explain why they considered B2 as easier probably.. and dp is not what most div 2 contestants used.
•  » » 20 months ago, # ^ |   +19 Basically, we have to put together B1 and B2 due to contest restrictions due to which we are not able to swap B2 and C. But we have provided the scoring according to difficulty B(750+1500) total 2250 and C only 1500.
•  » » » 20 months ago, # ^ |   +3 In problem D's editorial shouldn't it be "We will always break before or on reaching root" instead of "Note that we will always break at the root as it is marked visited in the initial step."
 » 20 months ago, # |   -13 The term "Contiguous Subarray" is much more quicker to grasp than "Subsegment".Hope future authors see this :) Nice Contest btw
•  » » 20 months ago, # ^ |   +4 but you know what you got something to learn.
•  » » 20 months ago, # ^ |   0 I mixed subsegment with subsquence, and it wasted me lots of time to solve this problem in a wrong way
 » 20 months ago, # |   0 Could someone please write a simpler edit for Problem-C, I have gone over it a lot of times but am still confused as to why the question creator went for:value[a[i]] += (i + 1);please help me out with the logic. I understood the task but couldn't implement it that well and now I'm even more confused.
•  » » 20 months ago, # ^ |   0 I think (i+1) refers to the total number of subarrays ending at i. Since we are using 0 indexing, so +1 for the adjustment.
•  » » 20 months ago, # ^ |   0 Consider and element i. Now if take an element j such that a[i]==a[j] and j < i then subarray a[j-i] will occur as part of all subarray's from  i = 0 to j i.e j + 1 times. So  value[a[i]] += i + 1
•  » » » 20 months ago, # ^ |   0 Oh now i get it, thank you so much
•  » » » » 20 months ago, # ^ |   0 Happy to help
•  » » » 12 months ago, # ^ |   0 I was confused over this part. Thanks now,I got it.
 » 20 months ago, # |   0 Can someone help me with my solution :My idea: Maintain a path and its endpoints. Maintain a 'visited' array which denotes whether or not this node is in current path. Consider 0 as base case and mark it visited and initialize both the endpoints to 0. Iterate from 0 to i In order to find if there can be a path having all nodes [0, i] we just need to check if the endpoints of path having all of [0, i-1] can be extended to i, so move from ith node to its parent till we find a node that is in the path that includes the nodes [0, i-1], that is first visited node. If this node happens to be one of the endpoints then extend the path and update endpoints else there can be no such path that includes all of [0, i] nodes and we dont need to check this for following i's. I am unable to figure out why this gives TLE ! https://codeforces.com/contest/1527/submission/116924323
»
20 months ago, # |
-13

# include<bits/stdc++.h>

using namespace std;

int main(){

int t;
cin>>t;

while(t--){
int n;
cin>>n;
string s;
cin>>s;
int count=0;
if(n==1&&s[0]=='0'){
cout<<"BOB"<<'\n';
continue;
}
for(int i=0;i<s.length();i++){
s[i]=='0'?count+=1:count=count;
}
if(count%2==0){
cout<<"BOB"<<'\n';
}
else{
cout<<"ALICE"<<'\n';
}

}

}

•  » » 20 months ago, # ^ |   0 This is my logic, you can use to improve your code 116814530
•  » » » 4 months ago, # ^ |   0 Thanks
 » 20 months ago, # |   +9 Nice explanation of problem B2
 » 20 months ago, # | ← Rev. 2 →   +4 Can someone give a small test for those codes which fail test case 5 by printing 1 instead of 0 at 1923rd position for problem D? Submission
•  » » 20 months ago, # ^ | ← Rev. 2 →   0 I got that error by incorrectly calculating in the tree "0 -- 2 -- 1" the number of pairs with mex == 2 (there are 0).
 » 20 months ago, # |   0 In problem A I am getting wrong output format Expected integer, but "2.68435e+008" found Solutionlong long n;cin>>n; long long ans=log2(n); cout<
•  » » 20 months ago, # ^ |   0 pow() returns a double, while the expected output is an integer, hence the WA. Also as anubh4v stated, pow() (and log2() too) can be imprecise at times, leading to incorrect rounding of the number.
•  » » » 20 months ago, # ^ |   0 I will keep that in mind. Thanks
 » 20 months ago, # |   0 Can someone explain me how does the dynamic programming solution for B2 works?From my understanding of the problem when we consider alice we add positively, when we consider bob we add negatively. But how does that happen in code? How does the code distinguish bob from alice? And how does it simulate turns?In other words: can someone explain me how the simulation of the game occurs during the bottom up transitions of the editorial / given code?Thanks in advance.
•  » » 20 months ago, # ^ |   0 Because dp[i][j][k][l] is the optimal answer for a state where i is the number of 00, j is the number of 01 or 10, and k = 1 denotes if the middle position in case of odd length string is 0 and l = 1 denotes that in the last turn other person reversed the sting thus we can not reverse. For all the states, we will assume that the current turn is of Alice and to compute the answer for that state, we will add negative of the transition states, which will denote Bob's optimal score.
 » 20 months ago, # |   0 can anyone pls tell why i am getting time limit exceeded on test case 7 in problem D MEX TREE i am just doing a dfs traversal once to calculate subtree sizes and then iterating from 1 to n and marking not visited nodes as visited in my current path and calculating answer for each mex value.my submission link https://codeforces.com/contest/1527/submission/116996030
 » 20 months ago, # |   0 In Problem D: as mentioned in the editorial that we need to "update the subtree sizes as we move up to parent recursively", we don't need to do this. When (l!=r) we will always choose the other parent. Only when we are calculating MEX1 (the previous l was equal to r) so we have to update the size of 0 subtree only once.
•  » » 20 months ago, # ^ |   0 Yoir solution simplified the question mqnyfolds. Thanks anadii!
 » 20 months ago, # |   +8 I do not know if this approach has been covered for E using divide and conquer dp. To get cost of current interval, maintain global 2 pointers on the array, sum variable and array of deque. Fit the pointer for each query. Amortized complexity over per level of dp should be N*log(N). So with K layers it becomes K*N*log(N).
 » 20 months ago, # |   0 the_nightmare's solution for D will TLE for the following case. https://drive.google.com/file/d/1K-1sb5ls2PP0lKiGQf5dy9BVuqMBvReG/view?usp=sharing
 » 18 months ago, # |   0 Problem 1527C - Sequence Pair Weight could have been done greedily (and imo it's easier). Let $d(x, y)$ denote the number of segments which contain elements at indices $x$ and $y$ (indices start from 0 so $x,y \in {0, 1, 2, \dots, n-1}$). It is easy to see that if $y > x$ then $d(x, y) = (x+1)*(n-y)$. This allows obvious $O(n^2)$ solution, but it can be done faster in $O(n)$. Let's say we have a vector $v$ and we are at it's $i$-th element. Then, we can calculate the answer as: $d(v_0, v_i) + d(v_1, v_i) + \dots + d(v_{i-1}, v_i)$which is just $(v_0+1)*(n-v_i) + (v_1+1)*(n-v_i) + \dots + (v_{i-1}+1)*(n-v_i)$and this can be simplified to: $(v_1 + v_2 + v_3 + \dots + v_{i-1} + i-1) * (n-v_i)$Which you can easily calculate while iterating through the vector. Code: 124839948
•  » » 3 months ago, # ^ |   0 Shouldn't it be $(v_0 + v_1 + v_2 + v_3 + \text{...} + v_{i-1} + i) \times (n-v_i)$?
 » 17 months ago, # |   0 I think more short solution for A: https://codeforces.com/contest/1527/submission/127546267
 » 10 months ago, # |   0 in b1 can anyone explain output for this string 01011010 i am getting draw
 » 7 months ago, # | ← Rev. 3 →   0 in problem C let the array be, 1 3 1 2 1 when we take subarray , 1 3 1 2 1 weight will be 3 {(1,2),(2,5),(1,5)} 3 1 2 1 then weight will be 1 { (3,5) }but according to editorial the weight of second one will be 3.anyone please reply ?
•  » » 7 months ago, # ^ |   0 I ran the same code. Its giving correct answer : 7
•  » » » 7 months ago, # ^ |   0 can you please explain me ? how is it possible
•  » » » » 7 months ago, # ^ |   0 Check input format