### flamestorm's blog

By flamestorm, 3 weeks ago,

Thanks for participating!

1692A - Marathon

Idea: mesanu

Tutorial
Solution

1692B - All Distinct

Idea: mesanu

Tutorial
Solution

1692C - Where's the Bishop?

Idea: flamestorm

Tutorial
Solution

1692D - The Clock

Idea: SlavicG

Tutorial
Solution

1692E - Binary Deque

Idea: flamestorm

Tutorial
Solution

1692F - 3SUM

Idea: flamestorm

Tutorial
Solution

1692G - 2^Sort

Idea: flamestorm

Tutorial
Solution

1692H - Gambling

Idea: mesanu

Tutorial
Solution

• +48

 » 3 weeks ago, # |   +6 Thanks for the fast editorial
 » 3 weeks ago, # |   0 Perfect contest for me.
 » 3 weeks ago, # |   0 tns fr the tutorial
 » 3 weeks ago, # |   0 Thanks for the fast editorial
 » 3 weeks ago, # |   0 I loved the contest, especially problems F and G.
 » 3 weeks ago, # | ← Rev. 2 →   +3 whats wrong with this https://codeforces.com/contest/1692/submission/160657931 P.S. found.
 » 3 weeks ago, # |   0 How to solve H without using trees?
•  » » 3 weeks ago, # ^ | ← Rev. 3 →   +7 For example using a Kadane's algorithm. My submission that using it is here.
•  » » » 3 weeks ago, # ^ |   +1 But if we apply Kadane's algorithm for every possible value of a wouldn't the complexity will be O(n^2)?
•  » » » » 3 weeks ago, # ^ | ← Rev. 2 →   +1 for each a, store positions of a[i] = a then the complexity is O(nlogn) because use have to compress value of a[i] or store them in map
•  » » 3 weeks ago, # ^ |   +6 I used Kadane's algorithm, where I kept track of the indices of each number, and inserted a certain amount of negative 1's between each index.
•  » » 3 weeks ago, # ^ |   +5 As explained in the editorial, for a fixed $a$, you can replace $arr[i]=1$ if $arr[i]==a$ else $arr[i]=-1$.Now, you can find max sum subarray in this newly obtained array without using Segment Tree.Observe that the max sum subarray will always start and end at indices where $arr[i]==1$.Using this observation you can just iterate over indices where $arr[i]==1$ and use Kadane's Algorithm to find max sum subarray.Since, there are only $n$ distinct values of $a$ possible, overall time complexity will be $O(n)$.For more clarification, you can see my submission
•  » » » 3 weeks ago, # ^ |   0 I don't understand how you found the constant a in this case
•  » » » » 3 weeks ago, # ^ |   +1 No, I am not saying $a$ is constant but $a$ will definitely appear in the given array. So we can just iterate over all the distinct elements of the array and then considering current element as $a$ calculate the answer. Maximum over all elements will be the final answer.
•  » » » » » 2 weeks ago, # ^ |   0 but wouldn't that be O(n^2) because for every distinct a[i] you need to check the maximum subarray?
•  » » » » » » 2 weeks ago, # ^ |   0 For every distinct $a_i$ we will iterate over indices of its occurrence. Now sum of number of indices for all distinct $a_i$ will be $n$(obviously). So inner loop will run only n times for the whole array.Time complexity: $O(n+n)=O(n)$
•  » » » » » » » 2 weeks ago, # ^ |   0 again, there can be n distinct a[i] numbers, and if you iterate each one that would be n*n?
•  » » » » » » » » 2 weeks ago, # ^ |   0 Think about it. If there are n distinct numbers then number of occurrences of every number will be 1. So inner loop will run only 1 time for every element not n times.
•  » » » » » » » » » 3 hours ago, # ^ |   +1 Great explanation , thanks!
•  » » 3 weeks ago, # ^ |   0 A solution using set Code
•  » » 3 weeks ago, # ^ |   0 A solution using map here 160632225
•  » » 3 weeks ago, # ^ |   0 Then it's just kandene my dear friend.
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   0 I used divide and conquer, here is my code
•  » » 3 weeks ago, # ^ |   0 I used DP
•  » » 3 weeks ago, # ^ |   0 I think this is easy solution using map 160604275
 » 3 weeks ago, # |   0 An elegant solution for GWas very painful as I submitted just a minute after the contest finished...
•  » » 3 weeks ago, # ^ |   0 nice one! sliding window is neat
•  » » 3 weeks ago, # ^ |   0 Hey could you help me in debugging my solution ?
 » 3 weeks ago, # |   +8 **problem H **, also can be solved useing kadane's algorithm.
 » 3 weeks ago, # |   +8 H can be solved in O(n) just iterating for indexes of each a that is in array. My solution is here
•  » » 3 weeks ago, # ^ |   +8 You're inserting into a map, so it's not $O(n)$
•  » » » 3 weeks ago, # ^ |   +1 We can utilize the map in a way that it never holds more than two elements, and hence works in O(1), which makes the solution O(n).see 160608733
•  » » » » 3 weeks ago, # ^ |   0 I think for an input like [1,1,1,1,2,2,3,...] the map would have to hold 3 elements. So in the worst case the map holds log(n) elements.
•  » » » » » 3 weeks ago, # ^ | ← Rev. 2 →   0 You are right, so it is O(n log(log(n)))
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   +3 Bro come Atleast do some research before claiming time complexity. It's not O(N) coz you are using map. But i guess the operations are quite low so we can it's O(N) but technically it's not.
•  » » 3 weeks ago, # ^ |   0 why my code gave wa on test case 3 pls check my code and tell the test case where my code fail 160727894 this is my soln link
 » 3 weeks ago, # | ← Rev. 3 →   0 Able to solve only problem A,B.For problem C got the approach but unable to implement it.
 » 3 weeks ago, # |   +4 Problem H can use Kadane in a nice way.Notice that the sum over frequencies of distinct elements is the size of our input array.We know any subarray endings should be at two elements that have the same value, otherwise, we are pointlessly reducing our answer and we can fix our boundaries to be on the same boundaries.say if we have XXXLXXXRXXX where [L, R] is the subarray we are considering then there is no point in having the left or right boundaries on Xs since we could improve the sum by restoring boundaries to L and R.We can make an array of positions for each distinct element and use the observation that the Xs can be compressed together, therefore for each distinct element, such a compression yields a size of array twice the array of all positions for that element.So our solution boils down to making a compressed array for each distinct element and then applying kadane to it, finally taking the one which gives the maximum answer.Time complexity is still the order of n as we are effectively applying kadane on twice the input size.
 » 3 weeks ago, # | ← Rev. 2 →   0 Problem F : O(10^3) preprocessing + O(n) approach Submission
•  » » 3 weeks ago, # ^ | ← Rev. 3 →   0 easier O(10³) + O(n): 160566183
 » 3 weeks ago, # |   0 https://codeforces.com/contest/1692/submission/160650500Here , is my submission for problem G, i didn't get how I am getting wrong answer on testcase 15, can someone please explain fault in my submission?. Thank you.
•  » » 3 weeks ago, # ^ |   0 probably a floating precision error cause you are using v[i]=log2f(x)
•  » » » 3 weeks ago, # ^ |   0 is there anyway we can store log base 2 of any number upto say 20 order of precision?
•  » » » » 3 weeks ago, # ^ |   0 Depends on what you mean by "20 order of precision". If you mean just 20 digits in the number then surely double-precision floating-point can hold that amount of digits. If you mean 20 digits in fractional part, then you can't have a guarantee since it depends on how big your number is
•  » » » » 3 weeks ago, # ^ | ← Rev. 2 →   0 Btw, you can use 63 - __buitin__clzll(x) instead of log2f(x) if you're using gcc/g++. Or you can use something like that (code below), but I'm not sure if it'll resolve your problem codeint my_log2(long long n) { int z = 1; int cnt = -1; while (z <= n) { z *= 2; ++cnt; } return cnt; } 
 » 3 weeks ago, # | ← Rev. 2 →   +22 Solution of H without treesInstead of doubling and halving, consider the score of a subarray to be the difference between the frequency of $a$ and the number of elements in the subarray not equal to $a$. This is done because we will double our score for $freq[a]$ times and halve it for the rest elements and hence we want this difference to be maximum. Notice that for the best subarray, $a = x_l = x_r$. The problem can now be solved with dp. Let $dp[i]$ be the best score for a subarray with $l = i$ and $a = x_i$. Then we can see that $dp[i] = 1 + max(0,dp[next[a[i]]] - (next[a[i]]-i-1))$, where $next[x]$ is used to store the next occurence of $x$ on iterating over the dp. The right end can be maintained similarly(if the dp is maximised then $right[i] = right[next[a[i]]]$.Submission at the time of the contest — https://codeforces.com/contest/1692/submission/160583853
 » 3 weeks ago, # |   0 how do you find the test cases of a question?????????
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   0 some of the test cases are shown in the status i.e your submission but sadly you can't see all of'em as they're truncated if they're too long. although some times what i do is (it's stupid :D) that for some value of test case i print it instead of printing the ans with all values in it but this too doesn't always work.
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   0 You can try something like this.' Change the values according to your question and change the "if" value to your testcase-1
 » 3 weeks ago, # |   0 can any one do the f question using bit manupulation
 » 3 weeks ago, # |   0 What is wrong with my code?? can anyone check for f question : please https://codeforces.com/contest/1692/submission/160683383 Link
•  » » 3 weeks ago, # ^ |   0 actually what you're doing is that you're checking for pair i,j,k but if u dont get it ur also removing it from the array by freq--.like: 1 2 5 6 you'll check for 1,2,5 = 9 so u remove 1 2 but 2 5 6 actually gives the ans i.e 13what you have to do is first check if we take 3 same values (if they're present) and get the ans. then again same if we take 2 same values and one any other value. then all 3 different values.in all these check if they're present with the required freq.if we want to take 3 same values.  for(int i = 0;i <= 9;i++){ if(arr[i] > 2 && ((3*i) % 10) == 3){ y = 1; break; } } if we want to take 2 same and 1 diff.  for(int i = 0;i <= 9 && !y;i++){ if(arr[i] == 0) continue; for(int j = 0;j <= 9;j++){ if(i == j || arr[j] < 2) continue; if((2*j + i) % 10 == 3){ y = 1; break; } } } and lastly all different  for(int i = 0;i <= 9 && !y;i++){ for(int j = 0;j <= 9 && !y;j++){ for(int k = 0;k <= 9;k++){ if(i != j && j != k && k != i){ if(arr[i] > 0 && arr[j] > 0 && arr[k] > 0){ if((i + j + k) % 10 == 3){ y = 1; break; } } } } } } 
•  » » » 3 weeks ago, # ^ |   0 got it.. Thanks man..
•  » » » 3 weeks ago, # ^ | ← Rev. 2 →   0 bool helper(vector freq, int target){ for(int i = 0; i<10; i++){ if(freq[i]<=0)continue; freq[i]--; for(int j = 0; j<10; j++){ if(freq[j]<=0)continue; freq[j]--; for(int k = 0; k<10; k++){ if(freq[k]<=0)continue; if((i+j+k)%10 == 3)return true; } freq[j]++; } freq[i]++; } return false; } Your code here... changed a little bit and got accepted thanks
 » 3 weeks ago, # |   0 Can we use sliding window for E by searching total sum minus reqd target in the sliding window?
•  » » 3 weeks ago, # ^ | ← Rev. 3 →   0 yup i did it this way only 160576052
•  » » » 3 weeks ago, # ^ |   0 Thank you so much :). And this is then O(n), right?
•  » » » » 3 weeks ago, # ^ |   0 i mean if u don't count the complexity of map :D yeah
•  » » » » » 3 weeks ago, # ^ |   0 Think O(n) is possible too right sliding window type of approach but without map https://codeforces.com/contest/1692/submission/160607428
•  » » » » » » 3 weeks ago, # ^ |   0 yeah you're right we could use the fact there's only 1s an 0s
•  » » » 3 weeks ago, # ^ |   0 can You explain it a bit further?
•  » » 3 weeks ago, # ^ |   0 This is my code to use a sliding window Your text to link here...
•  » » » 3 weeks ago, # ^ |   0 O(N)
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   0 https://codeforces.com/contest/1692/submission/160581476Easiest Approach Sliding Window O(N)
 » 3 weeks ago, # |   0  for(int i = 0; i<8; i++){ int c = 0; int k; string s = arr[i]; for(int j = 0; j<8; j++){ if(s[j] == '#'){ c++; k = j; } } if(c == 1 && i !=0 && k != 0){ cout<
•  » » 3 weeks ago, # ^ |   0 _ _ _ _ _ _ # __ _ _ _ _ # _ __ _ _ _ # _ _ __ _ _ # _ _ _ _# _ # _ _ _ _ __ # _ _ _ _ _ _# _ # _ _ _ _ __ _ _ # _ _ _ _consider this example i think so it's self explainatory.
•  » » » 3 weeks ago, # ^ |   0 yaa man,, i am such a noob
 » 3 weeks ago, # |   0 For problem D, A brute force approach I used — 160599689.I just kept adding the minutes to the time until the time was the same as starting and kept checking for palindrome.
 » 3 weeks ago, # |   0 can someone tell me what's wrong with this https://codeforces.com/contest/1692/submission/160689162 ??
 » 3 weeks ago, # |   0 Please add the proper tags for the problem and not the unnecessary one. Thanks.
 » 3 weeks ago, # |   0 Solution of H without using segment trees: https://codeforces.com/contest/1692/submission/160692781
 » 3 weeks ago, # |   0 Could anyone help me with problem E: Binary Deque? My solution is kind of greedy. I create an array with distances (in numbers of zero entries) between two conseсutive ones. For example, for the sequence: 0 0 1 0 0 0 1 1 0 1, my array is [2 3 0 1 0] (last zero means that we have tail of zeros with length 0). Next, I use two iterators for decreasing the sum of the array choosing at every step iether the head or the tail with zeros and errasing it from my array. I cannot find the mistake and 1st test is OK, but I get WA for the second. If this idea is correct then the solution is O(n). Here the link: 160692840
•  » » 3 weeks ago, # ^ |   0 I have a solution of E: You can define an array $k$, It is prefix sum of $a$, so $k_i$ is "How many ones in [1,i]". The range [l,r]'s sum is S when $k_r-k_{l-1}=S$, you can define an array $m$. $m_i$ is "The max index j that k[j]=i". Then enumerate $l$, and the max of $r$ is $m_{s+k[l-1]}$, the answer is the minimum of $n-(r-l+1)$. If every $r$ is $0$ ($s+k[l-1]>n$), Then output -1. My code: 160569001
•  » » 3 weeks ago, # ^ |   0 I think this logic is incorrect as this only accounts for the head or the tail. If the sum we needed was 5 and your array was [5 0 0 0 0 4 4 4 4 4], it would incorrectly assume that it should take from the tail every time instead of realizing that the 5 is better in the long run.
•  » » » 3 weeks ago, # ^ |   0 Great! Got it and fixed! Thanks a lot!
 » 3 weeks ago, # |   0 Can anyone explain the Kadane algorithm approach for the H problem? I am unable to understand what one is exactly trying to do.
•  » » 3 weeks ago, # ^ |   0 I've done something very similar to kadanes (with an early break when the sum reaches 0):For each position (starting from left) iterate through the array adding +1 for matches and -1 for other values. If the sum gets to 0 break. Any start positions that have previously added a +1 are skipped (as you've already calculated the sum from that position starting from at least 1).
•  » » » 3 weeks ago, # ^ |   0 ohk thanks:)
 » 3 weeks ago, # |   0 Thank you for fast editorial and interesting tasks.
 » 3 weeks ago, # |   0 Pretty sure we can do a O(n) solution for E right? I did this. Doesn't use map stl. https://codeforces.com/contest/1692/submission/160607428
•  » » 3 weeks ago, # ^ |   0 I did an O(n) with 2-pointers technique https://codeforces.com/contest/1692/submission/160604366
•  » » » 3 weeks ago, # ^ |   0 can you explain the logic behind this? I also tried a 2-pointers technique but I don't think my solution works. Also, how is your solution O(n) if it has nested loops?
•  » » » » 3 weeks ago, # ^ | ← Rev. 3 →   0 Ok I'll try.We're moving right pointer while sum in segment [l, r] is less/equal to s. If our sum is larger than s, we're moving the left pointer to reduce it. When sum is equal to s, update the answer. Thus, we'll find all subsegments with sum equal to s.Complexity is O(n) because we're moving the right pointer N times and left pointer no more than N times.Actually, my code is almost a copypaste of edus one (first block), there's a detailed explanation as well.
 » 3 weeks ago, # |   0 The tutorial is very good, I can understand it very clearly.
 » 3 weeks ago, # |   0 Two pointers Contest :)
 » 3 weeks ago, # |   0 I recently read a blog that said "Stop making Div4 contests". All I can say is, please DON'T stop making them, they are very helpful for people like me, and I'm sure a lot of others. I learnt a lot from this contest. Thank you..!!
 » 3 weeks ago, # |   0 Can we do E by 2 pointers, and a greedy approach? I thought of this — https://codeforces.com/contest/1692/submission/160617363 but it is giving a wrong answer :')Can anyone think where can it be wrong? because I am unable to find any case...
•  » » 2 weeks ago, # ^ |   0 Try this 10 2 0 0 1 1 0 1 0 1 0 0
•  » » » 2 weeks ago, # ^ |   0 Ohh I get it... Thanks!
 » 3 weeks ago, # | ← Rev. 2 →   0 The hacking phase is done why isn’t my rating updated?
•  » » 3 weeks ago, # ^ |   0 same question
 » 3 weeks ago, # |   0 anyone please tell me where am i wrong 160716728 problem D
•  » » 3 weeks ago, # ^ |   +16 I think line 74 should be "checkhour>=24" instead of "checkhour>24"Also line 41~44 is not correct because the clock never shows 24:00.
 » 3 weeks ago, # |   0 thanks for the editorial!
 » 3 weeks ago, # |   0 Good Contest for me
 » 3 weeks ago, # | ← Rev. 2 →   +24 I'm not too fond of the author's solution to problem H, though the problem was great. But no divisional 4 rounds should contain any Data Structure. Here's my solution — The basic observation is that the problem simply asks us to find, for each element(x) of the given array, the maximum subarray sum on another array(b) where b[i] = 1 if a[i] == x, -1 otherwise.We can apply Kadane's algorithm for each element lazily. In Kadane's algorithm, the important variables are current sum, current best, and where the best subarray is. So, as we iterate over the array, we evaluate elements one by one and remember these important variables for each element and their last positions, std::map works fine.And finally, we can just iterate over the map to find the best answer and the desired segment with the same asymptotic, O(n.log(n)). See my implementation if need be.
 » 3 weeks ago, # |   +1 What is the wrong on my solution Problem E ? im using two pointers to get the best distance from left and right.. 160727889
•  » » 3 weeks ago, # ^ |   0 1 12 1 0 0 1 0 0 1 1 1 1 0 0 0 You answer is 8 , but the correct answer should be 7. It is easy to know that this greed will not work...
 » 3 weeks ago, # |   0 Can anyone please point out the error in my code/logic (Problem G). I used a sliding window approach by making a window of size k. The variable bad used in my code tells the last index till the current index where the inequality does not hold. My submission https://codeforces.com/contest/1692/submission/160622089
 » 3 weeks ago, # | ← Rev. 3 →   0 can anyone tell whats wrong with this code for problem H? https://codeforces.com/contest/1692/submission/160731054
 » 3 weeks ago, # |   0 160731775 : An Elegant solution for Problem H:no segment tree required
 » 3 weeks ago, # | ← Rev. 2 →   0 why my code gave wa on test case 3 pls check my code and tell the test case where my code fail 160727894 this is my soln link
 » 3 weeks ago, # |   0 why my code gave wa on test case 3 pls check my code and tell the test case where my code fail 160732937
 » 3 weeks ago, # |   0 Where can I find a tutorial for the standard problem of finding the maximum sum of a subarray?
 » 3 weeks ago, # |   0 In D, why is the second for loop running for 2022 iterations?
•  » » 3 weeks ago, # ^ | ← Rev. 2 →   0 Any number greater than 1440 because after that the numbers will repeat again.
 » 3 weeks ago, # |   0 When does the rating update?
 » 3 weeks ago, # |   0
 » 3 weeks ago, # | ← Rev. 3 →   +16 An alternative solution to problem H without segment tree. Notice that if we choose $(a,l,r)$, the money we will get is $2^{x-(r-l+1-x)}\ =2^{2x-(r-l)+1}\$ ($x$ is the number of times that $a$ appears in round $l$ to round $r$). Thus our goal is maximizing $2x-(r-l)+1$. For a certain number $a'$, we can create an array $b$, where $b[i]=-i+$ $2\times$(the number of times that $a'$ appears in round $1$ to round $i$). And the money we will get by choosing $(a',l,r)$ is equal to $2^{2x-(r-l)+1}=2^{b[r]-b[l]+1}\$! On the other hand, it's obviously that if the answer is $(a,l,r)$, then the dice must show $a$ in both round $l$ and round $r$. Therefore, we only need to consider the rounds where dice show $a'$ for a certain number $a'$. Based on the above, the problem can now be solved with time complexity $\mathcal{O}(nlogn)$ by using maps and arrays. Checks this code for better comprehension:https://codeforces.com/contest/1692/submission/160769341
 » 3 weeks ago, # |   -9 when div5 begins to be held? I want a more easier division!
•  » » 3 weeks ago, # ^ |   0 kidding ?
•  » » » 3 weeks ago, # ^ |   -8 not kidding, i want a more happy contest, during which I can ak very fast
•  » » » » 3 weeks ago, # ^ |   0 But when sol hard prob, you feel happy more than it is easy
 » 3 weeks ago, # |   0 Great round!
 » 3 weeks ago, # | ← Rev. 2 →   0 In binary deque we can use deque to save index of 1s but it's giving WA can anyone help
•  » » 3 weeks ago, # ^ |   0
•  » » 2 weeks ago, # ^ |   0 The problem is that you do not know what the order of 1s is after you remove the i-th one. So for example; 0 0 1 1 0 1 0 1 0 0 There is a 1 equally distanced on each end of the string, but the 1 that is after it, is closer on the left than on the right. So this "greedy" approach doesn't work, since you would have to also consider all other 1s up to the next one, but that is too long. Try a different approach.
 » 3 weeks ago, # |   0 Why does binary search work in E? Doesn't the array need to be sorted for that?
•  » » 3 weeks ago, # ^ |   0 It's already sorted since we bs on the prefix sum array
 » 3 weeks ago, # |   0 mesanu For Problem H, can you explain how you are updating and calculating the maximum sum subarray having elements -1, 1, and like how you are updating it in log(N). Like in your code what does pref, suf, Val, and sum store, and how your modify function is working as I'm having trouble with these parts. Thank you.
 » 3 weeks ago, # | ← Rev. 2 →   0 Can someone tell me what am I doing wrong ?https://codeforces.com/contest/1692/submission/160812969
 » 3 weeks ago, # |   0 For Problem E, the tutorial says we need the "smallest" value of r. Shouldn't it say the "largest" value of r instead?
 » 3 weeks ago, # | ← Rev. 2 →   0 can i use mod when multiplying with a big power of 2 (G-2^sort) ?
 » 2 weeks ago, # |   0 https://codeforces.com/contest/1692/submission/160699527 Can Some help me to find the the error in my code. Why my code fails.
 » 2 weeks ago, # |   0 Here's another approach to problem H which is very surprising for me. We can just make store all positions of each distinct value. Then apply the Kadane's algorithm to find the leftmost and the rightmost of each local maximum power of 2. Lastly, just find the maximum of them. Here's the code to what I'm saying: https://codeforces.com/contest/1692/submission/161122830
 » 2 weeks ago, # |   0 for Find Bishop why doesn't this code work void solve() { act vector a(9); for (int i = 0; i < 8; ++i) { for (int j = 0; j < 8; ++j) { cin>>a[i][j]; } } int r(0),c(0); for (int i = 1; i < 7; ++i) { for (int j = 1; j < 7; ++j) { if(a[i][j] == '#'){ if(a[i-1][j-1] == '#' and a[i+1][j-1] == '#' and a[i+1][j-1] == '#' and a[i+1][j+1] == '#'){ r = i+1,c = j+1; break; } } } } cout<
•  » » 2 weeks ago, # ^ |   0 note ml is a macro for '\n' please help me
•  » » 2 weeks ago, # ^ |   0 because u have written one condition twice in the second if condition  this \/ and this \/ if(a[i-1][j-1] == '#' and a[i+1][j-1] == '#' and a[i+1][j-1] == '#' and a[i+1][j+1] == '#') Here is My Submission for the Same
•  » » » 2 weeks ago, # ^ |   0 thank you sir that was a big help
•  » » » » 7 days ago, # ^ |   0 i am glad to know that
 » 2 weeks ago, # |   0 Tutorial for 1692E - Binary Deque should be to "binary search on the largest value of $r$", and sum is "non-decreasing"?
 » 12 days ago, # |   0 for G, just apply log to all the array elements and then sum with each index. i.e [2^x . Ax ] will become x + log(Ax). then just iterate.
 » 12 days ago, # |   0 I love Div.4 very much
 » 12 days ago, # |   0 Thanks for the fast editorial
 » 7 days ago, # | ← Rev. 2 →   0 [Deleted]
 » 4 days ago, # |   0 Greedy Approach for last question : Gambling ->Suppose we have this example : 10 8 8 8 9 9 6 6 9 6 6 First make map< long long , vector> and insert key : [with vector of indexes] 8 : [0,1,2] 9 : [3,4,7] 6 : [5,6,8,9] then iterate over vector of indexes of each element Initialize res = 1 ( represent power of 2. Initially 1 because for one occurrence we have 1*2) for 8 : res = 1 ( index 0 ) start iterating from : index 1 if gap>0 subtract gap ( res -= gap (here gap is 0) ) and add +1 ( for current occurrence ) index 2 : gap is zero so just add +1 we get res = 3 this means final money is pow(2,3) = 8 if at any moment res become <= 0 then set res =1 again (as this is minimum value possible) Do this for all and record max res CODE #include using namespace std; using namespace chrono; #define ll long long #define pb push_back #define For(i,n) for(int i=0; i>n; vectorv(n); For(i,n)cin>>v[i]; mapm; For(i,n){ m[ v[i] ].pb(i); } ll a = v[0] , l = 1, r = 1; double sum = 1; for(auto itr : m){ vli temp = itr.second; ll siz = temp.size(); double res = 1; ll start = temp[0]; Fora(i,1,siz){ ll gap = (temp[i]-(temp[i-1]))-1 ; res -= gap; res ++ ; if(res>sum){ sum = res; a = itr.first; l = start+1; r = temp[i]+1; } if(res<1){ start = temp[i]; res = 1; } } } cout<>t; while(t--){ solve(); cout<
 » 4 days ago, # | ← Rev. 3 →   0 hi
 » 18 hours ago, # |   0 I want to share my program of task D-The Clock to other ones. Click here to see my programMy Solution: #include using namespace std; int t; int begin(string s) { string hour=s.substr(0,2); int ans=0; ans+=(hour[1]-'0'); ans+=(hour[0]-'0')*10; return ans; } int end(string s) { string minute=s.substr(3,2); int ans=0; ans+=(minute[1]-'0'); ans+=(minute[0]-'0')*10; return ans; } bool palindrome(int a,int b) { string s1=to_string(a); string s2=to_string(b); if(a<10) s1='0'+s1; if(b<10) s2='0'+s2; reverse(s1.begin(),s1.end()); if(s1==s2) return true; return false; } int main(void) { cin>>t; while(t--) { string time; int interval,hh,mm; cin>>time>>interval; //03:12 360 hh=begin(time); mm=end(time); //cout<=60) jw=xmm/60,xmm%=60; xhh+=whh+jw; if(xhh>=24) xhh%=24; //cout<
 » 4 hours ago, # |   0 Another div4 is coming soon
 » 81 minute(s) ago, # |   0 Ban div 4 contests.
 » 74 minutes ago, # |   0 pls someone give a hint for problem e unable to understand.