Vovuh's blog

By Vovuh, history, 6 months ago, , 1157A - Reachable Numbers

Idea: BledDest

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Solution

1157B - Long Number

Idea: BledDest

Tutorial
Solution

1157C1 - Increasing Subsequence (easy version)

Idea: MikeMirzayanov

Tutorial
Solution

1157C2 - Increasing Subsequence (hard version)

Idea: MikeMirzayanov

Tutorial
Solution

1157D - N Problems During K Days

Idea: MikeMirzayanov

Tutorial
Solution

1157E - Minimum Array

Idea: Vovuh

Tutorial
Solution

1157F - Maximum Balanced Circle

Idea: MikeMirzayanov

Tutorial
Solution

1157G - Inverse of Rows and Columns

Idea: Vovuh

This is the comment about the quadratic solution. Thank you so much for mentioning this fact, STommydx!

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Solution 555, Comments (37)
 » The editorial for Problem G looks like a little the editorial of Problem F
•  » » Thank you for mentioning, I just forgot to fix this place :)
 » it was a greedy contest :D Good round by the way
 » My solution problem D after contest(sad): 1)Find minimum first element with binary search 2) let's get maximum suffix, when we can add 1 for all numbers on segment, do it, while exist this is segment 3) end O(nLogn), and write easyhttps://codeforces.com/contest/1157/submission/53393728
 » 6 months ago, # | ← Rev. 5 →   In problem "A" we don't need to store reachable numbers ,As soon as we hit a number less than 10 we add 9 to our counter and stop Nice problems btw
 » 1157D — N Problems During K Days can be solved simply by putting $1,2,\dots,k$ initially and then trying to put $(n-(k*(k+1)/2))/k$ elements in each, and then from reverse order adding as many elements in each position as possible, without considering special cases separately.You can go through my submission for clear understanding.
 » In problem number B, why we are not checking the digits which comes after the first digit which become less after replacing. for example if a = 13373 and f[] = {1, 2, 5, 4, 6, 6, 3, 1, 9}; then the correct answer is 15573 but why not 15575;
•  » » The problem mentions a contiguous subsegment and the operation can be performed at most once.
 » Anyone help me in finding the cause of TLE in Problem D in my solution.
•  » » 6 months ago, # ^ | ← Rev. 2 →   "erase" for vector works in linear time of size of the whole vector
•  » » » And what about erase in multiset?Is there any way that I can still use vector and solve the problem. I know that using multiset will make it very easy but still as a matter of knowing.BTW, Thanks for the reply buddy !!
•  » » » » 6 months ago, # ^ | ← Rev. 2 →   You can’t use vector.erase for this problem. unless you want to code yourself a multiset libary for vectors, multiset is the only way to solve the problem in AC time, which erases elements from the container at log(n) time.
•  » » » » » Thanx for the reply mate !!
 » This problem also has a solution like D:https://www.codechef.com/SNCK1B19/problems/MAXPRODU
 » please help me...y its showing error in test case 15 always!!!https://codeforces.com/contest/1157/submission/53440011
•  » » make it a habit to write long codes, your is too short to get accepted
 » Does anyone have a solution for problem E which does not involve STL structures, specifically without using multiset,set,map or multimap ?
•  » » You can use segment trees.check this solution 53473452
 » 6 months ago, # | ← Rev. 2 →   Help me out in finding TLE in my Solution for Problem C
•  » » while(l <= r)
 » In problem E, why "lower_bound(m.begin(), m.end(), val)" gives TLE 53488500 and "m.lower_bound(val)" got Accepted 53488794 ?
•  » »
 » 6 months ago, # | ← Rev. 2 →   Problem D: "Then if nn != k — 1 or k = 1 then this answer is correct."Any explanation for this ?
 » For problem 1157c1, what if the input is7 7 4 5 6 3 2 1The solution code out puts 5 RRRRL But if I discard 7 and take the subsequence [4 5 6 3 2 1],it's possible to get a strictly increasing subsequence of size 6, right?
 » When is the next div3?
 » What's wrong with my solution for problem B? https://codeforces.com/contest/1157/submission/53566680 I am getting wrong answer on test case 7
•  » » Try for 5 43111 9 2 3 5 1 1 1 1 1 Answer should be : 53999, get it?
 » 6 months ago, # | ← Rev. 2 →   I solved D in a different way. Mine
 » 6 months ago, # | ← Rev. 2 →   I'm kind of late, but problem D can be easily (more or less) solved by binary searching the first element, and then binary searching every other element as well, from left to right, so that the minimum possible sum is <= n and the maximum possible sum is >= n. My submission
 » 6 months ago, # | ← Rev. 2 →   Nice Problem Set.
 » For the D problem, I just initialized array to 1,2,...,K then added (S — (K * (K + 1)) / 2) / N to every element in the array, and then just distributed (S — (K * (K + 1)) / 2) % N from the behind appropriately. Just wondering why there were so less submissions for this problem?
 » How to solve problem Minimum Array using segment trees??
•  » » 6 months ago, # ^ | ← Rev. 5 →   I solved with segment tree, so I will try to explain my solution:First we need to sort B array, to be able to find some index j for each $A_i$ such that $A_i + B_l \lt n$ for all $l \lt j$, and $A_i + B_r \ge n$ for all $r \ge j$. We'll have only one index j for each $A_i$, because $max(A) + max(B) < 2n$.After that we can divide B array in two parts and find the best among these greedly, using segment tree here to get minimum, and then updating the used element to not be taken more than once.Submission
 » can anyone please tell why my recursive solution for c2 gives tle ? I think my solution too has the same time complexity as the editorial's .
 » Guys can someone help me understand why http://codeforces.com/contest/1157/submission/53985650 gives wrong results on test case 9
»
2 weeks ago, # |
Rev. 2