Идеи всех задач принадлежат MikeMirzayanov. Все задачи также приготовил он. Я просто помог с тестированием, вычитыванием и написанием разборов.
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
vector<int> ans;
int power = 1;
while (n > 0) {
if (n % 10 > 0) {
ans.push_back((n % 10) * power);
}
n /= 10;
power *= 10;
}
cout << ans.size() << endl;
for (auto number : ans) cout << number << " ";
cout << endl;
}
}
1352B - Сумма слагаемых одинаковой четности
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int n, k;
cin >> n >> k;
int n1 = n - (k - 1);
if (n1 > 0 && n1 % 2 == 1) {
cout << "YES" << endl;
for (int i = 0; i < k - 1; ++i) cout << "1 ";
cout << n1 << endl;
continue;
}
int n2 = n - 2 * (k - 1);
if (n2 > 0 && n2 % 2 == 0) {
cout << "YES" << endl;
for (int i = 0; i < k - 1; ++i) cout << "2 ";
cout << n2 << endl;
continue;
}
cout << "NO" << endl;
}
}
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int n, k;
cin >> n >> k;
int need = (k - 1) / (n - 1);
cout << k + need << endl;
}
}
Разбор
Tutorial is loading...
Решение за O(n)
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
vector<int> a(n);
for (auto &it : a) cin >> it;
int l = 0, r = n - 1;
int suml = 0, sumr = 0;
int cnt = 0, ansl = 0, ansr = 0;
while (l <= r) {
if (cnt % 2 == 0) {
int nsuml = 0;
while (l <= r && nsuml <= sumr) {
nsuml += a[l++];
}
ansl += nsuml;
suml = nsuml;
} else {
int nsumr = 0;
while (l <= r && nsumr <= suml) {
nsumr += a[r--];
}
ansr += nsumr;
sumr = nsumr;
}
++cnt;
}
cout << cnt << " " << ansl << " " << ansr << endl;
}
}
Решение за O(n^2)
// Author: Ivan Kazmenko ([email protected])
import std.algorithm;
import std.conv;
import std.range;
import std.stdio;
import std.string;
void main ()
{
auto tests = readln.strip.to !(int);
foreach (test; 0..tests)
{
auto n = readln.strip.to !(int);
auto a = readln.splitter.map !(to !(int)).array;
int [2] total;
int prev = 0;
int moves = 0;
while (!a.empty)
{
moves += 1;
int cur = 0;
while (!a.empty && cur <= prev)
{
cur += a[0];
a = a[1..$];
}
total[moves % 2] += cur;
prev = cur;
reverse (a);
}
writeln (moves, " ", total[1], " ", total[0]);
}
}
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
vector<int> a(n);
vector<int> cnt(n + 1);
int ans = 0;
for (auto &it : a) {
cin >> it;
++cnt[it];
}
for (int l = 0; l < n; ++l) {
int sum = 0;
for (int r = l; r < n; ++r) {
sum += a[r];
if (l == r) continue;
if (sum <= n) {
ans += cnt[sum];
cnt[sum] = 0;
}
}
}
cout << ans << endl;
}
}
1352F - Восстановление 01-строки
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int n0, n1, n2;
cin >> n0 >> n1 >> n2;
if (n1 == 0) {
if (n0 != 0) {
cout << string(n0 + 1, '0') << endl;
} else {
cout << string(n2 + 1, '1') << endl;
}
continue;
}
string ans;
for (int i = 0; i < n1 + 1; ++i) {
if (i & 1) ans += "0";
else ans += "1";
}
ans.insert(1, string(n0, '0'));
ans.insert(0, string(n2, '1'));
cout << ans << endl;
}
}
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
if (n < 4) {
cout << -1 << endl;
continue;
}
for (int i = n; i >= 1; --i) {
if (i & 1) cout << i << " ";
}
cout << 4 << " " << 2 << " ";
for (int i = 6; i <= n; i += 2) {
cout << i << " ";
}
cout << endl;
}
}
На русском не прочитать разбор D (Unable to parse markup [type=CF_MATHJAX])
Исправлено еще до того, как опубликовали, пакет с полигона еще не подтянулся. Подождите немного.
so some of the participants are going to reach 1800+ ratings ? i believe this shouldn't be happening. CHECK THIS OUT
)
edit : this is what cf predictor is saying.
Obviously many experienced users made new accounts just to increase their ratings and Mike asked people not to do this but they won't listen. This is pretty bad and unintended.
The contest is rated for only those people who have given at least 2 rated rounds I guess. Please correct me If I am wrong
It is rated for unrated people too
bro but i have not been rated and iu solved 1 problem
The ratings haven't been updated yet
I think there should be some limit on the max increase possible for this round. Otherwise, it is very unfair for participants with rating >= 1400.
Removed, I realize I am wrong
"unrated" starts off with 1500, then he gets 375, Add them up, what do you think?
Removed, I realized rating matters.
"Regardless of whether you are a trusted participant of the fourth division or not, if your rating is less than 1400 (or you are a newcomer/unrated), then the round will be rated for you." - mike
i agree that the big boost is bad but i don't think anyone who didn't deserve at least 1600-1700 rating could solve all the questions in like 30 mins. but 1800 rating is a bit yikes (which is the +300 rating), so i agree that the max gain should be lowered to around 250 at MOST (for when you solve it as fast as the reds). the problem lies more in smurfing though
Also an important thing to notice is that out of the top hundred official participants, there are more than 60 users who have created a new account and these users increase as we go for further ranks, so this makes it even more difficult for the green/gray coders to increase their ratings.
extension not working
Just wanna know, how accurate this CF PREDICTOR have been in the past ??
Completely!
VERY. like only a difference of <=+5 or >=-5 is what I observed from the actual rating change in every contest I sit in.
But even if someone with a low rating became Expert after this round, their rating will fall again drastically after a div 2 round due to the number of problems they will solve. So it doesn't make that much of a difference I guess!
I agree with you,rating is not all,what the most important is we can improve ourselves by the codeforces
Can anyone explain the binary search solution for problem C?
The number of integers not divisible by $$$n$$$ among numbers $$$1, 2, \dots, x$$$ is $$$x - \lfloor\frac{x}{n}\rfloor$$$, so you can do binary search to find suitable value of $$$x$$$ (such smallest $$$x$$$ that $$$x - \lfloor\frac{x}{n}\rfloor \lt k$$$).
You may see my binary search implemented solution. 79500682
...
Hi, I have solved the ques in O(1) time without binary search. You can see here 79680203
80552242
In problem C, how need=(k-1)/(n-1) ?
This can be explained more easily if you understand that before each multiplier of $$$n$$$ there are $$$n-1$$$ integers not divisible by $$$n$$$, so we need to "skip" some number of such blocks. $$$\lfloor\frac{k}{n-1}\rfloor$$$ doesn't match because when $$$k$$$ is divisible by $$$n-1$$$ then we don't need to count the last block.
won't number of such multipliers be k / n ? if n = 3 and k = 8 there are 8 / 3 multipliers of 3
Yeah, but we want the numbers that are NOT divisible by n.
for n=3, k=8
1,2,4,5,7,8,10,11 k-1/n-1 +k gives Ans=8+3=11
However, a simple k/n gives Ans=8+2=10
I guess, there is a more complex logic of using k-1/n-1. Can anyone explain it in a simple way? Stuck at this for quite some time.
okay so idk how simple this is but like n = 3 k = 9 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
so k/(n-1) is kinda like accounting for the extra cycle, because you need 1 less N to traverse a cycle (a cycle is 3 numbers, you only need 2 indices in this case to traverse the cycle). then k/(n-1) accounts for that extra N every time it cycles through (if u write it out on paper it makes sense lol)
and if the cycle perfectly lands (when k%(n-1) ==0) then you gotta subtract one because you overcounted a cycle. for example, in the n= 3 and k = 8 case, if you did this cycle counting normally you'll end up with 12, but because the cycle cleanly divides you gotta subtract 1. at least i think that's how you do it idk it made sense when i wrote it out and went through cases
and of course catch edge cases k == n and k < n
From 1 to N, there are N-1 numbers not divisible by N. In every block of N numbers, N-1 are not divisible by N. So if we want K numbers, there must be at least $$$\frac{K}{N-1}$$$ blocks.
Each block with N numbers. Let the number of blocks be B. This will take you to $$$B\times N$$$.
So, from 1 to $$$BN$$$, you have $$$B\times(N-1)$$$ bad numbers. Count how many more do you need and add that to your current value. If you are already complete, then $$$BN-1$$$ is your $$$K^{th}$$$ Number.
I hope I'm not making this complex. Code For Reference
i wonder how vovuhs solution is correct :)
nice dude, beautiful explaination...thank you!!!
I got stuck at this for an hour and ruined my contest. D was definitely easier than C and it is unfair to put it after C.
I think the difficulty is subjective. I couldn't do B for a long time, switched to C, and I felt it was easier. I totally agree that D was easy to understand but the implementation can get a bit complicated. C was thinking based and if you get the trick, it becomes really easy to implement.
"Each multiplier of n shifts our answer by 1. The number of such multipliers is need = k — 1 / n — 1" so he finds numbers such that are divisible by n
I attempted to solve F using dp[n1][n2][n3][ends] : answer to the original problem when you have n1,n2,n3 and the resulting string ends in "ends" ("0" or "1"). My Python solution seems to TLE. Is it because this solution is basically O(10^6) and the number of test cases <= O(10^3), overall O(10^9)? or am I doing something silly
It's even worse than that, because there can be $$$c_0+c_1+c_2$$$ digits in each dp state, so you get something like $$$O(T \cdot (c_0+c_1+c_2) \cdot c_0c_1c_2) \approx 3 \cdot 10^{11}$$$, which will very clearly not pass.
it can be done using DP, see this: 79628831
nice solution, just one doubt, inside the main function, how do I know whether to call
solve(n0,n1,n2,1)orsolve(n0,n1,n2,0), I mean both are equally likely to happen(except when we only want 0s or only want 1s in our desired string). Sometimes, I stuck with such doubts and hence unable to decide which way to call recursive dp functions. Can you explain why inmain(), you calledsolve(n0,n1,n2,0)first and then when s.length() was 0, then you decided to callsolve(n0,n1,n2,1). Rest is fine to understand.I assume you understand DP state, using 0 means string with ending at 0 which has n0,n1 and n2 count, but if there is no such string let's calculate one that ends with 1
ok, so you're first guessing whether it can be done with string ending at 0, if not then you try it with string ending with 1. Right?
this round was so good and i had so much fun as a pupil and learnt new things i hope u do more div4 as u can <3
Can we do Div4-E with segment trees?
You dont need to use it and it will most probably give MLE . Segment trees take huge amount of space and building and quering them would take additional logn time which would prove to be costly .
Segment trees take O(n) space. MLE won't be an issue, I guess. It is just that you don't wanna use a sword for cutting lemons.
But TLE might be an issue for sure . MLE not sure but still a bad choice for 64MB question XD.
Segment trees take 4*n space. So it won't be a problem with MLE.
How to solve problem G using DFS (Depth First Search) ? I tried by joining numbers whose absolute difference satisfies the given condition and then traversing. How can I modify to make it work.?
Checking all possible paths would lead to an exponential solution I think.
There can be cycles too [2->4->6->2] (>_<).
Maybe as someone mentioned above, don't cut lemons with swords.
removed
Could someone try to hack 79593227? It does O(n * n * log(n)), which should be cut by the strict time limit but I couldn't hack it.
Can anyone explain the tutorial of the B problem in a much simpler way?
essentially, odd numbers can only be made with odd numbers, and an odd amount of numbers at that.
even numbers can be made with any amount of even numbers and an even amount of odd numbers.
then you set all the numbers to 1 or 2 (depending on if you want the array to be even or odd based on those rules) and then the last number is the amount that ur left with after subtracting all of the 1s and 2s
I think my solution to G is a bit interesting. For all quadruples of natural numbers till the floor(n/4)th tuple can be arranged in order : 2 4 1 3, eg. 2 4 1 3 6 8 5 7 10 12.... and for the last incomplete tuple, it is easy enough to insert them in between elements of the last complete tuple.
can anyone tell me, how to solve G using bfs/dfs?
The vertex set can be represented as the set of all numbers from 1 to n. The adjacent numbers for a node 'x' would be {x-4,x-3,x-2,x+2,x+3,x+4} provided they are in between 1 and n inclusive. Now you can perform DFS. The answer would be the DFS which satisfies the given condition in the problem. (NOTE: Edited: It will not be the DFS which visits all vertices...since the graph formed is connected, each DFS will visit all vertices) Even if you do a DFS from each number, the complexity would be O(n^2) (this include the time for checking the validity of the current DFS), which I guess would pass the problem constraints.
i tried it earlier but its not working.
Here is my DFS solution
Saw your code, inside the dfs() for a given node, you are try all possible relevant nodes to jump. I saw that you are only pushing nodes into your path array. Why ain't you popping the elements off, if you don't find the relevant path? Also, why in your main function are you checking val<2 && val>4, ain't you pushing the relevant nodes into you path array. You yourself first built the graph with reachable nodes, why is that condition checking even necessary?
Kindly Help.
I am not trying all possible nodes to jump. Take for example, N = 4, if you draw the graph the edges will be 1-3, 1-4, 2-4. If I start a DFS from node 1, the path array will contain elements in the following order {1,3,4,2}, which is not a valid answer. The correct answer would be the DFS that starts from node 2, {2,4,1,3}. A DFS generates a Spanning Tree, not a path. The path array is just storing the sequence in which the depth first traversal occurs and it may or may not be a path. In order to check whether it is a valid path (both in terms of the definition of being a path and the given condition of the problem), by just checking the given condition of the problem, I would be able to arrive at the solution.
Wouldn't the if(val<2 || val>4) always hold false because you built the graph in way that only relevant nodes are traversed and after the dfs() call is over, you are again checking the same thing. Shouldn't you rather be checking if the lenght of path array is n or not. Because if it is n, then we always know that the path will be relevant(again because that's how you built the
grapharray and we will never be able to traverse irrelevant nodes(that do not hold the condition given in the question)). If you agree with me, then I still do not understand why without checking the path length that your solution is passing plus why that rechecking of(val<2 || val>4)condition when that's how your dfs function will traverse the nodes? Kindly explain.Please refer to my previous reply. In the example I have given, both have a length of n, but only DFS from node 2 is a valid answer. Refer to this link for better understanding.
You didn't get me, you stored nodes in the adjacency list such that they atleast differ by 2 and atmost by 4, then why are you doing checking it again after the dfs call. The dfs traversal will take care of jumping only to the relevant nodes. You said that the path array may contain {1,3,2,4} can you tell me how did you path store 2 after 3, when you don't have 2 in the adjacency list for 3?
Check it out! 79806425
In the tutorial of problem D it is mentioned that we could solve the problem in O(n^2) but isn't the required complexity O(n) ? Correct me if i am wrong.
O(n^2) is the time complexity and O(n) is space complexity. In this question space is unusually restricted so space complexity should not exceed O(n) but time complexity of O(n^2) works just fine
I am talking about problem D not E.
I think you overestimated the running time of O(n^2) in this problem.Simply (n^2)*T or (SUMn)^2 might not give the correct complexity in this problem. Just try using lagrange or some differential method and we get that worst case is when T=200 and all Ni = 1000, which can be optimized to 1e8 executions:Link
The solution of G in tutorial is very good but how to come up with this solution? I mean the intuition for the tutorial solution.
I played with small permutations, trying to find a working block small enough so I could replicate it, and built the big N-elements answer. For me, it turned out to be a block of the first 4 numbers:
2 4 1 3
For N = 4 this is good. So as I said, I tried to make it work like a "brick" to build the answer, making chains of this block. So, if you have N = 8, you can "link" two of these structures:
2 4 1 3 — 6 8 5 7
And so on. Then I worked out on the case when N = 5 (more generally, when N%4 = 1), and I noticed that simply adding n to the sequence is ok.
For N = 6, or N%4 = 2, you have to pop the last element (3) of your sequence, and mix the remaining two elements:
2 4 1 5 3 6
And lastly for N = 7 or N%4 = 3, an extra step, also removing the last element and mixing the remaining three elements:
2 4 1 5 7 3 6
Am I the only one who solved problem G by graph?
Can you explain your approach
Make a graph of N nodes numbered 1 to N. Then for each node i add bidirectional edges between node i and i+2,i+3,i+4. After that Run dfs from each node separately, and check only for the straight path going from the node to any leaf node, if that path covers all N nodes then answer is the node of the path in the same order, otherwise if don't get any such path at the end, then answer doesn't exist.
I saw in your code that in your dfs(), you have used break statement. If we don't get a relevant path by traversing from the given node then don't you think that then we should try for another node rather than giving yup our hopes that we won't find the right path if the very first node is unvisited. Moreover, in that case the
pathvector (in your code-v) will be not remain global but rather be passed in the function(in non-reference way)Kindly correct me, as to why your code used break?
I just saw your solution for G, where you visit a node if it's not visited and then break out of the loop. What if you start from Node 1, then visit Node 3 and find out that permutation starting from 1->3 is not possible but permutation starting from 1->4 is possible ? You are not visiting node 4 from 1 as you have a break statement.
What you are saying is theoretically not possible, can you give any test case?
Yes, it seems like it's not possible after checking few cases. But can you prove that your solution always finds a solution if it exists? Also why is it theoretically not possible ?
ok, nice, just one question, is that due to the order you store nodes in the adjacency list for a given node or what? Will that still hold if you change the order of placing elements in some random order inside the adjacency list?
Did you figure it out ?
Actually, changing the order of elements does not change the answer. But, I still didn't get the break; in the dfs in the for loop. If you got that, kindly reply.
is this contest unrated?
Rated for participants below 1400 rating
Can anyone explain in Problem E why am I getting TLE when i am using map to count the frequency and find them 79611958 and it accepts the solution when i m using unordered map 79613165 i think in both cases overall time complexity is O(n*n). vovuh?
Map has a time complexity of O(nlogn).
still smaller than n*n
NO it's O(n^2logn)...
The total complexity becomes $$$n^2*\log{n}$$$, which is higher than the expected complexity. The extra $$$\log{n}$$$ factor arises because of using
mapI used unordered_set instead of map. Which should be have constant Time Complexity for inserting and finding elements. Still TLE on subtask 3.
I think unordered_set has a very high constant time factor because it uses hash table.
Even though insertion could be O(1), the hash table updating is very slow. Therefore, the number of insertion operations that can be done in 1 second is lesser than that of typical constant time operations.
Unordered (hash-based) sets/maps have average constant time complexity, but since the standard hash function is deterministic, a test case can be crafted specifically to cause a high number of items to be placed in the same bucket, degenerating it to a linked list.
Can anyone explain the case when n1 = 0 for problem F. Weren't we supposed to output a SINGLE string??????? I assumed that n1 = 0 would implicitly mean that either n2 = 0 || n0 = 0. Wasn't that supposed to be the case?????????
can anyone explain how case n0=1 , n1=0 , n2=1 will be handled in question F ??
there is no valid answer for cases like n1=0 and n0 and n2 being both being non zero .your submission will not be evaluated on these type of cases.
when will the ratings be updated
After system testing is complete.
Is there a way of doing E using sliding windows that doesn't cause TLE? or particularly any other way of doing E, I don't like editorials' solution for E, and was thinking if there is some other way.
I got accepted with two pointers (which you can relate to sliding window). My approach was going through each element of the array and finding a subsegment with a sum equal to the element, as long as the difference between the two indices $$$l$$$ and $$$r$$$ is higher than 1. Two pointers has a complexity of $$$O(2n)$$$...so you can simplfy it to $$$O(n)$$$, and doing it for each element gives a complexity of $$$O(n^2)$$$.
Here's my submission: 79549243
Just to be sure, I also used printf and scanf instead of cin, cout and FAST I/O, as it can be even faster with large inputs. Also note that I could have also saved the result for previous numbers to make it faster when there are repetitions, but in the worst case scenario where that wouldn't happen, it wouldn't be of much help.
use this function for each elements of array ,increment ans by 1 if this function return true,take care of the function it return true for subarray of length one also but in our case it should be at least two. overall time complexity O(n*n),space complexity O(n). my sol
did not receive a rating even after solving a question in the contest div.4
Rating will be updated soon!
I can't find my username in the final standings list.
click on show unofficial
Not there either
You are ranked on position 9100.
Can someone please explain me why this submission is WA. This is the link of my submission It went wrong on testcase 111, I do not understand why?
1 1 8 1
What do you mean by this Eureka17?
Your code will output 110101010100. However there is 9 pairs that have 1 "1". The reason is that you probably didn't consider the case where there is even number of pair that has 1 "1". When this happens, before you add only zeros, the last digit of sequence will be 1 (the reason is you started with 0 and you add even number of digits). Because the last digit is 1, when you add zeros, one more pair that has 1 "1" appear.
For a case 1 2 0, your code is producing a output 10100. In your output n2 = 3, but it should be 2.
Can someone help me out, in problem A my code is working fine with testcase1 in Hackerrank editor and my terminal but it is giving different output when I submit it in CForces. Submission: https://codeforces.me/contest/1352/submission/79486521
submit GNU C++14 or GNU C++17
I had the same problem during the round. I fixed it by replacing pow function with a variable starting in 1 outside the loop, and multiplying by 10 each cycle.
I have used a hashmap in problem E but still TLE.Can anyone explain why i cant get it Accepted with this solution(https://codeforces.me/contest/1352/submission/79563004).I use java,can this be a factor as it is slower than c++.
Consider case n1=0 separately and print the sting of n0+1 zeros or n2+1 ones correspondingly.
won't this make a 01 appear in the string while there should be no such pair in the string?
(I kind of misunderstood what you said, so I'm rewriting my comment.)
That sentence means how to handle cases like
3 0 0or0 0 4.For
3 0 0you need0000, and for0 0 4you should output11111.Testcases such as
3 0 3will not appear at all, since there must exist "01" or "10" somewhere so there is no solution for this case. The problem statement says that impossible testcases will not be given.nice problems for rating<=1400....i learned a lot (mainly difference b/w sets and arrays)
Скажите пожалуйста почему в задаче C количество множителея не равно [n/k]?
Can anyone tell me the problem with my Codeforces Round 640 (Div. 4) Ques B. Here is my code:
include<bits/stdc++.h>
using namespace std; int main(){ int t; cin>>t; while(t-->0){ int n,k; cin>>n>>k; if(k==1){ cout<<"YES"<<endl; cout<<n; } else if(n==k){ cout<<"YES"<<endl; for(int i=1;i<=k;i++) cout<<'1'<<' '; } else{ if(n%2==0){ if(k<=n/2){ cout<<"YES"<<endl; for(int i=1;i<k;i++) cout<<'2'<<' '; int number = n — (k-1)*2; cout<<number; } else cout<<"NO"; } else{ if(k%2==0) cout<<"NO"; else if(k>n) cout<<"NO"; else{ cout<<"YES"<<endl; for(int i=1;i<k;i++) cout<<'1'<<' '; int number = n — (k-1)*1; cout<<number; } } } cout<<endl; } return 0; }
In your submission of that code, the judge says it failed in case 54 of the 2nd test, as you didn't give an answer to it, although it exists. Luckily that line of the test case is shown, and it seems that fails with 6 4. And as you can see, you can represent it with numbers 1 1 1 3.
Anyone still struggling about writing a good comment?
I do :(
My solution for 1352E had O(n^2) time and got TLE. Approach is quite similar to the one described in editorial. Perhaps the time limits are too strict
https://codeforces.me/contest/1352/submission/79526875__
The issue with your code is that you are using
map, and increases the complexity of your solution to $$$O(n^2*n log n)$$$, giving TLE.Can anyone tell me if there is any mathematical intuition behind the derivation of formula ⌊k−1/n−1⌋ in .1352C - K-th Not Divisible by n
for every n integers , there are n-1 integers before it which are not divisible by n; say n=4, then there are n-1=3 integers 1,2,3 which are not divisible by 4.
so for after every n-1 positions,there is multiple of n . So our ans is shifted by 1; then for k postions, your ans is shifted by k/(n-1);
but if k is divisble by n-1 , we dont need to count the final shift.
So we can check for only k-1 pos , i.e shift= k-1 / (n-1)
shift=(k-1)/(n-1); ans= k + shift ; print ans;
why this code is accepted? where i mark "here",sum is out of bounds.
problem E
Arrays in C or c++ are allocated in contiguous memory. Let your array size is 6 and you store 8 elements in the array. Then the two extra elements may or may not be stored in the array. It depends on your memory location. If the two extra elements stored lie in another's memory location (i.e. stored in those locations where memory is already allocated) then the compiler will show some error, but if those additional elements get some unallocated memory or free memory then there will be no problem while running the program.
In your case, the additional elements are getting free unallocated memory in the memory allocation (i.e. program's virtual memory), so your program is not showing any error. It varies compiler to compiler.
thank you very much
I feel div4B harder than 1329B - Dreamoon Likes Sequences, which is div1B. Div4B was too dirty kind of problem.
That depends on the perspective and the mathematical background. For me 1329B is more or less unsolvable. I remember having read the editorial some days ago but still could not instantly tell how it works.
I am getting WA on test case 2 of problem E Can someone help me know which test case can it go wrong? Link : https://codeforces.me/contest/1352/submission/79717403
say please, why the solution of C isn't k+[k/n]?
You Can Refer This Explaination
Thank you
Скажите пожалуйста, почему решение C не равно k+[k/n]?
In problem E. Special Elements, if I compute a prefix array and run two loops to take out sum and use hashing as used above to count the special elements, why does it gives TLE?
When this also takes O(n^2) and the above mentioned solution also takes O(n^2) ?
My Solution : https://codeforces.me/contest/1352/submission/79552941
Does it shows TLE even if memory limit exceeds ? If not then what does the judgement says ?
You did not check the sum
int k = pre[j]-pre[i];to be in bounds of the array, but then write at that position.h[k]=0;Thanks, got an AC.
Could you please state the reason as why does not checking the bounds gives TLE and not Runtime error ?
Accessing something out of bounds gives Runtime error , right? Please correct me.
Thats the undefined part of undefined behaviour
For Problem F .. The constrain (0≤n0,n1,n2≤100; n0+n1+n2>0 ) .
But n2 must be n2>=1 . if n0 = 1 n1 = 0 n2 = 2 output is = 00 . Why? and How !!!!!!!
if n2==0 there will no such solution that can satisfy the condition.
It is guaranteed that the answer for given n0,n1,n2 exists.
Can Anybody tell me the logic behind why l<=r and not l<r in alice,bob and candies sum??
Because either of them will eat the candy at the index l=r. If you use l<r then 1 candy will remain uneaten.
BhavyaBhatt Suppose if Alice has eaten that candy , and According to the loop's Condition(l<=r) bob can eat the candy but he shouldn't be eating that candy because Alice has already eaten it.
Am I right?
l is the index of leftmost uneaten candy and r is the index of rightmost uneaten candy. so when l=r happens, that means all candies from index 0 to index l-1 are eaten by alice and all candies from index r+1 to n-1 are eaten by bob. so candy at index l=r will be uneaten so in the loop we should proceed. Now if alice's turn is to eat candy then l++ will be done and all candies are eaten (l>r happens) and similarly if bob's turn is to eat candy then r-- will be done and again all candies are eaten (l>r happens).
Thank you for the reply BhavyaBhatt.I know the concept you are talking about , but Suppose Alice ate the candy at l=r then we should have not let bob eat that already eaten candy , because if bob also eats the candy then output will be wrong because it will be counted!
Yes, Bob should not be able to eat that candy and the code given in editorial won't allow Bob to eat that candy as there is (l<=r) condition in the while loops inside the if-else condition.
For F(Binary string construction), if n0=1, n1=0, n2=1, the editorial solution, gives output 00. Some other codes gives output 0011, which is incorrect. What should be the correct answer since no valid string is possible for this case.
Update- Missed the statement " It is guaranteed that a solution exists."
Can someone explain why am I getting TLE in question D, I know its an easy problem but I checked my while loops and none of them are recurring, idk why my code suddenly stops any bad practices you see,Thanks My Solution
This loop?
Problem 1352F Can someone please explain why am I getting TLE in this solution
This is my submission
Since maximum number of elements inside the string will be always less than 500 and number of independent test case is 1000 so we can solve the question in O(t*n) time right?? But why I am getting TLE here?
mysubmission
In problem C, why is $$$need = ⌊(k-1)÷(n-1)⌋$$$, and not $$$need = ⌊k÷n⌋$$$?
Shittiest constraints ever for problem E.
1352C - K-е неделящееся на n
The key observation is that in every block of n consecutive integers, exactly one number is divisible by n, and the remaining n — 1 numbers are valid.
Instead of calculating the kth number which is not divisible by the number n manully(that may give tle as k can go upto 1e9), we think in terms of blocks.
Every block of size n contributes n — 1 valid numbers, and that means for n — 1 valid numbers, we skipped exactly one multiple of n .
To reach the kth valid number, we must add the number of skipped valid blocks before k, in k such that we can reach the answer.
OR we can apply binary search also(easy approach but takes a little more time than other method ).
Why D questions O(n) solution also get 171ms but someone's solution can get 46ms. I'm confused
386984602 accidently solved this G question , first tried with possibility approch, but then it got accepted, thought of it why its happening and then later optimized it- 386986823 , the logic is , i ll check if we can form valid array by starting with i th element out of all n (1, 2, 3.. n) , and the way i was making array was — checking first valid element at current index which is not used till now. if all valid elements for current index (that is last index value — 4 to last index value — 2 and last index value + 2 to last index value + 4) is not possible to put in (either cause of going out of n or being lessor than 0 or is already in set) then it means its not possible to make valid array for current starting element, will then try for all other starting elements possibilities.