Идея: MikeMirzayanov
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int t;
cin >> t;
while (t--) {
int a, b;
cin >> a >> b;
if (a % b == 0) cout << 0 << endl;
else cout << b - a % b << endl;
}
return 0;
}
Идея: MikeMirzayanov
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
#define forn(i, n) for (int i = 0; i < int(n); i++)
int main() {
int t;
cin >> t;
forn(tt, t) {
int n, k;
cin >> n >> k;
string s(n, 'a');
for (int i = n - 2; i >= 0; i--) {
if (k <= (n - i - 1)) {
s[i] = 'b';
s[n - k] = 'b';
cout << s << endl;
break;
}
k -= (n - i - 1);
}
}
}
Идея: vovuh
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int t;
cin >> t;
while (t--) {
int n;
string x;
cin >> n >> x;
string a(n, '0'), b(n, '0');
for (int i = 0; i < n; ++i) {
if (x[i] == '1') {
a[i] = '1';
b[i] = '0';
for (int j = i + 1; j < n; ++j) {
b[j] = x[j];
}
break;
} else {
a[i] = b[i] = '0' + (x[i] - '0') / 2;
}
}
cout << a << endl << b << endl;
}
return 0;
}
1328D - Карусель, карусель --- это радость для нас
Идея: MikeMirzayanov
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
int solve() {
int n;
cin >> n;
vector<int> a(n);
for (int i = 0; i < n; ++i) {
cin >> a[i];
}
if (count(a.begin(), a.end(), a[0]) == n) {
cout << 1 << endl;
for (int i = 0; i < n; ++i) {
cout << 1 << " ";
}
cout << endl;
return 0;
}
if (n % 2 == 0) {
cout << 2 << endl;
for (int i = 0; i < n; ++i) {
cout << i % 2 + 1 << " ";
}
cout << endl;
return 0;
}
for (int i = 0; i < n; ++i) {
if (a[i] == a[(i + 1) % n]) {
vector<int> ans(n);
for (int j = 0, pos = i + 1; pos < n; ++pos, j ^= 1) {
ans[pos] = j + 1;
}
for (int j = 0, pos = i; pos >= 0; --pos, j ^= 1) {
ans[pos] = j + 1;
}
cout << 2 << endl;
for (int pos = 0; pos < n; ++pos) {
cout << ans[pos] << " ";
}
cout << endl;
return 0;
}
}
cout << 3 << endl;
for (int i = 0; i < n - 1; ++i) {
cout << i % 2 + 1 << " ";
}
cout << 3 << endl;
return 0;
}
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int q;
cin >> q;
for (int qq = 0; qq < q; qq++) {
solve();
}
return 0;
}
Идея: MikeMirzayanov и vovuh
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
int T;
vector<int> p, d;
vector<int> tin, tout;
vector<vector<int>> g;
void dfs(int v, int par = -1, int dep = 0) {
p[v] = par;
d[v] = dep;
tin[v] = T++;
for (auto to : g[v]) {
if (to == par) continue;
dfs(to, v, dep + 1);
}
tout[v] = T++;
}
bool isAnc(int v, int u) {
return tin[v] <= tin[u] && tout[u] <= tout[v];
}
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int n, m;
cin >> n >> m;
T = 0;
p = d = vector<int>(n);
tin = tout = vector<int>(n);
g = vector<vector<int>>(n);
for (int i = 0; i < n - 1; ++i) {
int x, y;
cin >> x >> y;
--x, --y;
g[x].push_back(y);
g[y].push_back(x);
}
dfs(0);
for (int i = 0; i < m; ++i) {
int k;
cin >> k;
vector<int> v(k);
for (auto &it : v) {
cin >> it;
--it;
}
int u = v[0];
for (auto it : v) if (d[u] < d[it]) u = it;
for (auto &it : v) {
if (it == u) continue;
if (p[it] != -1) it = p[it];
}
bool ok = true;
for (auto it : v) ok &= isAnc(it, u);
if (ok) cout << "YES" << endl;
else cout << "NO" << endl;
}
return 0;
}
Идея: MikeMirzayanov
Разбор
Tutorial is loading...
Решение
#include <bits/stdc++.h>
using namespace std;
const long long INF64 = 1e18;
int main() {
#ifdef _DEBUG
freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
int n, k;
cin >> n >> k;
vector<int> a(n);
for (auto &it : a) cin >> it;
sort(a.begin(), a.end());
vector<pair<int, int>> cnt;
for (auto it : a) {
if (cnt.empty() || cnt.back().first != it) {
cnt.push_back({it, 1});
} else {
++cnt.back().second;
}
}
vector<long long> prefsum, sufsum;
vector<int> prefcnt, sufcnt;
for (int i = 0; i < int(cnt.size()); ++i) {
long long cursum = cnt[i].first * 1ll * cnt[i].second;
int curcnt = cnt[i].second;
if (prefsum.empty()) {
prefsum.push_back(cursum);
prefcnt.push_back(curcnt);
} else {
prefsum.push_back(prefsum.back() + cursum);
prefcnt.push_back(prefcnt.back() + curcnt);
}
}
for (int i = int(cnt.size()) - 1; i >= 0; --i) {
long long cursum = cnt[i].first * 1ll * cnt[i].second;
int curcnt = cnt[i].second;
if (sufsum.empty()) {
sufsum.push_back(cursum);
sufcnt.push_back(curcnt);
} else {
sufsum.push_back(sufsum.back() + cursum);
sufcnt.push_back(sufcnt.back() + curcnt);
}
}
reverse(sufsum.begin(), sufsum.end());
reverse(sufcnt.begin(), sufcnt.end());
long long ans = INF64;
for (int i = 0; i < int(cnt.size()); ++i) {
int cur = max(0, k - cnt[i].second);
int needl = 0;
if (i > 0) needl = min(cur, prefcnt[i - 1]);
int needr = max(0, cur - needl);
long long res = 0;
if (i > 0 && needl > 0) {
res += prefcnt[i - 1] * 1ll * (cnt[i].first - 1) - prefsum[i - 1];
res += needl;
}
if (i + 1 < int(cnt.size()) && needr > 0) {
res += sufsum[i + 1] - sufcnt[i + 1] * 1ll * (cnt[i].first + 1);
res += needr;
}
ans = min(ans, res);
needr = 0;
if (i + 1 < int(cnt.size())) needr = min(cur, sufcnt[i + 1]);
needl = max(0, cur - needr);
res = 0;
if (i > 0 && needl > 0) {
res += prefcnt[i - 1] * 1ll * (cnt[i].first - 1) - prefsum[i - 1];
res += needl;
}
if (i + 1 < int(cnt.size()) && needr > 0) {
res += sufsum[i + 1] - sufcnt[i + 1] * 1ll * (cnt[i].first + 1);
res += needr;
}
ans = min(ans, res);
}
cout << ans << endl;
return 0;
}
Could someone please elaborate the tutorial for question B? Specifically, from where did we get the equations
k<=n-i-1andn-i-1?Starting at the end, for every index, let's imagine some string such that it looks like
aaaaaabb.Now imagine the left-most
bmoving to the left. For each move to the left, the right-mostbhas one more possible location. For example,aaaaaabbhas 1 position for the secondb,aaaaababhas 2, and so on.This remaining k value can also help us calculate the position of the right-most
bbecause of the fact that it's remainder once it is less thann-i-1hints at where it should go.Got your point. Thank you! :)
I still didnt got your point. How we are forming these equations k<=n-i-1 and n-i-1?
Consider a string s of length n. First, number the positions of the letters from 1(rightmost letter) to n(leftmost letter).
Lexicographically, the smallest string will have a 'b' in positions 2 and 1. It is easy to observe that there will be
i - 1strings having leftmost letter in theith position. So, all you need to do isk -= (i - 1)as long ask > 0andigoes from2ton. Suppose,jis the smallest value ofifor which you weren't able to perform this operation. That means the leftmostbwill be at positionj. Since, this isjfrom the right, it will be equivalent ton - j + 1from the left. This will bes[n - j].Now, if you would have
k = 1then rightmostbwill be at position of1. In general, for some remaining value ofk, the rightmostbwould be at positionk. So, that would meann - k + 1from the left. That would means[n - k].For eg:
n = 7andk = 5. You can dok -= 1andk -= 2to getk = 2.jwill be4. That meanss[7 - 4] = 'b'. Also, ifk = 2thens[7 - 2] = 'b'. Thus,s = "aaababa"fork = 5andn = 7.I hope this helps. It will be easier to understand it if you work it out a bit more on paper with examples of your own.
nice contest :)
I am unable to understand third part of D's code . Any hint .. it seemed very easy ,but still i got wa at tc 2.
you can always get an answer if you color every odd vertex with 1 and every even with 2 but when n is odd then first and last will have same color.so if you have a consecutive pair with same value then you can color both of them with same value which will change parity of all subsequent numbers.EX: 1,2,2,3,4 can be colored as 1,2,2,1,2 instead of 1,2,1,2,1. and if no such pair exist then you have to use 3rd color.
I had a different approach for 1st problem but it did not accept And when I compile on my idea it worked good.please help me
I checked your code, the idea is absolutely correct. However, you need to print a newline after every distinct answer otherwise answers for all test cases would be shown side by side. cout<<count_<<endl; Make this change in your code and it should work fine!
\n is faster than endl
Yes, but that's a different topic. Since he wrote the previous terms in CPP manner, I suggested using endl. And for this problem, this is not even a big deal!
Your solution will not work for large cases.
Let me expand the solution given, i.e how this equation came up
b - a%b— let's derive it.We know we can write
aasa = b * k + a%bwhere k is the quotient.kcan also be written ask = (a - a%b)/b. We are interested to find the nextkfor which remainder is zero, so we needk+1.So,
a + c = (k+1)b + 0where c is the count/number of steps (+1) to add toato makea%bzero. Substitutingkfrom above to find c:a+c = b * ((a- a%b)/b + 1)) = (a - a%b) + ba+c = a - a%b + bc = b - a%b. There it is!The constraints for problem B were weak many o(k) solutions passed. Example — https://codeforces.com/contest/1328/submission/74509641. I tried to hack it but can't
They'd probably FST.
Let's hope.
It passed o_0
This is actually unfair to those who actually solved it :(
I did B in O(1) per test case
Printing the string is O(n)
Beautiful approach for E!
Is there a name for the technique used in E? Or where can I read more on it (formal proofs, etc.)?
LCA using euler tour
Actually, you can solve some tasks about LCA such as https://codeforces.com/blog/entry/43917?locale=ru
I have a different approach for B.
We know that after n*(n-1)/2 permutations, both b's will be together. So we can basically make an array of all such numbers (1,3,6,10,..),and for a given k, we can find an n s.t n*(n-1)/2<=k (by binary search). Let j be the pos of that n. We can clearly see that for this value the no. of a's after the 2 b's will be exactly j (i.e, we can determine that particular permutation).
So, from that permutation, we can lexicographically check next permutations till we reach k and print that kth permutation.
C++ soln:74461692
Yes, I used binary search too.
" Write some LCA algorithms and other hard stuff or write about 15 lines of code and solve the problem " — One of the best editorial (E problem)!
I believe E can be solved by just maintain an array of vertex while do dfs. When you enter a vertex mark it and all it son in the maintained array, when you leave a vertex remove all it sons from the array. Now we can answer the queries in the order we visit the vertexes in the dfs. When we enter to a vertex v , we know exactly which vertexes appear on the path to it or in dist 1 and we can answer all the queries where u = v.
My mistake, I changed the problem in my head and made U alway the last vertex of the query.
I don't understand the solution for E after it explained how to figure out if vertex u is parent of vertex v. Can somebody explain it please ? (I don't want the explanation for checking if one vertex is ancestor of another, I want explanation for AFTER it).
if Vertex A is ancestor of Vertex B, on the way of reaching B we will cross A
In dfs(), we mark every vertex by its parent. (say u is parent of v, we mark v with u).
On the path to reach the deepest vertex(say vertex x) in each query:
3a. we mark all vertices with parent vertex 1(because vertex 1 is always on the path , therefore we can reach all of the immdediate children of vertex 1, since they all have distance of '1 unit' from the path )
3b. we mark all the ancestors of 'x'( cause we will cross them to reach 'x')
3c. we mark all the immdediate children of these ancestors we found( because we still can reach them as maximum distance from the path to x is '1')
My solution for Question D without using Graphs and DP.
https://codeforces.com/contest/1328/submission/74499296
The site is too f#cking slow today, it's taking minutes to logout/login, all links are loading really slow, is this normal or is this new??
because of the quarantine that is in place in a lot of countries a lot of people have more time to practice CP so the site gets a lot more people on it.
Anyone else used Quadratic equations to solve the main part of Question B in O(1) for each test case??
My solution :- https://codeforces.com/contest/1328/submission/74439092
Yeahh...me....that give the position of "b"'s in O(1)
Can someone suggest some good resources to learn how to make changes in DFS methods amd efficient graph algorithms...like many a times I see dfs1 and dfs2 kimd of functions.... Expecting genuine help!!
Dear MikeMirzayanov and vovuh, nice problems. But Someone stole my rating after this round. I was dark blue color, so i should be unrated for this Div 3 !
Thank MikeMirzayanov and vovuh for interesting problems and fast editorial!
I had used StringBuilders for the problem C. Here is my submission https://codeforces.com/contest/1328/submission/74463191 Can any java coder tell me why did it give a TLE?
Use PrintWriter for fast output. System.out.println() is too slow for large output. My submission: 74442693
this is my solution 74533033 to problem C in yesterday's contest. i dont understand what's wrong with it. it seems to work fine. can anyone help?
I have fixed your solution: 74544016. You need only to resize your strings such as x, a and b.
thanks:)
vaaven I generally take whole string as input at once and not by each character. So I have a doubt that while taking character wise input(cin>>s[i]), why resizing the string to n, was not needed for smaller inputs(n~5) as in test cases 1 and 2. Runtime error came in test case 3 for n=250.
Even I have the same doubt
For problem F, I sorted the array and used the difference between consecutive elements and the number of moves required to make all the numbers before a particular number to make all of them equal to that number. I did this for both prefix and suffix part and only counted the required number of moves. 74541622
how do you sure that you have " at least k equal elements in the array." in each step?
for initialization part i took first k elements that means that first k-1 elements are converted to the the kth element and subtracted the answer with the number of occurrences of the kth element. i also did the same thing for the kth element from the last.
Why you subtracted the number of occurrences of kth element?
Suppose
k = 5,
Ar = 2 4 5 6 6 6 7 8
m[6] = 3
dp[0] = 0 cost for all element to become 2
dp[1] = (4-2 * 1) + 0 = 2 cost for all element to become 4
dp[2] = (5-4 * 2) + 2 = 4 cost for all element to become 5
dp[3] = (6-5 * 3) + 4 = 7 cost for all element to become 6
dp[4] = (6-6 * 4) + 7 = 7 cost for all element to become 6
Now m[6] = 3-2 = 1
our answer is dp[4] — m[6] = 7 — (1) = 6
How answer is 6 ?
Keep in mind that we already have three 6's so we need not to convert 5 -> 6
Final answer = 6 , 6, 5, 6, 6, 6, 7, 8
Total cost = 6-2, 6-4, 5, 6, 6, 6, 7, 8
We only need to convert those which are required so if our Kth element is x and we are suxh more x already present ? we can use that too.
Similarly
k = 4
for ar = 1,1,1,2,3,3
3,3,2,2,3,3
answer = 5
We can do this for both sides and select the minimum one
Right. That's what i had in mind for this solution
Please explain what exactly is happening in the B problem?how are we knowing the position of b by decreasing by n — i — 1?
Can someone please help me figure out why am I getting timeout even if my complexity is O(N + sum(K)) for problem E? Here's a link to my solution. https://pastebin.com/y9Fx5day. Thanks in advance!!
Maxi can be equal n in worst case. So your solution is O(n^2)
Ohhh... Thanks a lot!!
can somebody please explain how the scoring works in codeforces? start off by explaining what's does it mean by +,+1,+2.
Number of attempts
Anyone who has done D using DP ? Please share your solution
I know that my code is not the most beautiful code :D, I would like if the tutorial had a DP solution for this problem.
my code
Here is my dp solution for proble D https://codeforces.com/contest/1328/submission/74586478
Hi Thanks for the perfect solution i was looking for. Thanks alot :)
This was my solution for B.
Solution- 74469154
i used the concept of "Sum of first n terms" So if the string size is n then the the total lexicographical strings produced will be sum of first n-1 terms.
for ex- n=5 so total number of lexicographical strings produced will be sum of first 4 numbers 1+2+3+4=10
(The concept is For every movement of leftmost b , Following will be the movement of rightmost b till the its position one more than the position of the leftmost b.
)
Complexity — O(n)
Me too
My Submission
Hello and thanks for contest. I was EXPERT before the contest DIV3 but my rate change -20, WHY? Thanks again.
Similarly I've seen an EXPERT coder get his rating increase for about 20 in the last div.3 round.Maybe it is a bug?
probably because you had registered yourself for this contest before rating change of last round (when you were<1600)
It says time limit exceeded but is the logic correct?
void solve()
{ int n, k;
}
Yes, logic is correct but it works very slow because you have O(k) where k can be equel 1e9.
Can anyone tell me what's the reason of getting TLE for the following solution of problem E? https://codeforces.com/contest/1328/submission/74550098
Max depth can be n. So you will make n iterations to check only 1 node. Your solution work in O(nm)
got it. thank you.
Because you iterate over the whole longest path every query, so the complexity may be O(mn) instead of O($$$\sum\limits_{k = 1}^mk_i$$$)
The testcase may be a linear chain tree with every query contains a far node from the root
got it. thank you!
Problem : E My submission : 74555362 Verdict: WA on TC 56 Can anyone tell what is the mistake? I cant figure it out. Used the same method as told in Editorial. My code is very short. Thanks a lot :)
I fixed your solution:74558555
I just add this:
and edit this:
Anyone who has done D using DP ? Please share your solution
C) Why if x[i]=2 or x[i]=0, then a[i]=b[i]=1 or a[i]=b[i]=0 relatively? Who can explain it? Please.
because max(1, 1) < max(2, 0) when x[i]=2 and max(0, 0) < max(1, 2) when x[i]=0
Can someone please explain this part Of problem E
It lies on this path if the root is the parent of u (it is always true) and u is the parent of fv. This approach can be used for each vertical path (such a path from x to y that lca(x,y) is either x or y).
And this code of problem E
int u = v[0]; for (auto it : v) if (d[u] < d[it]) u = it; for (auto &it : v) { if (it == u) continue; if (p[it] != -1) it = p[it]; } bool ok = true; for (auto it : v) ok &= isAnc(it, u);
See Mashup F: https://www.youtube.com/watch?v=mw2J6lvZZJ4&t=9712s. The explanations are really nice.
Help needed in E: My approach-: Check if there is path from root containing all the parents. Sort all the parents of given nodes according to their height. Then find lca of every adjacent pair, suppose x and y are adjacent , then if lca(x,y)==x then path exists else not. I am getting runtime error in test case 100. Can anyone help me out in this?
My submission 74567267
somebody plz explain B...
I also didn't know it before, got the solution by generating few cases while contest.Generate 6/7 cases and you will find the pattern.
Keep = (x)(x+1)/2 — k;
Editorial solution for E is nice, but there is a solution that requires much less thinking. Handle the queries offline, and then you simply need to do a dfs from the root and add all the neighbors you pass on the way down to your current set (to add to your current set, just mark them as visited in all the queries that contain this vertex. if a query has everything marked as visited, then it is good) and remove the neighbors on your way back up. This way you can just check all the paths in O(n+m) time (fairly high constant but still runs < 1s).
74575633
can you plz elaborate how you use add and remove in your code
He has made a vector of vectors qs where each entry(qs[i]) is a vector which stores queries in which i-th vertex has been mentioned.
Then in add function, for a vertex v, all the queries are traversed in which v-th vertex was mentioned and another vector qc is being increased for those queries (++qc[query];). When value of qc for a particular query reaches number of vertices in that query(denoted by qn[query]), it means that all the vertices in that query are neighbours of some path from root to some vertex. This count of vertices which are neighbours, was kept by qc[query].
Now remove fn is easy to understand as it is basically removing the neighbours count.
Yes, amangupta's description is correct. Just to clarify a bit more on the specific variables:
qn[i] — the total number of vertices in query i (qn = query number)
qc[i] — the number of vertices in query i that have been currently marked (qc = query current)
qg[i] — 1 if query i is achievable, 0 otherwise (qg = query good)
qs[i] — a list of queries that vertex i appears in (qs = queries)
My Submission
Thanks MikeMirzayanov and vovuh for awesome editorial specially problem E.
vovuh MikeMirzayanov how can we prove this fact for problem F.
** The second observation is that we first need to take elements from one end (only less or only greater) and only then from the other (if needed).**
why it will give correct answer and why we can't expand from both side. say if we need x, and we take y from begining first then we will take x-y from end. similarily for the same index, if we take a from beginning , we again take x — a from right.
I feel like C was too easy :/
CAN ANYONE EXPLAIN THE ERROR I HAVE REPLACED V[] WITH THEIR PARENT // TO OVERCOME THE DISTANCE FACTOR OF 1 I HAVE SORTED THE V[] ACCORDING TO THE LEVELS THEN CHECKED EVERY CONSECUTIVE WHETHER THEY LIE IN SAME PATH OR NOT
include<bits/stdc++.h>
using namespace std;
define ll int
define pb push_back
define fi first
define se second
define mp make_pair
ll in[200007],out[200007]; vector adj[200007]; ll level[200007]; ll par[200007]; ll k=0; ll k2; ll timer=0; int flag; pair<ll,ll> p[200007]; ll v[200007]; void dfs(ll u,ll lv) { level[u]=lv; ++timer; in[u]=timer; for(int i=0;i<adj[u].size();i++) { par[adj[u][i]]=u; dfs(adj[u][i],lv+1); } ++timer; out[u]=timer; } int main() { ios_base::sync_with_stdio(false); cin.tie(NULL); ll n,m; cin>>n>>m; level[0]=0; in[0]=INT_MIN; out[0]=INT_MAX; for(int i=0;i<n-1;i++) { ll x,y; cin>>x>>y; adj[x].pb(y); } dfs(1,1); par[1]=0; for(int idx=0;idx<m;idx++) {
}
Sharing your code via one of the text storage sites like pastebin or ubuntu paste might significantly increase your odds of getting a help
Or simply include the submission id like 74594978
This gives you TLE, wich means your solution runs out of time, hence is to slow. I think for every query you recreate the whole path from root to deepest node. This makes it basically $$$O(n^2)$$$ wich is to much for very deep trees.
hello i submitted a solution for kth beautiful string problem.i got runtime error for that during the contest.but today i submitted that same file from my pc without any edit and it got accepted.where can i complain regarding this? this is my first contest on codeforces.please help.
I think it is not the same code, you changed the types of your integers. Most likely the runtime error was caused by overflow.
Aside from that, would be useful for such question to include the both submission ids.
i didnot change anything bro...i just submitted the file that was there in my pc.without even opening it in the editor.and it got accepted this time
Your submission history does not look like that.
ok let it be
You're welcome.
Hi @vovuh, I have a question regarding 1328F and sorry if disturbing you.
I think I may have triggered some wired behavior of Java 11, with which Arrays.sort(int[]) becomes extremely slow. From my attempted submission https://codeforces.com/contest/1328/submission/74614186, you will see that it's TLE with Arrays.sort(a) at the top part of the solve(n,k,a). If I switched from Arrays.sort(a) to use an ArrayList of tmpA to sort and assign back to a, which becomes submission https://codeforces.com/contest/1328/submission/74613801, it's accepted.
I tried to reproduce this on my local machine by randomly generating 200000 numbers between 100000 and 200000 and failed, it's super fast with Arrays.sort(int[]). This may due my usage of openjdk 11, which differs from that used by codeforces, or due the data itself.
Could I have the 3rd testcase so I can verify it on my local machine?
Quick sort is known to have $$$O(n^2)$$$ in worst case.
see some link
Yeah I know that basic info.
But:
But you made your point, and I already suspect that this specific testcase may be one of the rare bad case for this implementation. That's why I need this case to experiment on my local machine. And if it is I may report this to jdk team as a bug.
Anyway, thanks for the reply.
Hi, MikeMirzayanov and vovuh, I did further experiment on this issue.
As in submission https://codeforces.com/contest/1328/submission/74698017, I deliberately force the 3rd test case to use ArrayList.sort(...) and let rest of the test cases use Arrays.sort(int[]), everything goes well and it's accepted. This makes me more confident that the 3rd test case happens to be one of the rare bad case for DualPivotQuicksort in Java 11's standard library, which actually has been there since Java 7.
It would be great if I can have the 3rd test case and confirm it on my local machine. Anyway I can have it?
Link to the generator.
So you deliberately made such a case, haha :P
Good to know.
Dude this is some next level stuff.
Could anyone tell me what's wrong with string
c_strin GNU C++?My two submissions of problem C:
https://codeforces.com/contest/1328/submission/74623565
https://codeforces.com/contest/1328/submission/74623526
They're exactly the same code with only difference is the lang(one is GNU C++17, the other is Clang++17 Diagnostics). I get WA with GNUC++, but passed with Clang++. Why did this happen?
And I get AC using cout to print
stringwith GNU C++:https://codeforces.com/contest/1328/submission/74623915
That's weird... at least to me...
74633006 TLE on tc 58 can someone help me
This loop could reach O(n) so your solution has a worst complexity of O(n*m)
Thanks
where are the ratings of this contest?? My rating was incresed by 57 but it is not showing now.
Don't worry it will be back soon
i need help in problem E. I'm getting wrong answer in test 43 and i don't know the reason why here is my code: https://codeforces.com/contest/1328/submission/74652500 my answer is 253243 but expected answer is 253245
if you know the problem can you tell me.
finally i found it after few hours! if you are having same problem as me you have to check this test my previvious solution was giving me answer 10 but answer is 11 test: 5 4 1 3 5 7 9 good luck!
hey, thanks! it helped me a lot <3
no problem good luck!
the same thing is happening to me :'v
Hello, this is my first message on Codeforces, so apologies if it violates any community guideline. I am getting wrong answer on test case 2 for Problem D. My submission id is 74659654. I have looked at it a lot but I am unable to find the problem. Can someone please help me. Thanks in advance. Happy coding !
drag you submission page to the buttom, there's checker comment to show which case you didn't pass.
In problem E i am getting runtime error on test 100. I am not able to find the mistake in my code. Can someone help me out with error. link to my code:-74663766
In question E why we are choosing depeest node among the tree please explain
You can Simply refer my approach if you want Solution linkSolution
I know my approach should result in TLE, but I am getting wrong answer in Problem E. Could someone please help. 74696152
My Fair simple approach for E Prerequisites:-LCA of Two nodes in log(N) time in a tree. Step1:- Running dfs traversal on tree and noting down level of each node in the tree and also the parent of each node in the tree and also obtaining the sequence of dfs traversal with which you can get LCA of two nodes tutorial link:LCA Finding tutorial
Step2:-Now for each query we have to answer in YES or NO .So what I did is for each node in a particular query find parent of that node and if node is 1 skip it simply. Store two things for a node particularly its parent number and its level . Sort the obtained vector accorsing to level of parents of all nodes in decreasing order.
Step 3:-Start from top and find lca of top 2 nodes and then with this obtained lca find lca of nodes successively coming and check if obtained lca value is similar to coming value and if it is then move forward till end and else break and print NO otherwise print "YES".
Solution link Implementation
Could you explain why did you skip the node if its parent is 1.
Because when parent is 1 after that LCA of every node is gonna 1 as 1 is root and also as we started in bottom up fashion
Could you please check what is wrong with this piece of code. I'm creating a dummy root, I expected the code to get TLE, but it's giving wrong answer. Thanks in advance. 74696152
Hello __Apocalypse__.
@Line 48:g[x].push_back(y);You forgot add this
g[y].push_back(x);.I added it here: 75222286 and now it gets a
TLEinstead of aWA.Hope this helps. :)
My submission Can u tell why this giving tle. I have sorted the vertices which are input in m queries according to the height of the tree given by 'dp[i]',which is the only thing different from the solution ,so my complexity is mlog(m),but still tle
Hi Chodermal1.
@Line 6:Here, you're passing
maby value although it is not being modified there. Hence, it is taking more time to copy the data.I changed it to:
Notice the
&used to passmaby reference to prevent copying of the data and it gets anAChere.Hope this helps. :)
Thanks bro new thing for me :)
I dont know whats wrong with the code,It passed all the testcases and verified the logic.It seems to give an error as unintialised value at line 56 due to which i am getting an wa.Please help!!!!
here is my soln: https://codeforces.com/contest/1328/submission/74730546
Initailise your array before using it..
Can anyone explain why this answer for test case in D(Carousel) is wrong? Input
4 5 1 2 1 2 2 6 1 2 2 1 2 2 5 1 2 1 2 3 3 10 10 10 my output 2 1 2 1 2 1 2 1 2 1 2 1 2 3 1 2 1 2 3 1 1 1 1 Jury's output 2 2 1 2 1 1 2 1 2 1 2 1 2 3 1 2 1 2 3 1 1 1 1
Input:
Your output:
These are the same but not in the input. Assume that after the $$$n$$$-th figure the figure $$$1$$$ goes.
In problem F (1328F - Make k Equal — Make k Equal), Can someone please explain why the answer to the following custom test case is 29 and not 21?
12 8
1 3 3 6 6 6 11 11 11 11 11 11
Answer as per given Editorial Code = 29. Possible answer = ( 1*(6-1) + 2*(6-3) + 2*(11-6) = 21 ). {by making all 8 elements equal to '11'}
After making all the elements 6, you can't directly increase two elements to 11, because once you increase it by 1, it becomes 7 and is no longer the smallest element in the array....
My bad! Thanks a lot for pointing out!
I have solved "Problem E" using the Lowest Common Ancestor Concept. Straight forward lowest common ancestor logic.
My submission Can u tell why this giving tle. I have sorted the vertices which are input in m queries according to the height of the tree given by 'dp[i]',which is the only thing different from the solution ,so my complexity is mlog(m),but still tle
In problem D. why if n is even coloring always can be done like [1,2,1,2,..2] For n = 6 , 1 2 2 1 2 3 should give 3 , 1 2 2 1 2 3 instead of 2 , 1 2 1 2 1 2
u can have different colours for same number. So when n is even just alternatively put 1 and 2 so each adjacent position will have different colours regardless of the fact that they are same or different
Can anybody please explain the reason for doing this Let's take every non-root vertex (except fv) and replace it with its parent. Thanks
There are two types of nodes in queries. Either a node can already exist in the simple path from fv to root or not. Let's see both cases -
1. Exist in simple path from fv to root — Since it's parent will also exist in the path we are not making any harm by replacing it with parent.
2. Doesn't exist in simple path from fv to root — Since at most distance of 1 is allowed, it's better to check whether parent exist in path or not.
Where could I find more information about the Technique used for E, such as proofs or different applications?
can anybody tell what is wrong in my code for problem D https://codeforces.com/contest/1328/submission/75753056
test
awesome contest...!!
Now we have a beautiful structure giving us so much information about the treeMike got emotional here xD.
On E editorial.
"Let's take every non-root vertex (except fv) and replace it with its parent."
I think that you mean: "Let's take the k vertices vi (except if vi is fv, and except if vi is the root) and replace it with its parent.".
In problem E, I created a set containing all the vertices from 1 to fv. Then I just check if parent[querypoints] are all in the set. I think this is sum(k_i)logn? Why would this give TLE while sum(k_i) and n are only 2e5
No, finding all values in set in not logn. In worst case which is when all node are connected in a line and leaf is fv mentioned in solution, it is O(n).
Why topic tags of D is dp ?
The editorial for F is way too complicated, the problem can be solved simply by observing that we either only increase the minimum and make it equal to (p[k] or p[k-1]), or we only decrease the maximum and we make it equal to (p[n — k + 1] or p[n — k + 1] + 1), or we increase minimum to (median or median -1) and decrease maximum to (median or median + 1).
(Median in the third case because in a sorted array, the median is the point at which sum of distances from all other points is minimised)
Implementation: https://codeforces.com/contest/1328/submission/193670280
https://codeforces.com/contest/1328/submission/224691582 Anyone please HELP!