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Idea: mesanu
Tutorial
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Solution
#include <bits/stdc++.h>
using namespace std;
void solve()
{
int a, b, c;
cin >> a >> b >> c;
cout << (a+b+c-min({a,b,c}) >= 10 ? "YES\n" : "NO\n");
}
int main()
{
int t;
cin >> t;
while(t--)
{
solve();
}
}
Idea: flamestorm
Tutorial
Tutorial is loading...
Solution
#include <bits/stdc++.h>
using namespace std;
const int MAX = 200007;
const int MOD = 1000000007;
void solve() {
int n;
cin >> n;
int winner = -1, best_score = 0;
for (int i = 1; i <= n; i++) {
int a, b;
cin >> a >> b;
if (b > best_score && a <= 10) {winner = i; best_score = b;}
}
cout << winner << '\n';
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int tt; cin >> tt; for (int i = 1; i <= tt; i++) {solve();}
// solve();
}
Idea: flamestorm
Tutorial
Tutorial is loading...
Solution
#include <bits/stdc++.h>
using namespace std;
const int MAX = 200007;
const int MOD = 1000000007;
void solve() {
for (int r = 0; r < 8; r++) {
for (int c = 0; c < 8; c++) {
char x;
cin >> x;
if (x != '.') {cout << x;}
}
}
cout << '\n';
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int tt; cin >> tt; for (int i = 1; i <= tt; i++) {solve();}
// solve();
}
Idea: SlavicG
Tutorial
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Solution
#include "bits/stdc++.h"
using namespace std;
#define ll long long
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(),v.rend()
#define pb push_back
#define sz(a) (int)a.size()
void solve() {
int n, k; cin >> n >> k;
vector<int> a(n);
for(int i = 0; i < n; ++i) cin >> a[i];
sort(all(a));
int cnt = 1, ans = 1;
for(int i = 1; i < n; ++i) {
if(a[i] - a[i - 1] > k) {
cnt = 1;
} else {
++cnt;
}
ans = max(ans, cnt);
}
cout << n - ans << '\n';
}
int32_t main() {
ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
int t = 1;
cin >> t;
while(t--) {
solve();
}
}
1850E - Cardboard for Pictures
Idea: flamestorm
Tutorial
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Solution
#include "bits/stdc++.h"
using namespace std;
#define ll long long
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(),v.rend()
#define pb push_back
#define sz(a) (int)a.size()
#define int long long
void solve() {
int n, c; cin >> n >> c;
vector<int> a(n);
for(int i = 0; i < n; ++i) cin >> a[i];
int l = 1, r = 1e9;
while(l <= r) {
int mid = l + (r - l) / 2;
int sumall = 0;
for(int i = 0; i < n; ++i) {
sumall += (a[i] + 2 * mid) * (a[i] + 2 * mid);
if(sumall > c) break;
}
if(sumall == c) {
cout << mid << "\n";
return;
}
if(sumall > c) {
r = mid - 1;
} else {
l = mid + 1;
}
}
}
int32_t main() {
ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
int t = 1;
cin >> t;
while(t--) {
solve();
}
}
Tutorial
Tutorial is loading...
Solution
#include "bits/stdc++.h"
using namespace std;
#define ll long long
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(),v.rend()
#define pb push_back
#define sz(a) (int)a.size()
void solve() {
int n; cin >> n;
vector<ll> cnt(n + 1, 0), mx(n + 1, 0);
for(int i = 0; i < n; ++i) {
int x; cin >> x;
if(x <= n) ++cnt[x];
}
for(int i = 1; i <= n; ++i) {
for(int j = i; j <= n; j += i) mx[j] += cnt[i];
}
cout << *max_element(all(mx)) << "\n";
}
int32_t main() {
ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
int t = 1;
cin >> t;
while(t--) {
solve();
}
}
Idea: flamestorm
Tutorial
Tutorial is loading...
Solution
#include <bits/stdc++.h>
#define startt ios_base::sync_with_stdio(false);cin.tie(0);
typedef long long ll;
using namespace std;
#define vint vector<int>
#define all(v) v.begin(), v.end()
#define int long long
void solve()
{
int n;
cin >> n;
map<int, int> up, side, diag1, diag2;
int ans = 0;
for(int i = 0; i < n; i++)
{
int x, y;
cin >> x >> y;
up[x]++;
side[y]++;
diag1[x-y]++;
diag2[x+y]++;
}
for(auto x : up)
{
ans+=x.second*(x.second-1);
}
for(auto x : side)
{
ans+=x.second*(x.second-1);
}
for(auto x : diag1)
{
ans+=x.second*(x.second-1);
}for(auto x : diag2)
{
ans+=x.second*(x.second-1);
}
cout << ans << endl;
}
int32_t main(){
startt
int t = 1;
cin >> t;
while (t--) {
solve();
}
}
Idea: SlavicG
Tutorial
Tutorial is loading...
Solution
#include "bits/stdc++.h"
using namespace std;
#define ll long long
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(),v.rend()
#define pb push_back
#define sz(a) (int)a.size()
#define int long long
const int N = 2e5 + 5;
vector<pair<int, int>> adj[N];
int val[N], vis[N];
void dfs(int u) {
vis[u] = 1;
for(auto x: adj[u]) {
int v = x.first, w = x.second;
if(!vis[v]) {
val[v] = val[u] + w;
dfs(v);
}
}
}
void solve() {
int n, m; cin >> n >> m;
for(int i = 1; i <= n; ++i) {
adj[i].clear();
vis[i] = 0, val[i] = 0;
}
vector<array<int, 3>> c;
for(int i = 1; i <= m; ++i) {
int a, b, d; cin >> a >> b >> d;
adj[a].pb({b, d});
adj[b].pb({a, -d});
c.pb({a, b, d});
}
for(int i = 1; i <= n; ++i) {
if(!vis[i]) dfs(i);
}
for(int i = 1; i <= m; ++i) {
int a = c[i - 1][0], b = c[i - 1][1], d = c[i - 1][2];
if(val[a] + d != val[b]) {
cout << "NO\n";
return;
}
}
cout << "YES\n";
}
int32_t main() {
ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
int t = 1;
cin >> t;
while(t--) {
solve();
}
}
It feels like Problem A was directed at someone 😶
It is easy to write verses
When you have not what to tell,
Stinging words and hollow phrases
In a gangling doggerel.
Thanks to authors! The problemset was very cool
This competition is the best for me.
For E wasn't it worth mentioning O(1) solution with quadratic formula if we ignore time for taking input
https://codeforces.me/contest/1850/submission/214907399 or https://codeforces.me/contest/1850/submission/214911845
sqrt is log(n), so in fact there are no O(1) solutions
heh) I saw the problem and immediately thought about binary search and solving the equation. I decided to choose the second one and as a result I first wrote the code for pluses, and then for python (I never used int128)
Hey, How come you have reached to this solution? Can you please explain about your quadratic formula approach
A painting with side s_i requires a cardboard square with side (s_i + 2*w).
The total area of the cardboard squares (the c input in the problem) is:
The last equation is quadratic in terms of w and we can solve it using the quadratic formula
EDIT: Thanks for the correction aryang22
I guess there should be a bit of correction in the last line:
Here, n = number of pictures. If you see this for 1st test case, 3 and 50 and {3,2,1}, so
sum(s_i^2) = 14,c = 50,sum(s_i) = 6, andn = 3. If you putw = 1, it satisfies only if you considernalso, that is,LHS:
14-50+4*6*w+4*w^2*3=>14-50+24*(1)+12(1)^2=>-36+36=>0 = RHSYeah, absolutely. That should have read sum(4*w^2) which is just equal to 4*w^2*n as you point out, since it does not depend on the summation index.
That’s what i did, had problems with the output for large numbers tho, but it’s O(n), not O(1), instead of O(nlogn)
For anyone using JAVA , never ever use Arrays.sort even after shuffling the array, I got unnecessary TLE and rectified using Collections.sort. The tests were designed carefully.
thanks
What is a failed test case for this solution for D?
214942704
I make the same code and i do not know where is the wrong
G Problem is kind of impossible with python
Even if we forget the fact that there are workarounds to the hash hack in python, we can still solve the problem using python quite easily.
Instead of using maps/dictionaries, we can just store all values of $$$x$$$, $$$y$$$, $$$x + y$$$ and $$$x - y$$$ in four separate arrays, sort each of them and check the number of equal elements that way. And as far as I know, there is no hack for sorting in python (unlike in java).
Code (pls forgive me for horrible python): 214952260
thanks got it now sorting also a good option
i got hacked on G (tle) because for some reason i thought i had to sort the points in order to get the intersection with y = x+c and y = -x + c lines.
It turns out this is unnecessary and you can solve the question with linear time O(N).
HACKED submission O(Nlog(N))
O(N) submission
Bro, it's TL unfortunately
https://codeforces.me/contest/1850/submission/214868771 This code look sus. Is it allowed?
That is (quite obivously) code obfuscation, which is not allowed. Hopefully they'll get skipped for breaking rules.
what does code obfuscation mean?
If only there was some magical place where you could just type the words "what does obfuscation mean" and you'd get an answer instantly...
But how many defines is code obfuscation then? Some c++ solutions are absolutely unreadable even though their authors had no intention to muddle their code.
I don't think there are any specific numbers I can give; I think the more important part is intention:
If the author didn't intend to make the code unreadable (and they can read it themselves), I wouldn't call it obfuscation.
If the author deliberately made the code unreadable (possibly using automated help), I would definitely call it obfuscation.
I think it should be clear that the aforementioned code falls under the second case.
Unfortunately I saw a lot of solutions with defines like this that make the code unreadable. But I don't really know what to do with that as I haven't found any flag/signal. button yet.
Please ban him MikeMirzayanov
Of course, this is an obvious violation of the rules, and the user will be punished.
Hello! This is my first contest and I submitted the problem A. I was registered but I do not see an increase in my rating, can someone tell me when will I get them?
check again tommorrow afternoon (indian time)
can someone try to hack my solutions?
I used binary search for problem-E but it has some error which I am unable to figure out. Can anyone please help me find error in my binary search logic for my submission 214950555. ? edit : when i set
r = 1e10the sample test-cases passed but on submission it fails again on some test case.it looks like the res function might overflow, I had that problem and dealt with it as explained in the editorial. you can also use __int128 instead of long long to avoid overflow.
try r=sqrt(c) , it's the smallest value of r that can be valid
true, an alternative is also to use r=1e9+1.
Why this solution TLE in problem F because It does unnecessary work when we have more occurrences of one elment. I think the TC Is exactly the same as the one proposed by the editorial in the end. Also if we have the test case 1,2,3....,2e5. The solution posted does the same number of operations as the one of the editorial. Am I wrong thinking that this has the same TC as editorial?
If all n values of ai are equal to 2, each one will need n/2 iterations, in total n*n/2 operations.
Therefore it is necessary to treat together the array values that are equal.
maybe f can be hacked, consider case {1,1,1...} * 10^5
this prevents it.
I liked problem H . Before reading editorial, i didn't though it will turn that easy.
can someone explain why my E solution is accepted here:
https://codeforces.me/contest/1850/submission/214991383
but it is failing here:
https://codeforces.me/contest/1850/submission/214991353
the only difference is in my accepted version, i did (sizes[i] + w * 2) * (sizes[i] + w * 2)
but in the failing version, i did pow((sizes[i] + w * 2), 2)
Check this blog: https://codeforces.me/blog/entry/21844
In this contest can my E be WA, G be TL?
My submission in E in contest: 214829184. I don't know how it's passed pretests. In code wrongly I did: x = s[i] + 2 * md + 1 and printed l but x should be s[i] + 2 * md and printed l — 1. Like this code: 214997775 which I submitted after contest. Please let me know if it's hackable.
My submission in G in contest: 214869824. Can it be TL or ML? It passed pretests with 1668 ms 76900 KB(Time limit is 2s). After contest I optimized it (removed set) and passed pretests with 904 ms 39300 KB(214997753). Please let me know it's hackable.
During the contest,I misunderstand Problem F,I think each second we can place a trap,so in this case,can we also calculate the maximum number of the frogs we can catch?I need help,thanks.
Can anyone help me my code has resulted in TLE in system testing?, it is having similar complexity as of author O(nlogn)
Your code is $$$O(n^2)$$$ because of unordered_map
The solution of F was already available on geeks for geeks and other various websites , seems to be a quite popular problem
I used an unordered map on F. I will never use an unordered map again.
Why does that give TLE if unordered_map is used instead of the normal map in C++? Can anyone tell me when to use unordered_map and normal map?
https://codeforces.me/blog/entry/62393
Can anyone explaine how Wrapper function works? This is my hacked submission: https://codeforces.me/contest/1850/submission/214906792 And then i fixed it by Wrapper function which make my submisson accept https://codeforces.me/contest/1850/submission/214984466 *sorry for bad english
can someone tell why Question 'G' Fails for an Unordered map, but runs absolutely fine for ordered map
void mr_kamran(){
ll n; cin>>n; ll ans = 0; unordered_map<ll,ll>m1,m2,m3,m4; for(int i = 0;i<n;++i) { ll x,y; cin>>x>>y; m1[x]++; m2[y]++; m3[y-x]++; m4[y+x]++; } for(auto it : m1) { ans+=(it.second*(it.second-1)); } for(auto it : m2) { ans+=(it.second*(it.second-1)); } for(auto it : m3) { ans+=(it.second*(it.second-1));
} for(auto it : m4) { ans+=(it.second*(it.second-1)); } cout<<ans<<an;
}
Check the editorial, this question has been answered a million times here.
https://codeforces.me/blog/entry/62393
Solved problem H using Union Find.
215052919
why F is nlog(n)? if there are 2e5 1 than is n*n
It looks like n*n but it's not. the time complexity is n+n/2+n/3+n/4+n/5 +... We can factor out n and it becomes n(1+1/2+1/3+1/4 + .... + 1/n) which you should be familiar with if you have taken calculus. It's called the harmonic series. If you aren't familiar with calculus, all you need to know is that 1+1/2+1/3+1/4 + ... + 1/n is approximately equal to log(n) where n is the number of terms
why in problem G it gives TLE error if we use unordered_map<int,int> instead of map<int,int>. The time complexity of unordered_map is O(1) whereas operations on map take O(log n) time.
https://codeforces.me/blog/entry/50626?#comment-345515
Thank you for the link! Got the same problem today.
Hey guys! Can somebody help me please why in task E — Cardboard for Pictures we should use mid = l + (r — l) / 2? I am new at this and would appreciate any answer)
It's same as mid=(l+r)/2 but it's recommended to use mid=l+(r—l)/2 to avoid overflow.
SlavicG Can you explain why u also pushed negative edges in the last problem
Can we solve H using different approach?: For all loops found in graph, I check if "cnt" of the loop is equal to 0. "cnt" of the loop is sum of all (-d)s for all adjacent vertexes in the loop. For example, if we have loop
(a, b, d)
1 2 1
2 3 2
3 1 -3
cnt of this loop is 0. (-1-2-(-3) = 0). If cnt of the loop is not zero, it means that every soldier in the loop belongs to 2 different camps. My code gives WA3: 216588864
can we do F by checking for cycle ?
Comment on G implementation
If you want to calculate $$$n(n-1)$$$, you can instead calculate $$$2(1+2+...+(n-1))$$$.
By using this, you can avoid the second calculation pass and have a cleaner solution (https://codeforces.me/contest/1850/submission/221488328)
can anybody please tell me a counter test case for F problem according to my code? please
Your code is
overcounting, say for 4 you have contributions from 1,2,4 as multiples but 2 has contributions from 2 as well as from 1. So, eventually your answer for 4 counts the contributions from 1 twice. To avoid this you can startfrom n and go till 1and do the same process, it would pass thenFor part E, f(x) = sum of squares of s + 4*w*(w*n+sum of s). You can verify this by expanding each bracket in f(x). Both the sum of squares of sum and the sum of s can be computed in O(n) time only once, and then used to calculate f(x) in each binary search iteration in O(1). So in total, you perform one O(n) operation and one O(log n) operation. If you want, you can view my submission at code 239192984.
I wonder why the G question times out with unordered_map
Here you go
Thank you! I get it.
Solved F
What a beautiful problem!
can Anyone SlavicG please tell why my solution for problem H is giving TLE
258450042
I have doubt on using map<pair<ll,ll>,ll> me;
I have been searching on codeforces blog for what may be the problem but unable to find
I tried searching for constant factor TLE blogs but unable to find so can anyone kindly tell what is wrong
According to me time complexity is O(nlogn) where n= m (the number of conditions given in problem)
I think my solution logic may be wrong but then also want to know what thing is the leading to TLE as I have encountered similar TLE problem previously in some different question also
Interestingly in G, if you were to make a map of ints and do diag1 [y-x]++, it will give you wrong answer at test case 12, but the same with diag1 [x-y]++ gives an accepted solution. Note that both these cases work when you make a map of long long.
I couldn't understand why you didn't initialize the ans variable with 0. And also whenever I am using 1 there the ans is coming right for the 2nd test case but when using zero that time it's not giving me 1 ....I don't know why