We are really sorry to make the round unrated. Anyway, we hope that you enjoyed the problems!

Idea: shishyando

**Tutorial**

Tutorial is loading...

**Implementation**

```
// один манул
#include <bits/stdc++.h>
using namespace std;
int main() {
int T;
cin >> T;
while (T --> 0) {
int n;
cin >> n;
vector<int> a(n);
for (int i = 0; i < n; ++i) cin >> a[i];
sort(a.begin(), a.end());
vector<int> b;
for (int i = 0; i < n; i++) {
if (i == 0 || a[i] != a[i - 1]) {
b.emplace_back(a[i]);
}
}
for (int i = 0; i < n; i++) {
if (i > 0 && a[i] == a[i - 1]) {
b.emplace_back(a[i]);
}
}
for (auto x : b) cout << x << ' ';
cout << '\n';
}
return 0;
cout << "amogus";
}
```

Idea: Artyom123

**Tutorial**

Tutorial is loading...

**Implementation**

```
//два манула
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin >> t;
while (t--) {
int n, m;
cin >> n >> m;
map<int, int> cnt;
while (n--) {
int x;
cin >> x;
cnt[x % m]++;
}
int ans = 0;
for (auto &c : cnt) {
if (c.first == 0) ans++;
else if (2 * c.first == m) {
ans++;
} else if (2 * c.first < m || cnt.find(m - c.first) == cnt.end()) {
int x = c.second, y = cnt[m - c.first];
ans += 1 + max(0, abs(x - y) - 1);
}
}
cout << ans << '\n';
}
return 0;
}
```

Idea: shishyando

**Tutorial**

Tutorial is loading...

**Implementation**

```
// три манула
#include <bits/stdc++.h>
using namespace std;
int main() {
int T;
cin >> T;
while (T --> 0) {
int n, k;
cin >> n >> k;
if (n % 2) cout << 1 << ' ' << n / 2 << ' ' << n / 2 << '\n';
else if (n % 2 == 0 && n % 4) cout << 2 << ' ' << n / 2 - 1 << ' ' << n / 2 - 1 << '\n';
else cout << n / 2 << ' ' << n / 4 << ' ' << n / 4 << '\n';
}
return 0;
}
```

Idea: isaf27

**Tutorial**

Tutorial is loading...

**Implementation**

```
// четыре манула
#include <bits/stdc++.h>
using namespace std;
vector<int> solve3(int n) {
if (n % 2 == 1) return {1, n / 2, n / 2};
if (n % 4 == 0) return {n / 2, n / 4, n / 4};
if (n % 2 == 0) return {2, n / 2 - 1, n / 2 - 1};
}
int main() {
int T;
cin >> T;
while (T --> 0) {
int n, k;
cin >> n >> k;
vector<int> added = solve3(n - k + 3);
for (int i = 0; i < k; ++i) {
cout << (i <3 ? added[i] : 1) << ' '; // <3
}
cout << '\n';
}
return 0;
}
```

Idea: shishyando

**Tutorial**

Tutorial is loading...

**Implementation**

```
// пять манулов
#include <bits/stdc++.h>
using namespace std;
int main() {
int T;
cin >> T;
while (T --> 0) {
int n;
cin >> n;
vector<long long> s(n), tag(n), dp(n, 0);
for (int i = 0; i < n; ++i) cin >> tag[i];
for (int i = 0; i < n; ++i) cin >> s[i];
for (int j = 1; j < n; ++j) {
for (int i = j - 1; i >= 0; --i) {
if (tag[i] == tag[j]) continue;
long long dpi = dp[i], dpj = dp[j], p = abs(s[i] - s[j]);
dp[i] = max(dp[i], dpj + p);
dp[j] = max(dp[j], dpi + p);
}
}
cout << *max_element(dp.begin(), dp.end()) << '\n';
}
return 0;
}
```

1497E1 - Square-free division (easy version)

Idea: Artyom123

**Tutorial**

Tutorial is loading...

**Implementation**

```
// шесть манулов
#include <bits/stdc++.h>
using namespace std;
const int MAXA = 1e7;
vector<int> primes;
int mind[MAXA + 1];
int main() {
for (int i = 2; i <= MAXA; ++i) {
if (mind[i] == 0) {
primes.emplace_back(i);
mind[i] = i;
}
for (auto &x : primes) {
if (x > mind[i] || x * i > MAXA) break;
mind[x * i] = x;
}
}
int T;
cin >> T;
while (T --> 0) {
int n, k;
cin >> n >> k;
vector<int> a(n, 1);
for (int i = 0; i < n; ++i) {
int x;
cin >> x;
int cnt = 0, last = 0;
while (x > 1) {
int p = mind[x];
if (last == p) {
++cnt;
} else {
if (cnt % 2 == 1) a[i] *= last;
last = p;
cnt = 1;
}
x /= p;
}
if (cnt % 2 == 1) a[i] *= last;
}
int L = 0, ans = 1;
map<int, int> last;
for (int i = 0; i < n; ++i) {
if (last.find(a[i]) != last.end() && last[a[i]] >= L) {
++ans;
L = i;
}
last[a[i]] = i;
}
cout << ans << '\n';
}
return 0;
}
```

1497E2 - Square-free division (hard version)

Idea: isaf27

**Tutorial**

Tutorial is loading...

**Implementation**

```
// семь манулов
#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9 + 1;
const int MAXA = 1e7;
vector<int> primes;
int mind[MAXA + 1];
int main() {
for (int i = 2; i <= MAXA; ++i) {
if (mind[i] == 0) {
primes.emplace_back(i);
mind[i] = i;
}
for (auto &x : primes) {
if (x > mind[i] || x * i > MAXA) break;
mind[x * i] = x;
}
}
vector<int> cnt(MAXA + 1);
int t;
cin >> t;
while (t--) {
int n, k;
cin >> n >> k; // уже k манулов
vector<int> a(n, 1);
for (int i = 0; i < n; ++i) {
int x;
cin >> x;
int cnt = 0, last = 0;
while (x > 1) {
int p = mind[x];
if (last == p) {
++cnt;
} else {
if (cnt % 2 == 1) a[i] *= last;
last = p;
cnt = 1;
}
x /= p;
}
if (cnt % 2 == 1) a[i] *= last;
}
vector<vector<int>> mnleft(n, vector<int>(k + 1));
for (int j = 0; j <= k; j++) {
int l = n, now = 0;
for (int i = n - 1; i >= 0; i--) {
while (l - 1 >= 0 && now + (cnt[a[l - 1]] > 0) <= j) {
l--;
now += (cnt[a[l]] > 0);
cnt[a[l]]++;
}
mnleft[i][j] = l;
if (cnt[a[i]] > 1) now--;
cnt[a[i]]--;
}
}
vector<vector<int>> dp(n + 1, vector<int>(k + 1, INF));
for (auto &c : dp[0]) c = 0;
for (int i = 1; i <= n; i++) {
for (int j = 0; j <= k; j++) {
if (j > 0) dp[i][j] = dp[i][j - 1];
for (int lst = 0; lst <= j; lst++) {
dp[i][j] = min(dp[i][j], dp[mnleft[i - 1][lst]][j - lst] + 1);
}
}
}
int ans = INF;
for (auto &c : dp.back()) ans = min(ans, c);
cout << ans << "\n";
}
return 0;
}
```

SpeedForces

Point distribution was weird imo but Questions were really good tho.

For C2 — legend+ary

Problem A — Implementation based

Problem B — Implementation | Constructive (Great Problem)

Problem C1,C2 — Intuitive | Constructive (Make Cases and you are done with both problems)

Problem E1 — Simply Math | Implementation (Prime factorization)

A great Round shishyando and Artyom123 with great and interesting problems. Hope to see your next round soon !!

E2 and D are interesting problems overall.

Lesser number of submissions doesn't make a problem interesting

I think so. I am not able solve the problems, but reading the resulotion helps me a lot in understanding usage of DP. Love this round so much!

Really liked the problems.

Though I'm a tester, I want to say that C1 is the very beautiful hint for C2! Without C1, C2 can be the hardest problem in the contest.

Agreed, I literally copied my code from C1, made minor changes, then submitted C2.

Yes, The magic number of the problem,

3, seems rather arbitrary and unintuitive at first.lmao i gave 1.5 hours in upsolving yesterday and couldn't get C2 as i didn't look at C1 First. Now after reading the editorial i can see how intuitive it is.

Can someone explain in detail math behind C1?

It was simple first if n is

oddthen just print1 n/2 n/2if n is

eventhen check for ifn/2 is eventhen printn/2 n/4 n/4ifn/2 is oddthen print(n/2)-1 (n/2)-1 2we can extend this same logic in

C2by simply printing1,(k-3)times then takingn=n-(k-3)in C1No math just you have to figure out basic logic that's it the one who did it can easily get through

n can be of the form 4*k , 4*k+1 , 4*k+2 , 4*k+3 (for some k>=0) case 1: n = 4*k ans = k ,k, 2*k LCM = 2*k

case 2: n = 4*k+1 ans = 1 ,2k ,2k LCM = 2k

case 3: n= 4*k+2 ans = 2 , 2k , 2k LCM = 2k

case 4: n = 4k+3 ans = 1 , 2k+1 , 2k+1 LCM = 2k+1

I went straight to C(hard) without looking at C(easy) since I didn't want to be misled by some simple solution that only worked for simple testcases but not hard ones... instead I fumbled around for an hour looking for a general solution without realizing the k=3 case is all you need. That really backfired on me.

The scoring distribution: 500 — 750 — (750 + 500) — 1750 — (1500 + 1500) It was pretty much clear, C1 was supposed to be harder

At the risk of my submission soon being systested, I think E2 can be solved greedily.

Construct the greedy intervals from E1. Then, at most K times, consider all pairs of adjacent intervals. Compute the number of changes needed to merge all pairs of adjacent intervals in $$$O(n)$$$ and then just merge the best one, and update the array so that excess is put at impossible values.

It works because in the solution you will never remove a value completely from an interval. So taking suboptimal merger will never help since it isn't actually going to reduce a merging cost.

Since you always make at least one change, you repeat at most $$$k$$$ times so it is $$$O(nk)$$$. I think it can be optimised to $$$O(n)$$$.

UPD: I think it cannot be optimised to $$$O(n)$$$, sorry. I thought that maybe you could save computation by only calculating the effect of the merger rather than recalculating every cost, but that's still $$$O(n)$$$.

UPD2: WA47, I wonder if greedy is just wrong here or it's a problem with the code...

UPD3: Assuming that you are merging intervals, the solution is optimal. However, it is also possible to split up an interval, so it does not work.

Can you explain a bit more in detail about splitting intervals? I had the same reasoning and (theoretical) solution as you as well as WA47 and would like to know what the logical error is.

Say there are just three intervals originally. The greedy solution wants to subsume the middle interval into either the left or the right interval. However, it is also possible to subsume the left part of the middle interval into the left interval and the right part of the middle interval into the right interval. It is possible that this is less costly.

That's perfect, thank you!

what do you mean by best one? does it mean that bigger the interval, the better it is? or anything else? btw approach is nice. Thank you !!

Thank you for the problems!

even despite unrated, this was the best round ive taken in a while. clean and concise problem statements. thank you so much for the round. :D

Checkout this problem for a variant on problem A.

I recalled the problem while giving the round today, lovely coincidence.

Video Tutorial B:https://www.youtube.com/watch?v=l9RQpnyIOXw

Video Tutorial A:https://www.youtube.com/watch?v=-CK_6lqs1hc&t=5s

What are "constructive" problems? can anyone tell. how to become good at these?

Basically, it is related to problems which is asking you to find any answer(of possibly many) that satisfies the constraints of the question. You can practice these here

Thanks dude! I'll save this reply.

Constructive Problems asks you to build something like array, graph or matrix which satisfies the given condition in problem. Usually these type of problems have multiple correct solutions and asks you to output any one. For problems you can refer this and this blog.

Even though I am just starting out, I must say I loved this contest! Great work shishyando

"...it is easy to see that..."

Writing that sentence is like begging for downvotes. Remember, editorials are written for those not able to solve the problem. Telling them how easy it is is no good idea.

From problem D: "Then in binary form weight has its k-th bit set true if and only if j < k < i" Shouldn't it be j <= k < i ? Because we have for example 10000 — 00100 = 01100

Seems true, thank you. Fixed now.

Thanks for the round, although I couldn't participate.

Problem D is beautiful.

Is it ok that i got TL in E2 with O(nk log n + max(a_i)) complexity ?

Is your max(a_i) per test case, or in total? If it's per test case, then it would be O(nklogn+max(a_i)t) which is too much

it's in total, I need it to precalculate prime numbers which are less than 1e7, I passed E1 that way.

Help needed 110230087 for problem 1497B - M-arrays

Your Solution is failing for testcase:

Corrected 110257624,

Just changed m/2 to (m+1)/2 in FOR loop

thanku

You also have one more mistake.

If i=0 => m-i=m, but m can not be a reminder by mod m. So you have to check v[i] and v[(m-i)%m].

no, I started i from 1.

For me C2 strikes so fast even I hadn't thought of it...it just dropped from the sky :)

Simply, considering 1 1 1 1 ... upto k-3 and the C2 becomes C1

I think if there was no C1 then people may have a hard time coming to this solution for C2.

yes :)))

I used binary search for C1 & C2, didn't think it can be done in O(1) too :( My solution Great contest though! <3

I am curious while doing binary search how you got the idea that low is the ans instead of mid ??

See, the thing about binary search is that is always drifts towards the right answer. Similar thing is happening here.

Initially, high = ⌈n/2⌉-1, low = 1 and during first iteration, mid = ⌈n/2⌉/2.

Now if you look carefully, for first case (when n is odd) and for second case (when n is even but not divisible by 4) the answer is actually, (n-2*(⌈n/2⌉-1), ⌈n/2⌉-1, ⌈n/2⌉-1) and for third case(when n is even as well as divisible by 4) the ans is (n-⌈n/2⌉, ⌈n/2⌉/2, ⌈n/2⌉/2).

So you must have already observed that for 1st & 2nd case, the ans is initial value of high, (n-2*high, high, high) and for third case ans is the initial value of mid, (n-2*mid, mid, mid)

For the first two cases, the if condition will execute only once when mid=high, i.e., low=high=mid, and when this happens high will become less than low and it'll come out of the loop, so the ans will be low, [n-2*low, low, low] and for the third case, the if condition will execute only once, when mid is the ans, let's say value of initial mid is mid1, so for all further iterations, the if condition will never execute since high = mid1-1 and the value of low will keep on increasing until it becomes equal to high+1, i.e., low = high+1 = mid1, so again the ans is [n-2*low, low, low].

The dp of E2 can be solved in O(nk). https://codeforces.com/contest/1497/submission/110252728

What is the reason for the

unusual memory limitin Problem D?I'm assuming the reason was to prevent the more obvious solution with a dp[n][n] table, which would need to store up to 25,000,000 ints = 100 megabytes.

igz Can you please explain the dp[n][n] solution?

Why it is necessary to use map instead of unordered_map in 1497B - M-arrays? I submitted both versions but only map got AC but unordered_map one got WA on test 3. map version-> 110262091 unordered_map version-> 110262159

It's happening mostly because of collision, refer this to use a custom hash for unordered map. I've used the same in my submission.

Using cnt[i] may create a new element (cnt[i]=0) in map. And it may cause rehashing of unordered_map, iterators are invalidated (see "Iterator invalidation" here), and for-loop continues working with invalid iterators, which causes undefined behaviour. In usual map creating new element does not invalidate iterators, so UB won't happen. But u still got lucky that these extra zeroes in your map don't change result of your program.

thanks for the above link I learned a new thing...

My solution by using unordered_map 110274693. I just avoid new insertion and deletion or such kind of invalidate iterator operations that mentioned in this Your text to link here... given by KostasKostil.

I really like how C1 is a hint for C2.

C1 750pts

C2 500pts

Problem E is similar to this problem.

If Only I submitted C1 after My C2 was accepted!

i am curious while doing C1,C2 problem how people were able to come up with the idea of for n%2 == 0 ans is n/2,n/2,2 and for n%4 == 0 ans is n/2-1,n/2-1,2 and so on.During the contest I made a pattern for 1 to 11 but i was unable to intutively guess this solution .Can anyone recommend the problems that i can do for increasing my intution for doing this type of problems thanks

sorry for late response.I made a patter up to 23 before getting the right answer. My cases was if n is a multiple of 3 we done (n / 3, n / 3, n / 3), and it was difficult after that. I sais to myself it the limit is n / 2 i will try to be near to n / 2. and for each number i put n / 2 and see what can be done. I think the best is to go up to a limit (say 50) and if you don't see a pattern maybe it's not the right solution.

There is an $$$O(nk)$$$ dp solution in Problme E2.

At first, we can let $$$a[i]=mask(a[i])$$$. Assume $$$dp[i][j]$$$ denotes we consider the first $$$i$$$ positions, after making $$$j$$$ changes, the minimum answer we can get and the maximum position where the begining of the last segment at. So we can just maintain $$$last[i]$$$ which denotes the maximum position $$$j$$$ where $$$a_j=a_i$$$ (expecially, if $$$a_i=0,last[i]=i-1$$$) and make transfers.

The key observation in the solution is that if we change the number $$$a_i=x$$$, and for the first $$$j>i,a_j=x$$$, we should let $$$last[j]=last[i]$$$. But actually, there is no need to do that, we can still get the right answer.

See the code for details.

Stuck in E1. In sample test case 1, should't answer be 2 ? Isn't it valid -> (18, 6, 4), (2, 1) ?

We can't change the order of the sequence while dividing.

Can anyone tell me how to reach to the solution for $$$C2$$$. I am asking this because $$$C1$$$ made a path for $$$C2$$$. So, can someone give some mathematical proof or logical way to approach such problem

Achieving lcm of n / 2 lead intuitively (to me) to take n / 2 everywhere. now we can see that we can put n / 2 at most 2 time and i will remain a 1 case left.(that leads to k = 3 case). you can also think i put the same number some number of time and i put 1 everywhere since 1 does not increase the lcm. But i think it's very difficult to figure out alone without a hint.

for the tutorial for c2,

shouldn't it be (k-3) + n-(k-3) or (k-3) + n — k + 3?

Here is another $$$\mathcal{O}(nk)$$$ solution to 1497E2 - Square-free division (hard version).

For the first part, after normalization, instead of

`left[i][j]`

which is a bit bothering, we only need to find`pre[i]`

, which is the rightmost index to the left of the current index that shares the same value as the current index. This is very simple.Then for the second part, for each change number

`j`

, we store`best[j]`

denoting the minimal pair`(segs,-last_start)`

, and`second_best[j]`

denoting the second minimal pair`(segs,-last_start)`

. Here`last_start`

is the start index of the last open segment, and`segs`

is the number of segments.During the DP, if the last open segment can contain the new element, everything remains the same. Otherwise, we choose either to start a new segment (so that

`j`

does not change), or to change the current element (so that`j`

becomes`j+1`

).Note that there would be cases that

`second_best`

becomes better than`best`

, in which they should be swapped.Code: 110279809

Can we solve it using binary search + greedy?

I do not know whether it is possible, but I think that would be hard.

Like we will fix the max number of elements per segment and then try to put the elements greedily. Whenever we find a repeated value we will replace it with a very big prime number.(of course considering the restriction of k)

If it is possible then we set lo=mid+1 else hi=mid-1 Then the answer would be ceil(n/hi)

You may solve our problem here using your observation.

Can someone explain how left[i][j] is calculated in E2. I am not too comfortable with two pointers, perhaps there is any other way to calculate it? I understood the second part of the solution, the first part bothers me.

Umm.............anyone? It will be really helpful if anyone could.

is there any way to solve E1 problem using binary search??

No need of binary search just fill the elements greedily, you will get the answer

BhanuPratap Stuck in E1. In sample test case 1, should't answer be 2 ? Isn't it valid -> (18, 6, 4), (2, 1) ?

You cannot reorder the elements. The first test case is 18 6 2 4 1

For C1 wasn't there suppose to be a case when n % 3 == 0.The answear would be then n / 3 , n / 3, n / 3?

You could analyze that, but that is not needed. The solution explained in the editorial says if n is odd you do A and if n is even you do B, no need to add more cases. This is a typical bad practice from beginners.

If n is a multiple of 13 you could answer 6*n/13, 6*n/13, n/13. But it would be silly to add such a case

Now let's relax dp values. When we consider an edge {i,j},tagi≠tagj we try to solve problem i after solving dpj problems ending with j, and problem i after solving dpi problems ending with i. It means that dpi=max(dpi,dpj+p), dpj=max(dpj,dpi+p) at the same time, where p=|si−sj|.

Can anyone explain this part of the editorial of problem D?

An important part that the editorial misses is that we can only jump back once that is from j to i ;j>i, try to prove it yourself.

Even i am not able to get this part....

I am getting wrong answer for E1 in test 3. I cannot find out where I am making a mistake. Can someone please help? my submission is : 110286186

Can you tell me why (18, 6, 4), (2, 1) are not valid sequence for first test case of E1??

Thank you for C1 and E1 which give me some ideas about hard version

RadestionAdtinium Can you tell me why (18, 6, 4), (2, 1) are not valid sequence for first test case of E1??

You can't change the array

## include<bits/stdc++.h>

using namespace std; int main(){ int t; cin>>t; while(t--){ int n; cin>>n; int tag[n]; int i,j; for(i=0;i<n;i++){ cin>>tag[i]; } int s[n]; for(i=0;i<n;i++){ cin>>s[i]; }

} return 0; }

//can some one explain why logic is not true. for the same test case i am attaching a picture

if j1 is greater than j2 then 2^j1-i is always greater than 2^j2-k for all i and k less than j2 .if someone still didn't get the logic you can message me but please help me

Guys I'm really sorry I know posting your code and asking why it's giving me an error is a sure-fire way to get downvoted but I really have no choice,I have commented neatly and explained the code for problem D, but I keep getting a run-time error on TC-2, can anyone please help me outHere is the code : Python 3

Thank you in advance :)

it's because in your code first of all you didn't memoized all the states and 2ndly recursion tree for your code does not form a DAG(because yours is cyclic) which is necessary condition for any dp/recursion problem. In simpler words you end up calling a state which initially made you call these states.

For a case where n = 3z ,z being an odd natural number, isn't the answear (z, z ,z), not (1, n / 2, n/ 2) as it is specified in the editorial ?

Guys is there an algorithm involved in B problem. Or is it an only math based problem? What should be said ?

Both i guess

which algorithm ?

I can't understand what is the wrong here can you help me 110316296

Hello Everyone, Can some one tell me, where I am getting wrong in problem E2? I am getting WA on 8th testcase.

My solution.

Thank you in advance !!

Can someone provide better explanation for D and E2?

C1, C2 sucks :((

Hi, guys! May u help pls find out why do I have TL in E1 ?

I run the sieve to get least devisors of all numbers from 1 to 1e7. Which is 1e7 * log(log(1e7)) Then for each a[i] I get mask whish is 2*1e5 * log(1e7) in worst case.

So totally it is less then 1e9. But what's wrong?

My submission: https://codeforces.com/contest/1497/submission/110540450

Can anyone explain the reason why am i getting TLE on test case 102 on this submission but the exact same code gets AC here when i change 'long long' to 'int' with no other changes for problem E1.

Can anyone explain this line in editorial of div2 B "In this array the amount of x and the amount of m−x should differ not more than by 1, that's why we need to make max(0,|cntx−cntm−x|−1) arrays, containing a single number (x or m−x) that is more common."

In div2B why the counterpart of x is m-x. If m=5 and x=13 then counterpart of x will be 5-13 = -8 but we know that we need only 2 to add to 13 to make their sum divisible by 5. Instead it should be m-(x%m). Can anyone tell whether I'm correct or going wrong ??

its actually m-(x%m),that's why it is written that take each number modulo m.

Omg. I was amazed when I read the tutorial for Problem C2. I accepted C1. After that, I tried to think and find solution for C2 for 1 hour before I gave up. That's amazing. That's unbelievable. That's incredible.

Can somebody tell why my submission for problem E1 fails on test case 2:

https://codeforces.com/contest/1497/submission/113704368

Thank you!

Edit: I found my mistake. The function returning the reduced integer was wrong. A silly mistake!

For B, why is it $$$max(0, |cnt_x-cnt_{m-x}|-1)$$$? If the counts differ by 1, won't we have one additional array? Shouldn't this be $$$max(1, |cnt_x-cnt_{m-x}|)$$$?

If the counts differ by 1 we can always include that extra element in our array. It will not violate the rules since we need adjacent sum to be divisible

in problem {1497B — M-arrays} I think unordered_map works well instead of map but it gives wrong answer , don't understand why?

The idea behind E is https://www.geeksforgeeks.org/count-of-pairs-in-an-array-whose-product-is-a-perfect-square/

1497C1 — k-LCM (easy version) shishyando I tried problem and found a solution which is

bestand easy to understand## include<bits/stdc++.h>

using namespace std;

## define ll long long int

## define rep(i,a,b) for(int i = a; i < b; i++)

void solve() { int t; cin>>t;

}

int main() { solve(); return 0; }

Best solution indeed!

Why my code is giving TLE, I think it is completely right Time Comlecity is O(n*sqrt(a[i])) -><=10power 9 Please Help me

Can someone please help me understand why this solution for E2 works on all tests except 47?