Author: MikeMirzayanov

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
#include "random"
using namespace std;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using cd = complex<ld>;
void solve() {
int n;
cin >> n;
vector<int> v(n);
for (int &e : v) {
cin >> e;
}
int maxPos = max_element(v.begin(), v.end()) - v.begin();
int minPos = min_element(v.begin(), v.end()) - v.begin();
cout << min({
max(maxPos, minPos) + 1,
(n - 1) - min(maxPos, minPos) + 1,
(n - 1) - maxPos + minPos + 2,
(n - 1) - minPos + maxPos + 2
}) << "\n";
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
int t;
cin >> t;
while (t--) {
solve();
}
}
```

Author: MikeMirzayanov

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using cd = complex<ld>;
void solve() {
int n;
cin >> n;
vector<int> a(n);
int s = 0;
for (int i = 0; i < n; i++) {
cin >> a[i];
s += a[i];
}
if (s % n != 0) {
cout << "-1" << endl;
return;
}
s /= n;
int res = 0;
for (int i = 0; i < n; i++) {
if (s < a[i]) {
res++;
}
}
cout << res << endl;
}
int main() {
int t;
cin >> t;
while (t--) {
solve();
}
return 0;
}
```

Author: MikeMirzayanov

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
#include "random"
using namespace std;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using cd = complex<ld>;
void solve() {
int n, l, r;
cin >> n >> l >> r;
vector<int> v(n);
for (int &e : v) {
cin >> e;
}
sort(v.begin(), v.end());
ll ans = 0;
for (int i = 0; i < n; i++) {
ans += upper_bound(v.begin(), v.end(), r - v[i]) - v.begin();
ans -= lower_bound(v.begin(), v.end(), l - v[i]) - v.begin();
if (l <= 2 * v[i] && 2 * v[i] <= r) {
ans--;
}
}
cout << ans / 2 << "\n";
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
int t;
cin >> t;
while (t--) {
solve();
}
}
```

1538D - Another Problem About Dividing Numbers

Author: MikeMirzayanov

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using cd = complex<ld>;
const int N = 50'000;
bool isPrime[N];
vector<int> primes;
void precalc() {
fill(isPrime + 2, isPrime + N, true);
for (int i = 2; i * i < N; i++) {
for (int j = i * i; j < N; j += i) {
isPrime[j] = false;
}
}
for (int i = 2; i < N; i++) {
if (isPrime[i]) {
primes.push_back(i);
}
}
}
int calcPrime(int n) {
int res = 0;
for (int i : primes) {
if (i * i > n) {
break;
}
while (n % i == 0) {
n /= i;
res++;
}
}
if (n > 1) {
res++;
}
return res;
}
map<int, int> decompose(int n) {
map<int, int> a;
for (int i : primes) {
if (i * i > n) {
break;
}
int p = 0;
while (n % i == 0) {
n /= i;
p++;
}
if (p > 0) {
a[i] = p;
}
}
if (n > 1) {
a[n] = 1;
}
return a;
}
bool check(const map<int, int> &divs,
map<int, int>::const_iterator it,
map<int, int> &divsA,
map<int, int> &divsB,
int low,
int high,
int k) {
if (it == divs.end()) {
return __builtin_popcount(low) <= k && k <= high;
}
for (int p = 0; p <= it->second; p++) {
int pa = divsA[it->first];
int pb = divsB[it->first];
int nextLow = low;
if (p != pa) {
nextLow |= 1;
}
if (p != pb) {
nextLow |= 2;
}
if (check(divs, next(it), divsA, divsB, nextLow, high + pa + pb - 2 * p, k)) {
return true;
}
}
return false;
}
void solve() {
int a, b, k;
cin >> a >> b >> k;
int g = __gcd(a, b);
int low = 0;
int high = 0;
{
int t;
int ta = 1;
while ((t = __gcd(a, g)) != 1) {
a /= t;
ta *= t;
}
high += calcPrime(a);
if (a != 1) {
low |= 1;
}
a = ta;
}
{
int t;
int tb = 1;
while ((t = __gcd(b, g)) != 1) {
b /= t;
tb *= t;
}
high += calcPrime(b);
if (b != 1) {
low |= 2;
}
b = tb;
}
auto divs = decompose(g);
auto divsA = decompose(a);
auto divsB = decompose(b);
cout << (check(divs, divs.begin(), divsA, divsB, low, high, k) ? "YES" : "NO") << endl;
}
int main() {
precalc();
int t;
cin >> t;
while (t--) {
solve();
}
return 0;
}
```

Author: MikeMirzayanov

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
#include "random"
using namespace std;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using cd = complex<ld>;
vector<string> split(const string& s, char p) {
vector<string> res(1);
for (char c : s) {
if (c == p) {
res.emplace_back();
} else {
res.back() += c;
}
}
return res;
}
struct Word {
ll len;
ll cnt;
string s;
};
string getFirst(string s) {
if (s.size() < 3) {
return s;
}
return s.substr(0, 3);
}
string getLast(string s) {
if (s.size() < 3) {
return s;
}
return s.substr(s.size() - 3, 3);
}
int count(const string& s, const string& p) {
int cnt = 0;
for (int i = 0; i + p.size() <= s.size(); i++) {
if (s.substr(i, p.size()) == p) {
cnt++;
}
}
return cnt;
}
Word merge(const Word& a, const Word& b) {
Word c;
c.len = a.len + b.len;
c.s = a.s + b.s;
c.cnt = a.cnt + b.cnt + count(getLast(a.s) + getFirst(b.s), "haha");
if (c.s.size() >= 7) {
c.s = getFirst(c.s) + "@" + getLast(c.s);
}
return c;
}
void solve() {
int n;
cin >> n;
map<string, Word> vars;
ll ans = 0;
for (int i = 0; i < n; i++) {
string var;
cin >> var;
string opr;
cin >> opr;
if (opr == "=") {
string a, plus, b;
cin >> a >> plus >> b;
vars[var] = merge(vars[a], vars[b]);
} else {
string str;
cin >> str;
Word word;
word.len = str.length();
word.cnt = count(str, "haha");
word.s = str;
vars[var] = word;
}
ans = vars[var].cnt;
}
cout << ans << "\n";
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
int t;
cin >> t;
while (t--) {
solve();
}
}
```

Author: Supermagzzz, Stepavly

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <iostream>
using namespace std;
void solve () {
int L, R;
cin >> L >> R;
int ans = 0;
while (L != 0 || R != 0) {
ans += R - L;
L /= 10;
R /= 10;
}
cout << ans << '\n';
}
int main () {
ios::sync_with_stdio(false);
cin.tie(0);
int testc;
cin >> testc;
for (int i = 0; i < testc; i++) {
solve();
}
}
```

Author: MikeMirzayanov

**Tutorial**

Tutorial is loading...

**Solution**

```
#include <bits/stdc++.h>
#include "random"
using namespace std;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using cd = complex<ld>;
void solve() {
ll x, y, a, b;
cin >> x >> y >> a >> b;
ll l = 0, r = 1e9 + 100;
if (a == b) {
cout << min(x, y) / a << "\n";
return;
}
if (a < b) {
swap(a, b);
}
while (r - l > 1) {
ll m = (l + r) / 2;
ll right = floorl((x - m * b) * 1.0l / (a - b));
ll left = ceill((y - m * a) * 1.0l / (b - a));
if (max(left, 0ll) <= min(right, m)) {
l = m;
} else {
r = m;
}
}
cout << l << "\n";
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
int t;
cin >> t;
while (t--) {
solve();
}
}
```

thanks for quick editorial

MikeMirzayanov is that your short solution for D? Lol

This is my code. https://codeforces.com/contest/1538/submission/119791902

Hey can you please explain your logic

SpoilerI did not got the editorial

Have a look, if you want. https://codeforces.com/contest/1538/submission/120844896

I saw F so late ugh, sucks to miss on easy points

add high rated friends xd

how will that help?

high rated people are fast, so you can get an idea of which question to solve next. you can also check problem page for no. of solves but I check standings more so..

Great contest, I solved problem C with two pointers. Calculate the total number of pairs = (n *(n-1))//2, now calculate the pairs sum < L (using two pointers) let's call it A and calculate the pairs sum > R with the same approach let's call it B. So our answer will be total pairs — (A+B). Submission

can you please help me why is my code giving tle. i implemented lower and upper(well basically upper--) and from i=0 to i=n i calculated how many to right satisfy the criteria.submission edit: this is the submission not the above

Comments in my solution may help :) submission

Dude link you gave opens my submissions page wth?

sorry , please have a look now

Instead of passing vector in function

either make it global and use it's value

or

put your lower and upper function inside solve.

I think you are getting tle since your code copies whole vector every time you call lower() and upper() function.

I edited your code, it runs fine now.

Link to edited code

can you help me with c#?

im using two pointer technique for problem C but im get TLE

this my link submission https://codeforces.com/contest/1538/submission/119312558

logic is wrong your code is running n^2

if array is : 3 3 3 3 3 3 and l is 1 and r is 100 then your code will run n^2 times

thanks, i Will fix my code

Passing vector to a function

This link would help you.

god that trivial mistake. just missed it. thanx man

I did solve with two pointers directly (almost). I count all valid pairs (i, j) and (j, i) then subtract valid (i,i) and divide by 2. 118994288

How does the condition l<=2*v && 2*v>=r help in finding valid(i,i)??

Can you please elaborate.

Oh, never mind. Got it! Sorry for the trouble.

what is L and R?

l is current index i and r= n-1 since i am searching for all numbers which can pair up from i+1 to n

Thanks [user:_PeakyBlinder_]your two pointer solution was great .

thanks mate u r amazing

Amazing problem set

I like the solution to E, looks pretty simple, I thougt it would be much more complecated.

But binary search in G is unclear, I cannot see the trick from the formulars. Why a ternary search does not work? And how does the function work on thats result we can binary search?

Pretty sure a ternary search doesn't work because function isn't strictly increasing then decreasing, but feel free to correct me if I'm wrong.

In my understanding there are two lines, and the optimum is reached where they cross. So it is non decreasing until that point, and non increasing afterwards.

u can binary search for the maximum number of gift sets. Let a < b and diff = b-a . Let the number of gift sets equals n, n is valid if and only if we can put n items which size = diff into 2 knapsacks which sizes = x-(a*n) and y-(a*n) https://codeforces.com/contest/1538/submission/119046681

I did read some texts, watched a video... but still do not get how/why binary search works here. The code inside the

`while(l<r)`

loop looks completly arbitrary to me.Can you explain why it works, or how it works?

I misread the input and used a, b as the total number of candies and x, y as the number of candies per gift set . But the logic remains the same

ternary search works but you have to use real numbers https://codeforces.com/contest/1538/submission/119051522

That is what I tried to implement in my first submission Seems there is some implementation issue, maybe caused by rounding problems.

I should have been clearer what I extended was the codomain, not the domain.

I just noticed by the way

Div3 is supposed to be unrated for anyone who "have a point of 1900 or higher in the rating."

You do have such a point. But you gained rating in round 719

Nah, you mix up two different things.

It is rated for people up to 1600, current rating. And you are trusted up to 1900. Also current rating.

Yeah... my bad

For some reason I believed that it is unrated for anyone above 1600 now or 1900 any time in the past.

You can check, just go below 1600 once... ;)

It has a greedy solution if you're interested. first take the max(a,b) from max(x,y), and min from min. once they cross each other, or are equal,take sizes alternately like (b,a) (a,b). the submission is not clean, but it works. link

clickhttps://codeforces.com/contest/1538/submission/119207131

First of all Thank You for the Editorial !!!

I am facing difficulty in understanding Problem C

`Number of Pairs`

Editorial. In the code attached I could not get the significance of the following lines`if (l <= 2 * v[i] && 2 * v[i] <= r)`

Why we are checking for this particular condition ?`cout << ans / 2 << "\n";`

Why are we dividing our final answer by 2 ?division is done because the same pair will be counted twice and not sure but the condition checks that if we should not count the same index value as we should have two value from different index

2)This is done because in the question it is mentioned that i<j i.e i is always less than j , however in our code we are unable to check for this condition so for values of i>=j also the pairs are calculated. This means that all pairs are counted twice. To remedy this we divide our final ans by 2.

Got it !!! Thanks a lot :)

Thanks

Everything makes sense, but theres just one thing I don't get and it would be a great help if you would explain it to me: Taking the example of v=1 1 2 5 5 10 and v[i] = 5 and i = 3, and l = 9, r = 13, We have r-v[i] = 8 & l-v[i] = 4, and the upper_bound(r-v[i]) = 5 and lower_bound(l-v[i]) = 3. So instead of adding in 3 elements in the range of 4 — 8, why are we adding 2 elements? And how does this work?

thanks! :)

This means that all pairs are counted twice.But how can we be sure that pairs are counted twice..i mean instead of dividing by 2 , we could possibly divide by any no.(eg:1/x of segment is counted repeatedly).

learn about lower_bound and upper_bound !! great stuff imo!

Basically in Problem C we have to find all the pairs in range L to R whose sum is in this range .

So what we can do is that we can find all the pair till L-1 and all pair till R.

now we can subtract the pair found and get our desired result

in order to find the pair till sum x we can follow a two pointer technique whose algo is :

step 1) assign low = 0 , high = n-1 step 2) while ( low < = high ) will be our bounding condition step 3) if sum at a [ low ] + a[ high ] < x then pairs between low and high would be our ans now we increment our low pointer step 4) in case sum at a[low] + a[high ] > x we decrement our high pointer

the simplified solution is here : 119093049

please see : you can use PBDS for the same

I can't understand it either, but I wrote it myself according to the explanation:

Look at the complete code

To better understand this try thinking like this:

Sort the array

Now for each index, think of two pointers on the right:

to get left limit substract

`l - v[i]`

, similarly to get right limit`r - v[i]`

. (imagine to get sum as`l`

we have to get numbers greater than or equal to left limit and to get sum as`r`

we have to get numbers upto right limit.lower_bound and upper_bound for perfect for this as they get these values in

`log(N)`

time.after getting values check for extreme cases and also we dont want to check for previously calculated values, (keep between i+1 and n-1)

count numbers in the range of left limit and right limit by substracting their indices.

sum up all to get final answer.

Submission link

check this link to understand more about upper and lower bounds.

Now answer to your 1st Question:During lower_bound we may also count the current number as`2*v[i]`

as lower limit but the question says`i not equal to j (take different indexes)`

. Same applies to upper limitNow answer to your 2nd Question:In my solution I have discarded previously calculated values, by keeping lower limit as`i+1`

. But in the editorial OG has calculated all the numbers (all the pairs) for each index. That means they have been counted twice. Hence divide by 2.Hope its worth a vote :)

Thanks a lot pratyushgguptaa, looeyWei , beast_on_joB , MrBing and 18o3 for helping me out in understanding the problem as well as the editorial. Thank You !!!

In fact [user:pratyushggupta] , you could just simply do this

We are checking if the pair (v[i],v[i]) is valid I.e v[i]+v[i] lies in the range.If it is then answer must be decreased by 1 because we would have included that case when we are finding the upper bound. And ans/2 is because we would have included both the pairs (a,b) and (b,a). So the actual answer is 1/2 of what we have obtained in the for loop.

Problem F is arguably easier than problem A Got stuck at A for a long time :(

My solution to C using policy-based ds. code

Thank you so much, I was searching for this

Editorial for D is wrong, $$$n$$$ should be the sum of exponents of prime divisors instead of the number of prime divisors.

That's the same thing, since we count

ALLprime divisors, not just distinct ones.Well, probably you can argue that we define prime divisors of a number as a multiset, but it is really weird (at least for me).

I fully agree to you.

One thing seems weird in D. if a==b then shouldn't k be equal to 0 ? so like since it is not mentioned that when they are equal we stop... maybe this is the reason, but dont we sort of generally stop when we reach the condition a==b? in any other q for eg

No k shouldn't be equal to 0 , as you can still divide no.s from a and b if a==b

I thought the same way

deleted

Massively overcomplicated solution for F. This passes.

hey just wanted to know whether problem F was a digit dp problem ??

I complicated it so much and solved it using digit dp

Why is the provided implementation for F so long and far-removed from the description? For example, my short implementation is here, and is immediate from the description in the editorial: 119035674

We think alike :)

Same Here 119015785

Can someone help me find the testcase where this code 119549893 fails for problem F. What am I doing wrong? Upd: Got it.

Another solution was inserted into the analysis. Fixed now

I think you only need

`R != 0`

because $$$R \geq L$$$ so $$$R$$$ can only be $$$0$$$ when $$$L$$$ is $$$0$$$I feel dumb for getting WA 4x, only because of forgetting to use long long instead of int at problem C :v

Hope it helps :p

Its the same, either use "#define int long long" to stop getting WA but then you will get TLE on particular questions or try to remember using long long according to constraints and get WA sometimes. I prefer second, improves "attention" for other twists in other type of constraints too.

it help in 1e9 integer overflow

did the same mistake ;;

Less cumbersome implementation for D. code

Are you asking for it? here

I had a doubt- how to decide the upper limit of the prime numbers we need? (1000009) in your case.

Its sqrt(10^9) in this case, I took 10^5 just to be safe.

anyone can point out mistake in F ? 119076961

Problem F solution with simple explanation: https://www.youtube.com/watch?v=8c_dVBCBJSo&t=312s

i think naitikvarshney use invalid input for hacking solution of problem C. he have already hacked 95+ solution(including me). :+(

Your code got hacked due to Arrays.sort() function which in worst case have O(n^2) time complexity.

Another problem with Java

`Arrays.sort()`

:/Is it really necessary to hack all your victims during the hacking stage to disqualify their solutions? Aren't successful hacks used as a part of system tests anyway? Or am I missing something?

But hacking others gives a sadistic pleasure.

Profile picture checks out.

What kind of sort in Java is guaranteed n log n?

Use Collections.sort(), it will work

buy a new laptop and start with C++

Problem C video Editorial (using Binary Search) link : https://youtu.be/_2-iretTWqc

Thanks for such an interesting contest!

Looks like the code for problem D is wrong , Please correct it , the main issues are in Solve function where there are brackets but no for loops . MikeMirzayanov

Here is an easy to understand solution 119087228.

Thanks , I too had O( T * sqroot( Max(a,b) ) solution but it TLEd as it had a bit bigger constant factor.

I was initially doing same as yours but getting TLE(on test 6). Later came to know we have to use prime seive to get max count in O(T*log(max(a,b)) Otherwise, time complexity would be O(T*sqrt(max(a,b)). I wonder how yours got accepted. XD

I kind of applied the same logic in D, but not able to pass the tests. Can anyone tell me what am I missing link

Your code marked all of the cases with $$$a$$$ equals $$$b$$$ as a no. But actually, there may be some cases when $$$a$$$ = $$$b$$$ and the answer is yes. For example:

$$$1$$$

$$$2$$$ $$$2$$$ $$$2$$$

Thanks, missed that condition, just now realized that if K>1 and a and b are equal then also the answer is yes.

We can also use linear programming optimization for problem G to solve it in O(1).

For 1538G - Gift Set following solution pass tests. We have following set of restrictions

Forget for a moment about last requirement (n, m are integers). Then, first two restrictions is basically region on n,m coordinate plane. And n + m is cost of each point within allowed region. Then, we can draw lines of fixed cost D: it's all point with cost D = n + m. Cost = 1 is line 1 = n + m. Cost 2 is line 2 = n + m and so on. It's easy to see that all of them are parallel, and when you increase D you actually move line perpendicular to it. Given line with cost 0 is 0 = n + m we know that cost of any point is actually distance to this line (n + m = 0).

So, we want to find point within region with highest distance to the line n + m = 0. What about region we have? It's convex polygon. Thus, largest distance to the line is equal to distance to some vertex of our polygon. So, if we forget about integer n, m, we can pick this vertex as answer. This is actually explanation how linear programming works in 2D space.

What can we do with integer n and m? Well, I just did hack: round n in arbitrary direction and tried few other integers around and this passed. 119052049 The question is: is it valid solution or is there counter test?

Integer Programming in general is NP-hard. Just searching few points around the linear programming solution does not guarantee (integral) optimal solution.

This should be correct. In this particular case, as we move away from the intersection point, we get closer to the $$$n + m = 0$$$ line, since one of the lines has a slope less than $$$n + m = 0$$$ and the other has a slope greater.

It's more like intuition instead of formal proof. I would like to have formal proof, and here it is (at least some sort of).

Messy long proofThere are two big cases:

I'll focus on first case. Suppose we have intersection at coordinate (n, m), then if we round down both we will get $$$n_1 = \lfloor n \rfloor, m_1 = \lfloor m \rfloor$$$. Upper bound on answer is $$$\lfloor n + m \rfloor$$$, and after throwing of fractional part it's easy to see $$$\lfloor n + m \rfloor - (n_1 + m_1) \leq 1$$$. In other words, either we have optimal answer or we're off exactly by one. If this difference is zero, then we already have optimal answer. If this difference is 1 let's check points $$$(n_1+1, m_1)$$$ and $$$(n_1, m_1+1)$$$. If any of them fits our requirements then this point is optimal answer. If none of them fits our restriction, then I claim there is no point $$$(u, v)$$$ within region with integer coordinates and $$$u + v = n_1+m_1+1$$$.

Here is proof. Notice that fractional part of n plus fractional part of m is greater or equal to one. In other words $$$n - n_1 + m - m_1 \geq 1$$$. It's region with point $$$(n, m)$$$. Let's find out what region of plane it is. Border line of this region goes through two points we checked: $$$(n_1+1, m_1)$$$ and $$$(n_1, m_1+1)$$$ (for verification just put this points into inequality), and also this is the line where all points with our goal: potentially better answer $$$n_1 + m_1 + 1$$$. Let's call this line 'goal line'.

From vertex $$$(n, m)$$$ there are two outgoing lines of convex polygon. In case if $$$(n_1+1, m_1)$$$ doesn't fit our restrictions this means that one of lines goes through some point $$$(n_2, m_1)$$$ on side of cell where $$$n_1 < n_2 < n_1 + 1$$$. Left side of inequality because $$$(n_1, m_1)$$$ within region, and right side of inequality holds because otherwise $$$(n_1+1, m_1)$$$ would fit our requirements. But this point $$$(n_2, m_1)$$$ on side of cell is outside of our region of point $$$(n, m)$$$ which means this line of polygon crosses our 'goal line' with all potentially better points. Because 'goal line' is border line of region where $$$(n, m)$$$ located. And, two intersecting lines may intersect only once, thus only 'half' of our 'goal-line' (with n coordinate $$$ \leq n_1$$$) potentially may fit our restrictions. (In some sense we cut off that part of goal-line which goes out of convex polygon)

If we look at other polygon line outgoing from vertex $$$(n, m)$$$ it should go through $$$(n_1, m_2)$$$ with $$$m_1 < m_2 < m_1 + 1$$$. And argument repeats. Thus, only 'half' of our 'goal-line' (with m coordinate $$$\leq m_1$$$) potentially may fit our restrictions. And if we combine two restrictions, we get, that points $$$(u, v)$$$ on 'goal line' that might be better should satisfy $$$u \leq n_1$$$ and $$$v \leq m_1$$$ but none of points on 'goal line' fit this requirements because if we add two inequalities we get $$$u + v \leq n_1 + m_1$$$ but all points on 'goal line' is $$$u + v = n_1 + m_1 + 1$$$. So, there is no point on 'goal line' that fits our restrictions.

What if region is triangle? Well, I don't know, probably something similar applies.

With this proof code becomes a bit easier: 119096819 (three points enough)

IN G, shouldn't the inequalities be x >= a*k + b*(n-k) and y >= a*(n-k) + b*k. Also, the second equation in the four-set of equations should have y and not x?

I think there's no need of keeping length in E, because we can handle special cases by checking if the prefix or suffix's length is less than 3.

E problem seemed very tricky

For C. I use Java, and write my own binary search, but got TLE?

Update: Got it.

The sort() is O(n^2).

I put numbers in List and use Collections.sort(). it works.

Thx! I have the same problem

In G tutorial it should be, x >= a⋅k+b⋅(n−k) y >= a⋅(n−k)+b⋅k and similarly, the next two equations will be changed accordingly. Supermagzzz correct it.

Can someone tell me why my code is failing for problem C. It is the same logic as the editorial only with an extra condition. 119012335

Could someone plz tell me as of why this submission of mine getting TLE whereas this one is passing?

The second submission uses sieve and map to store the values, whereas mine uses normal sieve. Still contrary to what should have occured, he received AC.

I have even seen submissions having the same implementation as of mine, but passing the constraints easily.. Why is this happening?

Now what i did is- changed my long long to int data type and got AC plus a near about 1/3 execution time.. I have never seen so much of a difference just because of the change of data types. :(

Could someone please give an insight on the same.

Did you try it with a c++17 64 bit compiler? It is available on codeforces. Calculations with long long or long double data types are a lot slower.

Ya I did so

Result remains the same but.

This is the submission

Editorial's code for D is an eyesore

D has lot of corner cases

My solution to D is simpler I think

sadly in contest I used long long it barely passed so I'm a little scared

thanks for telling, I will try to hack:)

UPD : I cannot, it works fine.You're welcome it's better than waiting for tomorrow to know that it's TLE :P

Can you explain this line: if(x > 1) cnt++ Is this to include 1 as a divisor

No in any number there could be one prime factor bigger than sqrt like 11 without this line you won't count 11 as a factor

Got thrown into another dimension while solving E.

Here D editorial solution is very complex.....

Over the head... Btw It's not so complex how he show it!

For problem G I have some $$$O(1)$$$ solution.

First, think about how we can solve the problem easily when constraints are $$$10^5$$$.

($$$a <= b$$$, otherwise swap them) ($$$x <= y$$$, otherwise swap them)

To solve this version we can easily brute force step by step and while we can create a new gift we will decrease $$$b$$$ from the higher one and $$$a$$$ from the remaining one this is optimal and proof left as an exercise to the reader.

But when constraints are $$$10^9$$$ we can't brute force so let's look at this solution and see what happens.

At first, let's assume $$$y - x <= b-a$$$ in this situation if we apply our brute force algorithm at each step higher one will change because $$$y-b <= x - a$$$ and their difference also won't exceed $$$b-a$$$ because $$$x-a-(y-b) <= b-a$$$,$$$x-y <= 0$$$ is true so for this type $$$x$$$ and $$$y$$$ we can solve problem with

$$$(x/(a+b))*2 + val$$$, $$$val$$$ is 1 if at the last we can't decrase $$$(a+b)$$$ from $$$x$$$ and $$$x >=a, y >= b$$$

And while $$$y-x > b - a$$$ this means we always decrease $$$b$$$ from $$$y$$$ so we can handle this until $$$y - x <= b-a$$$

Here is my implementation 119099472

Can someone please prove it? I've done the same thing but couldn't prove it.

EDIT: Never mind, got it now.

I don't understand the official solution, but your method is really simple and easy to understand.That’s amazing:)

Is 10^4 * sqrt(10^9) complexity code acceptable everytime?

I guess it's $$$10^4 \cdot \dfrac{\sqrt{10^9}}{\log(10^9)} \approx 30\ 000\ 000$$$ if you only check divisibility from primes.

Actually, $$$10^4 \cdot \sqrt{10^9}$$$ also passes 119014291 if you don't do anything extra.

Yeah, my code has passed too, but with 1544 ms runtime. I am little scared that will it pass on system tests or not :(

It got passed, yay :)

For problem G, my solution is transfer the problem into an ILP problem, that is:

Treat it as a LP problem and then use Simplex to get the value of $$$x_1$$$ and $$$x_2$$$ when $$$z$$$ is maximized. Since $$$x_1, x_2$$$ could be float numbers, and I guess the answer for ILP problem will be around $$$(x_1, x_2)$$$, so I search a few integer points around $$$(x_1, x_2)$$$.

(FST Warning

For bugaboo D, we can find all the primes till 1e5. There are less than 1e4 prime numbers in this range. Now, for each a and b, we check the divisibility with this list of primes. If none of the primes till 1e5 divide a, it means that 'a' itself is a prime number. Because any factor more than 1e5 would need a smaller factor less than 1e5, because 1e5*1e5 = 1e10, which exceeds the constrains on a and b. UPD — My code 119080458

how to find the sequence of primes <1e5 : for(i -> 1e9)????

We only need primes till 1e5 not 1e9. You can use sieve() and then store the primes in some vector.

i did that but wrong on test 3 , I don't understand why I'm wrong, you can check help me, please

You may have a look at MyCode (warning — it's messy)

Thank bro, maybe i should have done the F problem first, probably raised the rate <33

119103856

Can any body hack my submission for Problem G ?

I didn't use binary search.

I think the problems should be sorted by increasing difficulty

Supermagzzz, MikeMirzayanov, I think there is a typo in the tutorial of problem G: Maybe it should be $$$\frac{(y−a⋅n)}{b - a} ≤ k$$$ rather than $$$\frac{(x−a⋅n)}{b−a} ≥ k$$$, and $$$\frac{(x−a⋅n)}{a - b} ≥ k$$$ rather than $$$\frac{(x−a⋅n)}{a−b} ≤ k$$$.

I can hardly believe this is the difficulty of div3, but the problem is very good, I like it very much, thank you for your tutorial.

Is there a wrong about problem G in Codeforces Round #725 (Div. 3) Editorial. In the tutorial (x−a⋅n)b−a≥k may be (y−a⋅n)b−a≥k

Can anyone tell which corner case I am missing in the solution to problem D? I am getting WA on token 1021 of test case 2. Here's my link 119065848. Thanks.

Bro, I covered all corner cases still getting WA, c=this case also I have covered. Can you point out any possible case I am missing. Thank you.

Submission link: https://codeforces.com/contest/1538/submission/119430768

In your factors function you should write while(a%2==0) instead of while(a%2!=0) so it's just a typo not logic mistake and you may get tle because in for loop you are writing i++ instead of i+=2 this i++ makes dividing a by 2 in the while loop pointless

Thanks a lot.

I solved C using 2 hash maps . One that stores the difference of the minimum sum and arr[i] and second that stores the difference of maximum sum and arr[i]. and then for each arr[i]. Found the number of integers in the array so that they are between low[arr[i]] and high[arr[i]]. But I was not sure if it was fast enough to pass the system tests. I would be grateful if anyone can hack it!

[EDIT] here is the link to the solution 119064844

if(k==1) ans can be YES if one of n,m is 1. Also tha ans is not always no for n==m.

I didnot get you... What is k?

also k can be 1 if n%m ==0

use Java in C, about

`Arrays.sort()`

AC

TLE

Can anyone tell, why it happens. even I am facing same issue, why its TLE while using long[] and passes on using Long[] ?

Confused. Need help

You can take a look at the Java source code for Array.sort(). You will find that when using long[] it is DualPivotQuicksort, and Long[] is TimSort. In the worst case, DualPivotQuicksort is still O(n^2), while TimSort is O(nlogn).

Can anyone please point out mistakes in D's code? 119088308 It's failing on the 5053rd token in test case 2.

UPD: Got it, thank you!Can you tell me why it was failing ? I'm getting same error.

UPD: I GOT ITI got a tle in D because of using long long instead of int.

exactly, me too!

but while practice, just figured it out!

Converted it to int and it got AC

Why, using long long would increase memory only and TLE could be understandable when recursions were used, but here no such cases are there. Please explain... I also submitted with long long only but it passed...119115670

Such a nice contest with excellent editorial.

Can someone provide me the test case of

Gwhere my solution fails 119117297https://codeforces.com/contest/1538/submission/119118611

This is my dp solution of problem G . It is giving runtime error on testcase 5 that is for large x ,y and small a,b . could someone please tell the possible reasons for the error so that I do not repeat the same mistake again in future contests . Please .. Thanks in advance..

I think it is giving Runtime error because for large values of x and y and a and b being small, the size of map would be greater than INT_MAX so you would not be able to store that many values and it is giving RE.

is there a size limit for storing elements in a map ? also , could you please tell is there a way to fix this problem?

Seriously I didn't liked the implementation of D given in the editorial. I simply used some tricks to speed up the sqrt(max(a,b)) solution and it was good enough to pass the tests. :)

Here's the link of my submission : 119025249

Lightning fast editorial.Kudos to the team

In problem F, I think solve two related problem which the number of changed digits in $$$[1, l]$$$ and $$$[1, r]$$$ is better. Than, we can use a simple subtraction to get the answer.

If we focus on problem as $$$[1, x]$$$, we can find a way to calculate the ans: first, each digit can provide [ $$$11 \dots 11$$$ (the number of $$$1$$$ is $$$b$$$) $$$ * 10^b - 1$$$ ] changes ($$$b$$$ mean the current number of digits), than, we add $$$n - 1$$$ to the result, n is n-digits, finally, we get the correct answer.

For example, to problem $$$[1, 5678]$$$, calculate detail is: $$$(5 * 1111 - 1) + (6 * 111 - 1) + (7 * 11 - 1) + (8 * 1 - 1) + (4 - 1)$$$.

The logic of this way is the same as editorial. Here is my code, this may be clearer than my comment.My Code

I think that the implementation of the problem D solution is complicated! I have implemented it in a simpler way. Here is my code:119017802

Supermagzzz why are taking only a<b in 1538G - Gift Set ?

can anyone tell whats wrong in my code for C in given input for 4th test case it is giving ans 0.

here is my code link https://codeforces.com/contest/1538/submission/119128758

You forgot to sort the array. Submitted your solution after sorting and it got accepted. submission

thnks

In problem G, why both of the equation is 1. x≤a⋅k+b⋅(n−k) 2. y≤a⋅(n−k)+b⋅k instead of x>=a⋅k+b⋅(n−k)

y>=a⋅(n−k)+b⋅k ? MikeMirzayanov

The "x" on the G is wrong. It should be "y".

Great contest!!

I solved problem D using sieve of Eratosthenes, hope it helps!!

Also, any suggestion for optimization in the code would be appreciated!!

SubmissionID

In problem G's solution, if I don't write the floor function while calculating ll right and instead write it as ll right=((x — m * b) / (a — b)), then why am I getting wrong answer. The integer division should give me the floor value, then why are we using floor function explicitly. Can someone help. Thank you.

Let's say we get the range [3.5,5.5] because we're getting integers

So the left interval is going to be 4, and the right interval is going to be 5

Casting to an int will truncate toward zero. $$$floor()$$$ will truncate toward negative infinite. So, $$$int(-0.9) = 0$$$ and $$$floor(-0.9) = -1$$$

Ok. So, can we say that, since the value here can be negative as well that is why we need to use floor. If it was guaranteed that the result will always be positive then we don't need to use floor function.

Questions about D floor and floorl What's the difference and 1.0l?

I am a begineer on codeforces so forgive me if i have done a blunder . can anyone help to tell me why i am getting tle for d which i code today as practice on test case 2 here is the link to my solution

https://codeforces.com/contest/1538/submission/119155153

Luckily exact same code got accepted in pypy3 in just around 800 ms while the same in python 3.9.1 gave tle on only test case 2

why am i getting a tle for D. I did save the results from previous states to avoid tle https://codeforces.com/contest/1538/submission/119065595

In G, it will be better if taken r=(x+y)/(a+b)

Can anyone tell me why this 119163197 for D is giving tle and not this 119163156

It was my code and for python 3.9.1 it gave tle on test case 2 and i got ac in pypy3 with same code instantly .

Can any body tell me how k>=(y-a*n)/(b-a) while given equation was y>=(n-k)*a+kb I guess equation should be k<=(y-a*n)/(b-a).

The solution of problem C can be written more efficiently as you are going through the already visited pairs again and again in each iteration of loop so i slightly modified the code to be more efficient do see my code for the problem thanks :) here's my submission link

The editorial for G seems to be incorrect. The second reordered equation is given as

`(x−a⋅n)b−a>=k`

but it should be`(y−a⋅n)b−a<=k`

in my opinion.Also, I can't understand why we have used greater than or equal to in the first two equations. Can someone explain this part to me please? I think the equations should be :

`x>=a⋅k+b⋅(n−k))`

and`y>=a⋅(n−k)+b⋅k`

Can someone explain me what is going on with the solution code for problem D? I understood the editorial but didn't understand the code.

Here is a list of the things I

thinkI've understood: - precalc() generates a sieve. - calcPrime(n) returns the size of the "largest prime factorization" from which we can obtain n. - decompose(n) decomposes n into the "largest prime factorization" from which we can obtain n.But what do these parts do?

This part of the main function:

The entire check function:

Thanks in advance.

Can you please explain O(1) solution to Problem G? (easy to understand if possible)

Wished the editorial explained the "if (l <= 2 * v[i] && 2 * v[i] <= r) "... the edge cases should be included in the editorial too!

anyone can give me any good video tutorial about porbem E.

I am not getting editorial of D question. Is there any theory I need to refer or if someone has any other solution, pls help.

What is the logic behind this line for problem C?

Our condition is that

`l<= a[i] + a[j] <=r`

, where`i`

and`j`

are two pairs(i<j). It can occur that the`a[i]`

itself can satisfy this condition and we may pair`a[i]`

with itself. This can occur because every time we check suitable pairs for`a[i]`

,a[i]itself is also included in the array. If this happens then it means`a[i]=a[j]`

so,`l<= 2*a[i] <=r`

. Since we want different pairs, to avoid possibility of an element pairing to itself and satisfying the condition, we check for that and reduce answer by one.For example ,

`a = [1,2,3,4,5,6]`

,`l = 5`

,`r = 10`

Suppose we are at

`a[4]=5`

. Now here, lower bound will take us to`0th`

index and upper bound will take us to`5th`

index.. Giving us total of 5 pairs for`a[4]`

.But, correct number of pairs would be 4.Thanks, Brother.

In problem G editorial i feel it should be x>= a*k + b*(n-k) similarly for y y>=a*(n-k) + b*(k) bcs x and y should have sufficient candies.I would be very happy if someone correct my intuition.

Please can anybody find out my mistake in problem D:

My logic is to keep dividing a by 2 if it is even and increase count similarly keep dividing a by all odd numbers and increase count. Similar thing I will do for b. This count will be the maximum times I can divide a and b to get a=b=1.

In my code max==2 is for the case when both a and b are prime numbers.

My submission:https://codeforces.com/contest/1538/submission/119194346

If

`a == b`

and`a is not prime`

then you are setting`min = 1`

but atleast 2 division operations will be required in this case.Ya I miss that case. Thanks!

Very fast one line O(n log n) solution to problem C in Dyalog APL.

Try it online

This would be a great language to add.

I solved problem C using a data structure called ordered set in c++. It is like the normal set but has 2 more functions. one of its functions is a function called order_of_key() which returns the number of elements strictly less than a certain number in log(n) time so I used this function to calculate the number of elements less than 1 or (l — ai) in case ai is more than l and the number of elements less than (r — ai + 1) and then subtract these two numbers and add it to the answer.

I saw other solutions which are faster than mine but I thought that this approach is cool to share

submission: https://codeforces.com/contest/1538/submission/119195153 more about ordered set: https://www.geeksforgeeks.org/ordered-set-gnu-c-pbds/#:~:text=Ordered%20set%20is%20a%20policy,in%20log(n)%20complexity%20.

Here's a slightly different way to do G. Let's say we make $$$l$$$ gift boxes first by removing $$$la$$$ red and $$$lb$$$ blue candies from the piles. (Each of these boxes contains $$$a$$$ red and $$$b$$$ blue candies.) Then, of the remaining $$$x-la$$$ and $$$y-lb$$$ candies, we can make $$$\min \left( \lfloor \frac{x-la}{b} \rfloor , \lfloor \frac{y-lb}{a} \rfloor \right)$$$ more boxes each containing $$$b$$$ red and $$$a$$$ blue candies.

So in total, we have made $$$l + \min \left( \lfloor \frac{x-la}{b} \rfloor , \lfloor \frac{y-lb}{a} \rfloor \right)$$$ boxes. We want to maximize this, under the conditions $$$l \geq 0, x - la \geq 0, y - lb \geq 0.$$$

It's not hard to show that this function (on the integers) is non-decreasing and then non-increasing, and that no three consecutive integer inputs of $$$l$$$ yield the same value. So, it's easy to binary search on to maximize this function.

Problem G: Gift Set. My solution is wrong for testcase 2.1103 It says it should be 0, but my solution gives 1. Could you, please, provide me with a testcase when my code gives a wrong answer? Thanks in advance.

ll k = min(x,y)/min(a,b); ll v = (x+y)/(a+b);

if (v < k) { cout << v << endl; } else { if (max(x,y)/max(a,b) >= k) { cout << k << endl; } else { cout << 0 << endl; } }

//my soln for c void solve(){ int n,l,r; cin>>n>>l>>r; vi a(n); for(int i=0;i<n;i++) cin>>a[i]; sort(all(a)); int ans=0; for(int i=0;i<n;i++){ int x=a[i]; if(x>r) continue; if(x<l){ int lb=lower_bound(a.begin()+i+1,a.end(),l-x)-a.begin(); int ub=upper_bound(a.begin()+i+1,a.end(),r-x)-a.begin(); // cout<<ub<<" "<<lb<<endl; ans+=ub-lb;

} else if(x>=l){ int ub=upper_bound(a.begin()+i+1,a.end(),r-x)-a.begin(); ans+=ub-i-1; } } cout<<ans<<endl; }

Can anyone help me with task C: I'm stuck in understanding the problem of next code (just a brute force) as a central part of the solution:

it fails on test #2 with WA: wrong answer 1491st numbers differ — expected: '10', found: '0'

I stress-tested this solution against the cononical solution and didn't find wrong example.

I found a mistake :-)

before this code there was a construction:

cin >> n >> l >> r;

if(n == 1){ cout << "0\n"; } else{ for(int j = 0; j < n; j++){ cin >> A[j]; } // solution code }

so it didn't read input if n == 1 and therefore input for previous test was considered as input for next test.

thanks for quick editorial

For G I felt compelled to speak about it.

My idea for the problem is nearly the same till the part of binary search on the max answer.

What I did next was since $$$a \ge b$$$ therefore an expression like $$$ap + bq$$$ where $$$p + q$$$ is a fixed value would yield a greater result for higher value of $$$p$$$ and therefore I decided to do another binary search on finding out the pair value $$$(p,q)$$$ which would cause the expression $$$ap + bq$$$ to be just below $$$x$$$ inclusive (as mentioned in the problem). This would cause the expression $$$aq + bp$$$ to be least and that's pretty much what we want to do since $$$aq + bp \le y$$$.

My idea uses another log in time complexity due to running binary search within binary search but I found it to be a lot lesser troublesome in terms of implementation against using floor or ceilings and therefore I decided to comment on G.

Link to my solution:- https://codeforces.com/contest/1538/submission/119120664

cool idea...But I didn't understand how the idea of maximizing the value p(in second binary search) gives optimal answer?

You want to make sure the first expression is $$$ \le x $$$ and second expression is $$$ \le y $$$

so if u maximize the first expression as you can to be just below $$$x$$$ it would cause the second expression to go as low as possible.. and that's what we want because it's always better to take a lower value for any expression.

Actually you can also do that other way around maximize the $$$q$$$ but then you have to run binary search for maximizing the second expression. ($$$aq + bp$$$)

I don't understand why E had very few submissions. Wasn't it simple bruteforce and hashing? btw I did in python.

Mee too In Python, it is quite easy.

Submission : https://codeforces.com/contest/1538/submission/119243916

It seems like the editorial on problem F is incorrect. The inequalities should be: $$$ x \geq k.a + (n-k).b$$$ and $$$ y \geq k.b + (n-k).a$$$. Hope the author correct it.

Can someone tell me why does the binary search in problem G code work? I'm not very experienced with binary search. Suppose we only have three possibles values right now, [1,2,3], and 2 satisfies the if condition. Then, we update the interval to [2,3]. The code stops here returning 2 as the answer. Why doesn't it check 3 too? Can't it be a posible better solution?

In the editorial solution, l and r are taken as such that l <= n < r (where n is the req solution). So u don't need to check at r. U can look this

So updating r = m and not checking it as a solution is the same as updating r = m-1 and do check it. Got it. Thanks!

How can we sort the array in problem C, as we are distorting the given array indexes? Generally we don't sort in such pair of indexes problem right?

Can anyone tell me why i am getting memory limit exceeded with my code in python forgive me if i did a big mistake . https://codeforces.com/contest/1538/submission/119263670

Memory limit exceeded doesn't mean your code is wrong. The strings won't fit in some tests. Note that the input can be a first line with := and then 49 lines like x = x + x, so the string will need more than 2**49 bytes, and I believe 256MB to be 2**28 bytes, which is the limit for the problem.

so how could i make improvement to same program and is it possible this was the approach which first came to my mind .Now after posting this comment i saw the editorial and it is using some different approach which i am not able to understand that well .

One thing which i am thinking about is counting substring after every iteration in the string and keep adding the substring which will be added after every iteration.

plase easy write to problem D?i can't understand.

There are some observations.

From those rules above we can find a minimum number of operations, and a maximum number of operations. If k is in between them then ans="Yes".

Sorry, didn't get this part. Why is this true?

EDIT: I took one example and it's clear now.

Can someone please explain the logic behind the solution for F.

Hope this can be of help! :-)

Take l = 1, r = 900.

The lowest digit changes 900-1 = 899 times. See it as 001 -> 900. This is the number of 1's added!

The middle digit will change 90-00 = 90 times. See it as 00 -> 90. If we had l = 60, r = 900 the middle digit will change 30 times, 60 -> 90. This is the 10's place digit changes given the number of 1's added.

The highest digit will change 9-0 = 9 times. See it as 0 -> 9. If we had l = 600, r = 900 the highest digit will change 3 times, 6 -> 9. This is the 100's place digit changes given the number of 1's added.

A far simpler solution for Problem D:

Simply count the total no. of prime factor and how many times they are repeating for example.

48 = 2 * 2 * 2 * 2 * 3 => 2 ^

4* 3 ^1=> solve(48) => 5 (remember I added the power which is 4 and 1)36 = 2 * 2 * 3 * 3 => 2 ^

2* 3 ^2=> solve(36) => 4Maximum time we can divide both no. to make them equal (maxStep) => 4 + 5 = 9

The minimum times we can divide both no. to make them equal is calculated by below condition:-

if both no are not divisible by themselves and are not equal then the minimum step required to divided them and make them equal is 1;

else minimum steps are 2.

below is the code for this particular logic:-

`int minStep = 2;`

`if ((a % b == 0 || b % a == 0) && a != b) minStep = 1;`

Now, if the value of k lies between minStep and maxStep then print "YES" else print "NO";

Here's the code.

Can problem C be solved by (set.upper_bound(r-*it)-set.begin())-(set.lower_bound(l-*it) — set.begin())+1? Ik that set.upper_bound(r-*it)-set.begin() generally gives the index but here its giving me a compilation error.can someone help?

can anyone please explain in problem C this: if (l <= 2 * v[i] && 2 * v[i] <= r) { ans--; }

can anyone explain problem G please, I'm having hard time with it. I do understand that the solution is binary search on answer, but couldn't understand how do we know if some value x can be an answer or not.

I have tried writing solutions in Java for this(I'm not familiar with it, like I am with C++).

Could we have a list of tips&tricks to speed up our solutions? I have had many issues with TLE.

So far I found: - Scanner is not always fast enough to get Accepted - same with Arrays.sort - long is very slow

Experience: I wrote the same code for problem C like in the official solution, but apparently Arrays.sort gives TLE, even after I change the algorithm after sorting to an O(n) instead of O(nlogn).

In problem D, I managed to write a solution to get accepted, but after changing long to int the solution goes from ~1100 ms to ~500ms. Lots of TLE before. If I use int, even unoptimized solutions pass(reads with Scanner, factorization without precomputing primes or even going through all even numbers as possible factors).

Also, my lack of experience with Java showed in other ways: - I had to implement swap of two variables manually(I was not able to find the library version) - I had to implement upper bound and lower bound manually, again I could not find the library version - I do not have a fast, optimized read/write class to copy/paste in my solution(frankly, I don't even know which one would that be, there seem to be many options).

P.S.: Bulleted lists don't seem to work.

Can someone give some hint on how to approach the Number of pairs by two pointer method?

I solved Problem 1538G in constant time complexity Link to solution: https://codeforces.com/contest/1538/submission/119439571

I think we don't need to store the string length in E. It is not used neither in prefix/suffix computation, nor in count computation?

in problem c if we sort the array then aren't we working on a different array then what given in the input . I am bit confused about this please help

Because no matter where we put an element, the number of pairs that we can make with that element won't increase or reduce and addition operation doesn't care about order.

Suppose an original array of 1, 2, 3, 4, 5. Now, 1 can make pairs with 2, 3, 4, 5. (4 elements). Now we move 1 to the middle. 2, 3, 1, 4, 5. Now, 1 can make pairs with 4, 5. (2 elements). But now, 2 and 3 can also make pairs with 1 since it's move to their right (+2 elements) and we get 4 elements total with 1.

It holds true for every element.

got it thanks for clearing my doubt u r great cf community is great

Problem G

"the maximum number of gift sets that Polycarp can make.", I should think about binary search, but I didn`t, what a pity, "maximun of" is a symbol of binary search.

the inequality, we may have this:

(bn-x) / (b-a) <= k <= (y-an) / (b-a) [1]

or, may have this

(an-y) / (a-b) <= k <= (x-bn) / (a-b) [2]

[2] is wrong, and [1] is right.

because, we presume x < y, a < b, k need to be positive

something else is about floor( ) and ceil( ),

Hi folks, I keep getting TLE for my Java O(n*log(n)) solution for problem C: https://codeforces.com/contest/1538/submission/120169883

My solution and its time complexity seem to be in lieu of what the constraints and the tutorial say. Can someone please point me to what I'm missing?

I saw a lot of TLEs for Java for problem C, one accepted solution is https://codeforces.com/contest/1538/submission/120156779

How do you come up with that solution for F?

Problem F solution with simple explanation: https://www.youtube.com/watch?v=8c_dVBCBJSo&t=312s

.

How can I avoid TLE on my solution? I'm pretty sure it's correct otherwise. Would first checking if the number is prime before calculation help? My submission: 120410573

Can anyone explain what floorl and ceill are in the solution of problem G. And is 1.0l used to convert to float like 1LL/0LL is used to convert to long long. Thanks in advance!

Problem F solution with simple explanation: https://www.youtube.com/watch?v=8c_dVBCBJSo

where the wrong in that ??

128818547//D

found it.

I was solving problem C (Number of Pairs). I read the editorial. I had a feeling that some begginers like me may face some problems understanding the "if" part of the code. I used the same approach as discussed in the editorial but my implementation might be a bit easier to understand to logic behind the code. This is my submission. Hope it will be helpful:) https://codeforces.com/contest/1538/submission/132188971

Can Anyone tell me what is my mistake in problem 4 ?? - my solution — 134181698