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

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

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

Passing vector to a function

This link would help you.

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.

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?

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

N/A

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.

learn about lower_bound and upper_bound !! great stuff imo!

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 :)

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

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

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.

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.

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?

buy a new laptop and start with C++

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 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:

$$$1$$$

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

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.

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

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.

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.

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)

119103856

Can any body hack my submission for Problem G ?

I didn't use binary search.

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

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

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!I got a tle in D because of using long long instead of int.

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

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.

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

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'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?

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

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`

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.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.

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!

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.

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.

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 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!

Does anyone know why this (145191446) solution fails for problem G?

I implemented the solution for problem D, but I got WA. Could someone help me figure out where my code breaks?

145908397

In problem G I think Inequalties should be x≥a⋅k+b⋅(n−k) y≥a⋅(n−k)+b⋅k

.

I am facing an issue in D problem 1538D - Another Problem About Dividing Numbers when I use

intthe test case passes but on usingLong Longit givesTLEI don't understand how it is creating such a big difference.PS:

INT: passed test cases i.e.

Accepted(1559ms) 169586396LONG LONG:

TLEon6thtest case (2000ms) 169586970How to calculate

`sum of exponents of prime divisors of a`

in problem D ``