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pikmike's blog

By pikmike, history, 3 months ago, translation, In English,

1334A - Level Statistics

Idea: pikmike

Tutorial
Solution (pikmike)

1334B - Middle Class

Idea: adedalic

Tutorial
Solution (adedalic)

1334C - Circle of Monsters

Idea: BledDest

Tutorial
Solution (Ne0n25)

1334D - Minimum Euler Cycle

Idea: adedalic

Tutorial
Solution (adedalic)

1334E - Divisor Paths

Idea: BledDest

Tutorial
Solution (pikmike)

1334F - Strange Function

Idea: Roms и BledDest

Tutorial
Solution (BledDest)

1334G - Substring Search

Idea: BledDest

Tutorial
Solution (BledDest)
 
 
 
 
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3 months ago, # |
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Can someone give a proof why the path $$$x\rightarrow \gcd(x,y)\rightarrow y$$$ is shortest in problem E?

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    3 months ago, # ^ |
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    Anyway, interesting problem set although with some difficulties on system test.

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    3 months ago, # ^ |
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    Whenever you move along an edge, some numbers are either added to the set of divisors of the current vertex, or removed from it (and the cost of the edge is the number of affected elements). So, ideally you want your path to have length equal to $$$d(x) + d(y) - 2d(gcd(x, y))$$$, where $$$d(i)$$$ is the number of divisors of $$$i$$$.

    If you go through $$$gcd(x, y)$$$, then all divisors of $$$x$$$ which are not divisors of $$$y$$$ are removed, then all remaining divisors of $$$d(y)$$$ are added, so the path has length $$$d(x) + d(y) - 2d(gcd(x, y))$$$, which is optimal. But if you don't go through $$$gcd(x, y)$$$, then you either go through one of its divisors (thus removing $$$gcd(x, y)$$$ from the set, so the length is greater than optimal), or go through some number which is neither a divisor of $$$x$$$ nor a divisor of $$$y$$$.

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      3 months ago, # ^ |
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      How it compares to $$$x\rightarrow lcm(x,y)\rightarrow y$$$? The length is $$$2d(lcm(x,y))-d(x)-d(y)$$$, which for me is not easy to tell if it's shorter or longer.

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        3 months ago, # ^ |
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        Going through $$$lcm(x, y)$$$ obviously adds $$$lcm(x, y)$$$ to the list of divisors, and we have to remove it afterwards, that's why it's suboptimal.

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    3 months ago, # ^ |
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    Reformulate the query problem: we have two numbers $$$a$$$ and $$$b$$$ and want to get $$$b$$$ from $$$a$$$. We can do it only by modifying or dividing a by prime number. Every change costs difference between each number divisor's count.

    Calculate change's cost

    We could represent any number $$$n$$$ like $$$n=p_1^{l_1}p_2^{l_2}...p_n^{l_n}$$$ ($$$l_i$$$ can be 0, it means $$$p_i$$$ isn't a divisor of $$$n$$$)

    Count of divisors, let's name it $$$d(n)$$$, it will be $$$d(n)$$$ = $$$(l_1+1)(l_2+1)...(l_n+1)$$$. When we change $$$a$$$ to $$$b$$$ by modifying we waste $$$d(b) - d(a)$$$.

    $$$a=p_1^{l_1}p_2^{l_2}...p_i^{l_i}...p_n^{l_n}$$$

    $$$b=p_1^{l_1}p_2^{l_2}...p_i^{l_i+1}...p_n^{l_n}$$$

    As we have seen, we have modified by $$$p_i$$$. Calculate divisor's difference: $$$d(b) - d(a) = (l_1+1)(l_2+1)...(l_{i-1}+1)(l_{i+1}+1)... (l_n+1) = \frac{d(a)}{(l_i+1)}$$$ For dividing it works in opposite way: we waste $$$d(a) - d(b)$$$.

    $$$Observation 1$$$: if we during our changes have modified and divided by some prime number we could not do these two operations.

    $$$Proof$$$. Consider changes between these two changes. $$$x➔xq➔x_1q➔...➔x_nq➔x_n$$$, if we delete first and latest operations we have: $$$x➔x_1➔...➔x_n$$$.

    Compare costing changes $$$x_iq➔x_{i+1}q$$$ and $$$x_i➔x_{i+1}$$$ by modifying by $$$p≠q$$$.

    $$$d(x_{i+1}q)-d(x_iq) = \frac{d(x_iq)}{l+1}$$$

    $$$d(x_{i+1})-d(x_i) = \frac{d(x_i)}{l+1}$$$

    $$$l$$$ is a power of $$$p$$$ in canonical representation of number $$$x_i$$$ and $$$x_iq$$$, $$$l$$$ is the same because $$$l$$$ only changes if $$$p=q$$$ $$$\frac{d(x_iq)}{l+1} > \frac{d(x_i)}{l+1}$$$, so better do in second way.

    $$$Observation 2:$$$ it is better firstly to do dividing by prime numbers, which aren't divisors of $$$b$$$, then do modifying until we get $$$b$$$.

    $$$Proof.$$$ Consider the simplest example: $$$a = pc$$$, $$$b=qc$$$, where $$$p$$$ and $$$q$$$ are prime numbers. Notice $$$c = gcd(a,b)$$$. If we firstly modify and then divide our changes will be $$$a➔pcq➔b$$$. We waste $$$2*d(pcq) - d(a) - d(b)$$$.

    If we firstly divide and then modify our changes will be $$$a➔c➔b$$$. We waste $$$d(a) + d(b) + 2d(c)$$$.

    Subtract second equation from first one:

    $$$2d(pcq) - 2d(a) - 2d(b) + 2d(c)$$$

    $$$(d(pcq) - d(a)) - (d(b) - d(c))$$$, substitute $$$a=pc$$$ and $$$b=qc$$$

    $$$(d(pcq) - d(pc)) - (d(qc) - d(c))$$$

    $$$d(pcq) - d(pc) = \frac{d(pc)}{l_q+1}$$$, $$$d(qc) - d(c) = \frac{d(c)}{l_q+1}$$$, where $$$l_q$$$ is a power of $$$q$$$ in canonical representation of number $$$pcq$$$ and $$$qc$$$.

    Numbers $$$pc$$$ and $$$c$$$ have same $$$l_q$$$, so $$$d(pcq) - d(pc) - d(qc) + q(c) = \frac{d(pc) - d(c)}{l_q+1}$$$ and it is bigger then zero, so $$$d(a) + d(b) + 2d(c)$$$ is less then $$$2d(pcq) - d(a) - d(b)$$$, so better firstly to divide and then modify.

    $$$UPD: $$$

    $$$Obseravation3:$$$ if we during our change have divided and modified by some number we could not do these two operations.

    $$$Proof.$$$

    Consider changes between these two change: $$$xq➔x➔x_1➔...➔x_n➔x_nq$$$. If we detele first and last operations we have: $$$x_q➔x_1q➔x_2q➔...➔x_n➔x_nq$$$. Name them $$$changing$$$ $$$paths$$$.

    Using $$$obseravation 2$$$ we know to get optimal $$$changing$$$ $$$path$$$ firstly we do dividing, then do modifying. Our $$$changing$$$ $$$paths$$$ can be represented like:

    $$$xq➔x➔x_1➔...➔x_k➔..➔x_n➔x_nq$$$

    and $$$x_q➔x_1q➔x_2q➔...➔x_kq➔..➔x_{n}q➔x_nq$$$.

    We do only dividing to get $$$x_kq$$$ and $$$x_k$$$ and then only modifying.

    Calculate sum of changing costs in first $$$changing$$$ $$$path$$$:

    $$$(d(xq)-d(x))+(d(x)-d(x_1))+...+(d(x_{k-1})-d(x_k))+(d(x_{k+1})-d(x_k))+$$$ $$$+...+(d(x_n) -d(x_n-1))+(d(x_nq)-d(x_n))=$$$ $$$\frac{d(x)}{l+1}+\frac{d(x_1)}{l_1+1}+...+\frac{d(x_{k-1})}{l_{k-1}+1}+\frac{2d(x_k)}{l_k+1}+\frac{d(x_{k+1}}{l_{k+1}}+...+\frac{d(x_n-1)}{l_{n-1}}+ \frac{d(x_n)}{l+1}= \frac{d(x)}{l+1}+\frac{d(x_n)}{l+1}+\sum\limits_{i=1}^k\frac{d(x_i)}{l_i+1}+\sum\limits_{i=k}^{n-1}\frac{d(x_i)}{l_i+1}$$$

    Notice in first and last fractions the denominator is the same because when we change $$$xq➔x$$$ and $$$x_n➔x_nq$$$ we divide and modify by same prime number $$$q$$$. In general, when we change $$$x_i➔xq_i$$$ or $$$xq_i➔x_i$$$ we waste $$$\frac{d(x_i)}{l+1}$$$.

    Calculate sum of chaging costs in second $$$changing$$$ $$$path$$$: $$$(d(xq)-d(x_1q)) +...+(d(x_{k-1}q) - d(x_kq)) + (d(x_{k+1}q) - d(x_kq)) +...+(d(x_nq) -d(x_{n-1}q))=$$$ $$$\sum\limits_{i=1}^k\frac{d(x_iq)}{l_i+1}+\sum\limits_{i=k}^{n-1}\frac{d(x_iq)}{l_i+1}$$$

    Subtact first equation from second:

    $$$\frac{d(x)}{l+1}+\frac{d(x_n)}{l+1}-(\sum\limits_{i=1}^k\frac{d(x_iq)}{l_i+1}-\sum\limits_{i=1}^k\frac{d(x_i)}{l_i+1})-(\sum\limits_{i=k}^{n-1}\frac{d(x_iq)}{l_i+1}-\sum\limits_{i=k}^{n-1}\frac{d(x_i)}{l_i+1})$$$

    Since $$$d(x_iq)-d(x_i)=\frac{d(x_i)}{l+1}$$$, so we get $$$\frac{d(x)}{l+1}+\frac{d(x_n)}{l+1}-\sum\limits_{i=1}^k\frac{d(x_i)}{(l_i+1)(l+1)}-\sum\limits_{i=k}^{n-1}\frac{d(x_i)}{(l_i+1)(l+1)}$$$, by modifying by $$$l+1$$$ we get: $$$d(x)+d(x_n)-\sum\limits_{i=1}^k\frac{d(x_i)}{l_i+1}-\sum\limits_{i=k}^{n-1}\frac{d(x_i)}{l_i+1}$$$

    Separately calculate $$$d(x)-\sum\limits_{i=1}^k\frac{d(x_i)}{l_i+1}$$$ and $$$d(x_n)-\sum\limits_{i=k}^{n-1}\frac{d(x_i)}{l_i+1}$$$:

    Since $$$d(x_i)-d(x_{i-1}) = \frac{x_i}{l_i+1}$$$, so $$$d(x_i) = \frac{d(x_{i+1})(l_i+2)}{l_i+1}$$$

    $$$d(x)-\sum\limits_{i=1}^k\frac{d(x_i)}{l_i+1} = d(x)-\frac{d(x_1)}{l_1+1}-\sum\limits_{i=2}^k\frac{d(x_i)}{l_i+1}=\frac{d(x_1)(l_1+2)}{l_1+1}-\frac{d(x_1)}{l_1+1}-\sum\limits_{i=2}^k\frac{d(x_i)}{l_i+1}=$$$ $$$\frac{d(x_1)(l_i+1)}{l_i+1}$$$-$$$\sum\limits_{i=2}^k\frac{d(x_i)}{l_i+1}=$$$ $$$d(x_1)-\sum\limits_{i=2}^k\frac{d(x_i)}{l_i+1}$$$

    If we go on in the end we will get $$$d(x_k)$$$

    Similarly we calculate $$$d(x_n)-\sum\limits_{i=k}^{n-1}\frac{d(x_i)}{l_i+1}=$$$ $$$\frac{d(x_n)(l_{n-1}+2)}{l_{n-1}+1}-\frac{d(x_{n-1})}{l_{n-1}+1}-\sum\limits_{i=k}^{n-2}\frac{d(x_i)}{l_i+1}=$$$ $$$d(x_{n-1}-\sum\limits_{i=k}^{n-2}\frac{d(x_i)}{l_{i+1}}$$$ Finally we get $$$d(x_k)$$$

    In this way, the total sum will be $$$2d(x_k)$$$, $$$2d(x_k)$$$ is bigger than zero, so first equation is bigger then second one, so it isn't optimal to divide and then modify by some prime number $$$q$$$.

    Using first and third observation we know we only have to divide by prime number, which aren't divisors of $$$b$$$, and then do modify by prime numbers, which arent't divisors of $$$a$$$.

    Using second observation we know that firstly we do dividing then we do modifying.

    As we have seen, $$$changing$$$ $$$path$$$ $$$x➔gcd(x,y)➔y$$$ is only one, which fits the requirements, so it is optimal path.

    P.S. I think it could be proved much easier :)

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3 months ago, # |
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Nice problems and a great editorial!

In question C- Circle of Monsters, I used pair to store energy for each monster. After that I used 2 for loops to calculate the minimum bullets required. So, overall time complexity is 2*n, but still I got TLE for Test case 3. I don't know why this happened, am I wrong in calculating complexity or there is some other reason? Code

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    3 months ago, # ^ |
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    use scanf and printf

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      3 months ago, # ^ |
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      It worked. Thanks! But the constraints were pretty normal. I still can't understand why scanf and printf worked for this problem.

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        3 months ago, # ^ |
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        below lines is a bit different in your code, can you try with these? maybe sync_with_stdio() is not same as sync_with_stdio(0) though I haven't checked the docs. you should be always okay with cin, cout without endl

          ios_base::sync_with_stdio(0);
          cin.tie(0);
        
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          3 months ago, # ^ |
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          or you can define endl '\n'

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    3 months ago, # ^ |
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    Submitted your code again Use the fastio commands properly.

    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

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    3 months ago, # ^ |
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    i am unable to understand the problem plz someone give 5 mins and explain thoroughly

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3 months ago, # |
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Thank you for nice problems and fast editorial, pikmike, Adedalic, Roms, BledDest!!!

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3 months ago, # |
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Can anyone explain C more clearl?

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    3 months ago, # ^ |
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    have a look to my solution https://codeforces.com/contest/1334/submission/76178087

    What I did was calculated the minimum number of bullets required for each position, if started shooting from monster at that position.

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      3 months ago, # ^ |
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      How did u do that within given constraints??

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        3 months ago, # ^ |
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        My solution was this:

        When a monster explodes, there might be some "extra" work that you have to put in to kill the next monster along. This value is (health of next monster) - (bullets released by previous monster). Where this value is negative, then you have to put in no extra work (the extra bullets you need to shoot is 0). This means that you could use the formula max(0, a[i] - b[i-1]) after you account for the wrapping-around behaviour.

        Consider the sum, T, of all the extra work that you have to put in for each monster. You can find this by calculating it for every single monster, in O(n), and then adding together all of the values you get.

        Then, if you decide to start at a particular monster, this means that the extra work for that monster should not be considered, because the previous monster did not explode. Instead, you need to shoot the same number of bullets as the health of that monster. This means that for any given monster, you can find the number of bullets needed using the formula (T - (extra work for that monster) + (health of that monster)). We showed above how to quickly work out the extra work for that monster.

        This means that the solution is O(n) for finding T, and then O(n) for finding how many bullets are needed for each monster, using that value of T.

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        3 months ago, # ^ |
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        I maintained an array: ans[]. ans[i] has the minimum number of bullets req for the task if we start shooting from ith(0 indexing) monster. I precalculated the ans[0] value. Then for any i we can calculate any ans[i] using ans[i-1] carefully handling the exclusion and inclusion of values. Refer to my code for more, easy to understand.

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      3 months ago, # ^ |
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      Didn't Binary Search would work for Problem C ??

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    3 months ago, # ^ |
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    Think it that way: It there where no explosions, we would have to fire one shoot per lifepoint. So, we can earn points by usings explosions.

    How much can we earn? The maximum earnings per monster is limited as $$$min(a[i], b[i-1])$$$ This is the power of the explosion, or the total life points of the affected monster, whichever is less.

    Since the monsters are in a circle, we can kill on after the other, the explosion damage will be maximum possible on every explosion.

    But we have to start somewhere in the circle. The monster where we start does not get any demage by an explosion. As the starting point we use the monster where we would earn the least points, this is the one with minimum $$$min(a[i],b[i-1])$$$

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      3 months ago, # ^ |
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      will please explain a little more....how is maximum earnings per monster is limited as min(a[i],b[i−1])

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        3 months ago, # ^ |
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        By "earning bullets", we're saying that the previous monster shoots off bullets for free when it explodes, instead of us having to do it. This is very good for us because we want to minimise the number of bullets we shoot, so letting the exploding monsters shoot for us is helpful.

        Let's say that the previous monster gives off 5 bullets when it dies, but the next monster has 3 health points. Ordinarily, we would need to shoot 3 bullets to kill the next monster. However, we "earn" those 3 bullets because the previous monster takes care of them for us. Thus, when b[i-1] > a[i], we "earn" a[i] bullets because we didn't have to shoot them.

        If the previous monster instead only gives off 1 bullet, then when it dies it reduces the 3 health points of the next monster to just 2 health points. We would have needed to shoot 3 bullets to kill the next monster, but now we only need to shoot 2 bullets. Thus, the previous monster has "earned" us 1 bullet by exploding. Thus, when b[i-1] < a[i], we "earn" b[i-1] bullets.

        This means that the formula min(a[i], b[i-1]) is just telling us the bullets that we don't have to shoot if we rely on the previous monster exploding instead.

        You might also choose to think of this problem as how many bullets do you still have to shoot despite all the help from monsters exploding. And this is very similar: max(0, a[i] - b[i - 1])

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      3 months ago, # ^ |
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      great explaination

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    3 months ago, # ^ |
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    I have a solution like the following: https://codeforces.com/contest/1334/submission/76283580

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Why I got tle in this https://codeforces.com/contest/1334/submission/76133446 It is nlog n only!!!

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    3 months ago, # ^ |
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    Use ArrayList instead of long[]

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    3 months ago, # ^ |
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    Maybe problem in not fast Java io.

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    3 months ago, # ^ |
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    Arrays.sort() can get up to n^4 time in weird cases like for nearly sorted raw type arrays (e.g long[]). To combat this, use Long[] or an ArrayList instead when you plan to sort the contents of a long[] array. I guarantee this change will fix your solution.

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      3 months ago, # ^ |
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      thanks

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      3 months ago, # ^ |
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      I faced this exact same problem, and only after the contest did I switch the long[] to a Long[] and finally pass all test cases.

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    3 months ago, # ^ |
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    I used dynamic programming for C. I haven't thought a greedy would works. Can someone explains please, why I got WA, on this : https://codeforces.com/contest/1334/submission/76184081

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3 months ago, # |
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Can someone explain why were there so many successful hacks on problem B that result in the program getting Time Limit Exceeded? How do the hacks work? Is it because of a flaw in the programming language's sort function?

Also, are the weak tests for problem A intentional?

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    3 months ago, # ^ |
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    Hi, I did many of the hacks for B. You are correct. In java, sorting an array of primitives uses quicksort. This can be exploited by creating an anti-quicksort hack to force the sort to run in O(N^2).

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3 months ago, # |
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My solution this failed the system test case giving tle verdict. But when I submitted the same code this after the contest it was accepted. Please anyone can help me what could have happened wrong.

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3 months ago, # |
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You can simplify the solution of problem E by using the fact that the graph is somewhat self isomorphic. So if you want to count the paths from $$$x$$$ to $$$y$$$ you can also count the paths from $$$\frac{x}{\gcd(x,y)}$$$ to $$$\frac{y}{\gcd(x,y)}$$$. Each shortest path will now go through $$$1$$$, therefore you only need the number of paths from $$$1$$$ to $$$d$$$ for each divisor $$$d$$$ of $$$D$$$. This can be calculated before with a simple do. My implementation: 76229493

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Really liked problem D!

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3 months ago, # |
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Can anyone explain problem D??!

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    3 months ago, # ^ |
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    First try to write the whole lexicographically smallest cycle of the whole graph.

    For simplicity I will take a complete graph of size 4: ( the graph below should be bi-directional )

    The lexicographically smallest cycle of this graph is:

    1 2 1 3 1 4 2 3 2 4 3 4 1

    As you can see, there's a pattern here ( 1 2 1 3 1 4 ) ( 2 3 2 4 ) ( 3 4 ) ( 1 ) which can be modeled by this for loop:

    for i in 1..n-1
        for j in i+1 .. n
            print(i + " " + j + " ")
    print(1)
    

    Still this is O(n^2), to make it O(n) we need the element close enauph to l and complete the pattern and write the element between l and r, to so do we can simplify the search using this observation: in the pattern the first group starts with 1 and it's of size 2*(n-1), the second group starts with 2 and it's of size 2*(n-2) and so on, we can use this fact to loop through i and find and i such that 2*(n-i) is closest to l , then just print the elements between l and r.( don't forgot the 1 at the end).

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    3 months ago, # ^ |
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    Suppose n = 5. for now forget about l and r. Then the lexicographically smallest list will be:
    1 2 1 3 1 4 1 5 (1-5)
    2 3 2 4 2 5 (2-5)
    3 4 3 5 (3-5)
    4 5 (4-5)
    1.

    And what if n = 6? If you guess it right, the list will be:
    1 2 1 3 1 4 1 5 1 6 (1-6)
    2 3 2 4 2 5 2 6 (2-6)
    3 4 3 5 3 6 (3-6)
    4 5 4 6 (4-6)
    5 6 (5-6)
    1

    Try it with some other graph with different number of vertices. Keep sum of every list above in a prefix array. From there it will be easy to determine where l starts and where r ends.

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      3 months ago, # ^ |
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      I got that too during practice. But the ques is, we can't store anything because of the large value of l and r, how can we know what element we would get in specific index without storing, I can't get any mathemetical formula

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        3 months ago, # ^ |
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        Look at my example above. When n = 5, length of the first list is 8. Sec list is 6, third is 4 and so on. We can store only the length of each sequence in an array. We don't need to store the whole sequencs. Got it now? Think for sometime about what I just said.

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Can someone please explain the solution of D? I am not able to understand editorial clearly..

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    3 months ago, # ^ |
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    First of all lets try to construct the lexographically minimum cycle in the graph:

    We have to choose a starting node, we will choose node 1 otherwise any other other as starting node makes the cycle lexographically greater than {1 followed by some other nodes}.

    For example : {3 followed by some other nodes} > {1 followed by some other nodes}.

    Now we have chosen the starting node as node 1. Where should we go now? Given a complete graph you could go to any other node, so we choose 2.

    Because again {1 follow by 2 followed by some other nodes} is less than any other possibility.

    now we have 1->2, now where? Again choose 1(for similar reasons). so we have 1->2->1.

    now we are back at node 1 but now we can not choose to go to node 2 again(edge 1->2 is already traversed once). So we are forced to go to some node j ( j > 2).

    Again for similar reasons explained above, you will choose to go to node 3.

    1->2->1->3. Where now? from 3 you may go to any other node.

    So again go to 1 : 1->2->1->3->1 So if you see the pattern now, then we have :

    1->2->1->3->1 ... 1->n-1->1->n Now where? If I go back to 1 then I am trapped because I have already used every outward edge from 1. So I am forced to go to node 2 instead of 1.

    1->2->1->3->1 ... 1->n-1->1->n->2->3->2->4->2 ... 2->n-1->2->n

    So we get the lexographically smallest cycle as below : {1->2->1->3->1 ... 1->n-1->1}->n->{2->3->2->4->2 ... 2->n-1->2}->n->{3->4->3...->3->n-1->3}->n.....n->{n-1}->n->1.

    The thing inside the braces(call it Bi) is simply {i->i+1->i->i+2 and so on} and after that back to node n.

    Once done we complete the cycle by traversing the edge n->1.

    Now choose in which brace from 1 to N-1 your L falls and generate that brace content to print. Ofcourse L to R may span over more than one braces or no braces at all(last two or L falling right at the end of some brace).

    Hope this helps :)

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3 months ago, # |
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Please , have a look at my solution : https://codeforces.com/contest/1334/submission/76259835 Can anyone please help me out to get why I'm getting wrong answer . Thank You

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    3 months ago, # ^ |
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    I didn't understand what was your idea, but I think you should have a look at my code. Idea: lets sort our array, if at some position i we have that a[i] + a[i + 1] + ... + a[n — 2] + a[n — 1] >= x * (n — i) then (a[i] + a[i + 1] + ... + a[n — 2] + a[n — 1]) / (n — i) >= x. Thats why all this people are reach. Thats why we want to maximize n — i (minimize i) and loop goes while all people at suffix are reach.

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      3 months ago, # ^ |
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      Thank you for replying me back . I got your answer but I need to know the test case where my code fails to pass the test case . This code https://codeforces.com/contest/1334/submission/76145656 is almost same as mine but got the AC verdict . Please somebody help me out by finding such a case where i got stuck .

      Thank You

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        3 months ago, # ^ |
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        Try this:

        1
        5 4
        2 3 3 4 6
        
        Correct answer: 4
        Your answer: 3
        
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          3 months ago, # ^ |
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          Thank You very much . hope u will succeed in your life . But how could you arrive at this test case . Is there a way to find out such type of test cases in code forces website or just by hit and trial and practice .

          Thank You

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3 months ago, # |
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E was not only an interesting problem But also a fresh one ! Enjoyed it.

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3 months ago, # |
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Cleaner solution to problem F, without any kind of data structure:

You can add $$$0$$$ and $$$n+1$$$ to the beginning and end of the lists $$$a$$$ and $$$b$$$ to complete them. Transform the sequence $$$a_i$$$ into a simpler sequence to solve the problem. If $$$a_i = b_j$$$, then you rewrite $$$a_i := 2j$$$. However, if $$$b_j < a_i < b_{j+1}$$$, rewrite $$$a_i := 2j+1$$$ (throughout the article := means assign).

Thanks to this, we need to form a increasing (two by two) sequence of numbers from $$$a$$$ starting at $$$0$$$ and ending at $$$2m$$$. To do so, we will use the following idea:

  • Once we've decided which position $$$a_i = 2j$$$ will be chosen to stay, we must remove all the elements equal to $$$2j$$$ and $$$2j+1$$$ before $$$i$$$ (thanks to the compression, this can be done with prefix sums in $$$\log n$$$ without doing it for all elements $$$x, b_{j-1} < x < b_j$$$). This will ensure the $$$f$$$ of the chosen sequence is $$$b$$$.
  • Additionally, we would like to remove the positions with negative $$$p$$$. Therefore, what we'll do is only count values with positive $$$p$$$ in our prefix sums, and add to the solution the negative values of all positions which weren't used.

Now. Let's put it into a dp:

  • $$$dp(i)$$$ for $$$a_i = 2j$$$ is the least ammount of coins you have to pay in order to choose the value $$$b_j$$$ from $$$[0, i]$$$ in the final sequence.

In general: dp(i) := dp(x) + prefix-sum of values 2j in [0, i) + prefix-sum of values 2j+1 on [0, i) + negatives-sum on (x,i), where $$$x$$$ is the last position $$$a_x = 2j-2$$$.

If we forced to choose the position $$$i$$$. To take into account the previos positions:

dp(i) := min(dp(i), dp(x) + negatives-sum on (x, i], where $$$x$$$ is the previous position with $$$a_x = 2j$$$.

Using lower_bound to get the prefix sums and the indices $$$x$$$ makes the algorithm run in $$$\mathcal{O}(n \log n)$$$. However, using an array of size $$$n$$$ to compress the sequence (given that $$$b$$$ is sorted) and to keep the last position of each element when traversing from $$$i = 0$$$ to $$$i = n$$$ we can do it in $$$\mathcal{O}(n)$$$.

I wrote the $$$n \log n$$$ solution shortly after the contest: https://codeforces.com/contest/1334/submission/76218976

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    3 months ago, # ^ |
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    Do you mean you can write an $$$O(n)$$$ solution or only a part of it?

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    3 months ago, # ^ |
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    Cool explanation. Can you take a look at my code.I used dp[i][j] as minimum number of coins required to generate first j elements of array B using only the first i elements of A. Transitions are quiet straight forward. If next element of A is the next required element of B and I am considering that element dp[i+1][j+1]=min(dp[i+1][j+1],dp[i][j]). If I dont consider the next element ie I want to delete it,dp[i][j+1]=min(dp[i][j+1],dp[i][j]+a[j]) My answer would be dp[m][n]. I dont understand where I am going wrong. https://codeforces.com/contest/1334/submission/76300349 Thanks in advance:)

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      3 months ago, # ^ |
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      Will that fit into the time limit?

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        3 months ago, # ^ |
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        Dont know,but it is still failing in the pretests.All I need is that logic correct?

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3 months ago, # |
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1334C — Circle of Monsters BledDest does this problem can be solved using dynamic programming if we store that how much bullets it will take to destroy all monster start from first and when we go further remove the bullets used by previous one and reduce the damage caused the previous monster on blast and count bullets for each of this individually and minimum of it is answer.

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3 months ago, # |
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Can someone explain why C has to be done in order? Like why is it most efficient to shoot the monsters in the order that they are standing in?

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    3 months ago, # ^ |
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    If we remove the monster at a position i before removing the monster at position i-1 then we won't be able to utilize the explosion of the monster at position i-1. And since, we want to maximize the utilisation of explosions so as to reduce the number of bullets used, we remove the monsters in the order they are standing. In this way, only the explosion of the monster removed last will be wasted.

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3 months ago, # |
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Could someone explain this line to me from the tutorial for Problem C: mn = min(mn, val); Is this just the minimum damage dealt to any monster? If so, how is it related to "ans".

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    3 months ago, # ^ |
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    Because, you can imagine someone kill some player, and then he will damage the next monster, maybe he will kill him or not, but the amount of bullets needed to kill the monster will be less than the amount of bullets needed to kill it it means that: a_i > max(0, a_i-b_(i-1)) Therefore you need to start the cycle at some point, and the point or index where you need to start shooting is the one that is the 'i' such that ( a_i — max(0, a_i-(b_(i-1))) < ( a_k — max(0, a_k-(b_(k-1)) ) where k is another index in the array.

    Hope be useful.

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      3 months ago, # ^ |
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      Ok, this makes sense. However, I am still a bit confused about this line of code in the solution: mn = min(mn, val);

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3 months ago, # |
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Nice contest!

btw in problem E, why is “You can remove them in any order because the length of the path is always the same” true?

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    3 months ago, # ^ |
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    Every vertex of the graph stands for a list of primefactors. Each edge removes or adds one such primefactor depending on direction of traversal.

    So, to find the path from one list of primefactor to another one, we need to remove all primefactors which are in the first list but not in the second, and add all primefactors wich are in the second list but not in the first one.

    The order of removals and addings simply does not matter.

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3 months ago, # |
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Where can I find more information about string matching and FFT?, I still need more examples or info about what a cycle is, and more, thanks.

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3 months ago, # |
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I was able to solve problem F using straight forward DP O(NM) without any data structure but with simple heuristic (update only useful DP states). https://codeforces.com/contest/1334/submission/76289715

Is it hackable? (already hacked TL).

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3 months ago, # |
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Can anyone explain why I am getting TLE for this solution https://codeforces.com/contest/1334/submission/76287771 even for O(n) Complexity. The Problem is https://codeforces.com/contest/1334/problem/C (C. Circle of Monsters)

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3 months ago, # |
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Can someone explain me editorial solution of problem C?

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    3 months ago, # ^ |
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    Update: I understood solution. Anyone who still wants an explanation can ask.

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      3 months ago, # ^ |
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      explain please . The more I think about the problem , more lost this editorial sounds to me

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        3 months ago, # ^ |
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        Since I too faced some problem initially, let me try to explain now.

        In simple words add all bullets needed to kill a monster and subtract all explosion values of previous monster. Do this for all possible pairs, Now the best part, since the last monster to explode will not contribute to minimizing bullets(because all others monsters would be dead by then) hence we choose to select a monster with least contribution i.e min(Bullets needed for next monster,value of explosion(B)). BUT we already subtracted it's explosion value in above for loop so let's add it again.( because like I said last monster will not contribute to minimizing bullets)

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          3 months ago, # ^ |
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          let's take the editorial example :

          lets iterate

          i=0 ---> bullets = 7

          i=1 ---> bullets = (7+2)-min(15,2) = 7

          i=2 ---> bullets = (7+5)-min(14,5) = 7

          i=3 ---> bullets = (7+7)-min(3,7) = 11

          now let's see who can be potential last monster. lets see contribution of each and every monster. monster 0 contributed : min(15,2) = 2 monster 1 contributed : min(14,5) = 5 monster 2 contributed : min(3,7) = 3 so least contribution by monster 0 is least (2) so we will make it the last monster. :) hope that helps.

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3 months ago, # |
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Can someone help me where I went wrong? https://codeforces.com/contest/1334/submission/76150692

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3 months ago, # |
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I would like to know the code of solving f with O (nm) method

Who can show me code .

Thanks

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3 months ago, # |
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It looks that I have had a simpler solution for the problem G. I write it down here.

For each $$$d$$$, compute $$$\sum_{i = 0}^{n - 1} (t_{d + i} - s_i) (t_{d + i} - p(s_i)) = \sum_{i = 0}^{n - 1} t_{d + i}^2 + \sum_{i = 0}^{n - 1} s_i p(s_i) - \sum_{i = 0}^{n - 1} t_{d + i}(s_i + p(s_i))$$$, where the last term can be done with NNT.

Suppose that we do the arithmetic under a prime field of order $$$p$$$. If we randomly sample $$$26$$$ elements in $$$F_p$$$ to be a, b, ..., z, the computed sum will be zero with probability less than $$$\frac{1}{p}$$$ if not all $$$(t_{d + i} - s_i) (t_{d + i} - p(s_i)) = 0$$$.

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3 months ago, # |
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In problem G, consider this case,p1=2,p2=14,....s is aabbb and t is babbb.Is t an occurence of s? Thanks in advance:)

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3 months ago, # |
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Is there a simpler explanation for problem C?

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    3 months ago, # ^ |
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    There is one monster you need to shoot first. This monster will not be affected by the explosion of the previous monster, all other monsters will be affected by the explosion of the previous ones. So you just calculate the monster, for which the explosion will save you the least amount of shots and then simulate the shooting starting from this monster.

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    3 months ago, # ^ |
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    Since I too faced some problem initially, let me try to explain now.

    In simple words add all bullets needed to kill a monster and subtract all explosion values of previous monster. Do this for all possible pairs, Now the best part, since the last monster to explode will not contribute to minimizing bullets(because all others monsters would be dead by then) hence we choose to select a monster with least contribution i.e min(Bullets needed for next monster,value of explosion(B)). BUT we already subtracted it's explosion value in above for loop so let's add it again.( because like I said last monster will not contribute to minimizing bullets)

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3 months ago, # |
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Hi, I am unable to understand the problem c solution. Can anyone please help me understand it.

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    3 months ago, # ^ |
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    try to think the editorial again and again! I finally understood it! Suppose the case are:(i make it different to clear it to you)

    7 15,

    2 1,

    5 3 :) see me.

    here if i fire the smallest one, then result would be 2 + (5-1) + (7-3) = 2 + 4 + 4 = 10

    but not the right ans. because shooting from 5 in this case gives least ans. :) 5 + (7-3)= 5+4=9 So shooting from smallest value doesnt make sense.

    Editorial says:-

    suppose everyone is damaging shooting everyone in the circle, which one gets the least bullet to be fired.

    here 5 is the one. because 2 is damaged by 7's 15, but 5 isnt. 5 is reduced by 2's one(accoring to my test case), then 5 becomes 4, so if you start shooting from 5 you get the least answer.perhaps this is what seems to be in the editorial formula

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    3 months ago, # ^ |
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    lets iterate

    i=0 ---> bullets = 7 i=1 ---> bullets = (7+2)-min(15,2) = 7 i=2 ---> bullets = (7+5)-min(14,5) = 7 i=3 ---> bullets = (7+7)-min(3,7) = 11

    now let's see who can be potential last monster. lets see contribution of each and every monster. monster 0 contributed : min(15,2) = 2 monster 1 contributed : min(14,5) = 5 monster 2 contributed : min(3,7) = 3 so least contribution by monster 0 is least (2) so we will make it the last monster. :) hope that helps.

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      3 months ago, # ^ |
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      so if we take the following example:

      • 7 15 — (7-3 = 4)
      • 2 14 — (2-15 = 0, cause it should be >= 0 i suppose)
      • 5 3 — (5-14 = 0)

      then we should consider the (2 14) pair or (5 3) pair to be the last one. Then i choose (5 3) to be the last one. And then continue to model the killings like: kill (7 15) pair, which uses (7) bullets, kill (2 14) pair, which uses (0) bullets, and finally kill (5 1) which uses (0) bullets.

      In total thus using (7 + 0 + 0) bullets.....plz correct where i'm wrong

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        3 months ago, # ^ |
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        lets iterate

        i=0 ---> bullets = 7

        i=1 ---> bullets = (7+2)-min(15,2) = 7

        i=2 ---> bullets = (7+5)-min(14,5) = 7

        i=3 ---> bullets = (7+7)-min(3,7) = 11

        now let's see who can be potential last monster. lets see contribution of each and every monster. monster 0 contributed : min(15,2) = 2 monster 1 contributed : min(14,5) = 5 monster 2 contributed : min(3,7) = 3 so least contribution by monster 0 is least (2) so we will make it the last monster. :) hope that helps.

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3 months ago, # |
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Problem C logic is sooo legit! I mean a slight difference of thinking takes it wrong result . :)

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3 months ago, # |
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A part of proof for problem E

We will represent numbers as tuple.

$$$x=(a,b,c,d,...)$$$ means $$$x=2^a.3^b.5^c.7^d....$$$.

Let $$$(a,b,c,d,e,...)$$$ and $$$(a',b',c',d',e',...)$$$ be the tuple representations of x and y respectively.

In one step we can perform two types of operations:

(1) $$$(a,b,c,d,e,...)$$$ becomes $$$(a+1,b,c,d,e,...)$$$

(exactly one number among a,b,c,... can be increased by 1, for the sake of clarity a is increased by 1)

from now on we will call it increasing operation.

(2) $$$(a,b,c,d,e,...)$$$ becomes $$$(a-1,b,c,d,e,...)$$$ from now on we will call it decreasing operation.

The cost of both operations is $$$(b+1)(c+1)(d+1)(e+1)...$$$

Our goal is to go from $$$(a,b,c,d,e,...)$$$ to $$$(a',b',c',d',e',...)$$$ using such operations

Some pairs say $$$a,a'$$$ satisfy $$$a<a'$$$ while some others say $$$b,b'$$$ satisfy $$$b>b'$$$ and while some others say $$$c,c'$$$ satisfy $$$c=c'$$$.

Claim 1:

No operations must be performed on pairs of type $$$c,c'$$$ such that $$$c=c'$$$.

Proof:

If there are operations performed on $$$c$$$ removing all such operations does not increase the cost.

Claim 2:

All decreasing operations must be performed before increasing operations.

Proof:

Assume on the contradiction that a optimal way to go to y from x performs a decreasing operation after increasing operation.

For concreteness lets assume

$$$(a,b,c,d,e,...)\rightarrow (a+1,b,c,d,e,...) \rightarrow (a+1,b-1,c,d,e,...)$$$

it's cost is $$$(b+1)(c+1)(d+1)(e+1)... + (a+1)(c+1)(d+1)(e+1)...$$$

swapping both operations gives

$$$(a,b,c,d,e,...)\rightarrow (a,b-1,c,d,e,...) \rightarrow (a+1,b-1,c,d,e,...)$$$

it's cost is $$$(a+1)(c+1)(d+1)(e+1)... + b(c+1)(d+1)(e+1)...$$$ and it's less when compared to the above cost. So the assumed order of operations is not optimal.

(Notice that final state in both cases is $$$(a+1,b-1,c,d,e,...)$$$)

Claim 3:

Increasing operations can be performed in any order

Proof:

For concreteness assume two adjacent increasing operations to be

$$$(a,b,c,d,e,...)\rightarrow (a+1,b,c,d,e,...) \rightarrow (a+1,b+1,c,d,e,...)$$$

it's cost is $$$(b+1)(c+1)(d+1)(e+1)... + (a+2)(c+1)(d+1)(e+1)...$$$ which is equal to

$$$(a+b+3)(c+1)(d+1)(e+1)...$$$

swapping the two operations gives

$$$(a,b,c,d,e,...)\rightarrow (a,b+1,c,d,e,...) \rightarrow (a+1,b+1,c,d,e,...)$$$

and it's cost is $$$(a+1)(c+1)(d+1)(e+1)... + (b+2)(c+1)(d+1)(e+1)...$$$ which is again equal to

$$$(a+b+3)(c+1)(d+1)(e+1)...$$$.

Claim 4:

Decreasing operations can be performed in any order

Proof:

Goes on the same lines as above.

Putting all these claims together we can say that we must first perform all decreasing operations and they can be performed in any order and after performing all decreasing operations we reach $$$gcd(x,y)$$$ and then we must perform all increasing operations and they can be performed in any order.

Note:

(1) This proof is not complete.

(2) Please correct me if I am wrong and please comment if you have any queries with regard to this proof.

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3 months ago, # |
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adedalic, sorry, but I really don't know how you can write such a complicated code in c++, but thank you for kotlin sample, although maybe some people don't know this language

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what is the time complexity of editorial solution of B?pls tell I cannot understand the solution .I did in o(n).

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https://codeforces.com/contest/1334/submission/76169596 can someone tell me the problem in my code for question C Circle of Monster I am trying to iterate over all the starting positions to kill the monster and finding the minimum bullets used.

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Does anyone solved F using Persistence Segment trees? I am getting MLE. BledDest pikmike is it intentional? If not can you please suggest me how to optimize with the memory? My Submission Link

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    2 months ago, # ^ |
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    It is possible to solve it using just Segment trees since b1...bn is in increasing order.

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Can someone please explain to me the Tutorial for the problem in 1334F — Strange Function in simple terms its very difficult to understand. I am have only done basic DP problems and want to learn but its going way over my head, not to able to grasp much.

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shouldn't a 26 * (|s|+|t|)* (log((|s|+|t|)) work for G

my approach was for each character i was creating a bit array for both t and s also i was considering the mapping of the indexes's given to be 1 in the bit array of t

for example if 1 is mapped to 20, then I put one int all those indexes that was having characters as a and t. and then performing a fft with a bit array of s reversed (using only a's for masking).

similarly for all characters. and then adding all the results of fft's ,if the index has value == size(s) then that index matches the substring.

please correct me where i am wrong. Or this approach shouldn't pass this constraint?

@pikmike @BledDest

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3 months ago, # |
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I can't solve problem C in Kotlin. At first, I did almost exact rewrite of the solution from editorial: https://codeforces.com/contest/1334/submission/76724818 And I got TLE. I suspected it might be slow i/o, so I tried to rewrite it: https://codeforces.com/contest/1334/submission/76725696

Still TLE

Then I searched and found a different template for Kotlin: https://codeforces.com/blog/entry/71089 So I tried that one: https://codeforces.com/contest/1334/submission/76726621

But I still get TLE. Is there anyone who managed to submit successfully in Kotlin? Can you please give me a hint how to read input or point out some error in my code that I am missing?

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For $$$F.$$$ Why no-one is talking about easy sgmt tree soln. . Editorial for F is tough to understand . lazy tree updates from 1 to x and another from x to m (if<0) , and updating dp as min(dpi,dp_i-1+p_i-1) . PLEASE tell me if its correct or not . As i have not coded yet!!

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3 months ago, # |
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Can anyone help me with 1334B - Middle Class?

My idea for solving is you find out if there are any wealthy people already and if no then answer is surely 0 else we sort the array and sum the total wealth. If total wealth/total people is larger than or equal to x then return total people(i.e n) else subtract wealth of first person(then second,third....) from total wealth, also subtract n and try again. 

Here is my submission: 76957152. My solution is failing at test 2. Can someone please point out the problem in the code or the approach? adedalic pikmike

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3 months ago, # |
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I got WA on C due to using LONG_MAX Why LONG_MAX is same as INT_MAX. I couldn't find answer on google...!

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    2 months ago, # ^ |
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    Because 'long int' is same as 'int' in 32-bit compilers and CF has a 32-bit compiler. Check this out. You can use 'long long int' for 64-bit integers.

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3 months ago, # |
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In B(Middle Class), I have first sorted the array and then I will taking value from end of the array(like a[i]) and if it is greater then x or equal to x,then I will take difference(a[i]-x) and I am counting all these for whom the value greater then or equal to x and storing it in some variable.After that when values lesser then x are coming I will give previous counted difference values and give it and in this way if that value can become greater then x then I will also count it for answer.This approach is failing.But I don't know why?Plz anyone help me. this is my submission[submission:77213843].

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2 months ago, # |
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For Prob. C, can we find the $$$i$$$ such that $$$a_{i+1}-\min(a_i,b_{i-1})$$$ is minimum possible, and break the circle between the monsters $$$i$$$ and $$$i+1$$$?

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2 months ago, # |
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i am getting wrong ans in D can anyone plz check my solution https://codeforces.com/contest/1334/submission/79674936

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5 weeks ago, # |
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For problem A level statistics why delta p should be greater than delta c? for eg (4 1), (5 5) This is still valid, I guess because in the second moment the number of play is 5 and the number of clear is also 5. Why the answer should be no for this. I do not know whether I am saying is correct or not. Plz help me.

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5 days ago, # |
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can someone explain why i m gettibg a tle for question 3 at test 3 in o(n). https://codeforces.com/contest/1334/submission/85927573