chokudai's blog

By chokudai, history, 4 years ago, In English

We will hold AtCoder Beginner Contest 179.

The point values will be 100-200-300-400-500-600.

We are looking forward to your participation!

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4 years ago, # |
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"We are looking forward to your participation!"

We are looking for quick editorials.

P.S. I love ABC's.

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4 years ago, # |
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What are the ALC contests of Atcoder?

Edit: It's is described in this blog

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    4 years ago, # ^ |
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    No, it is me after solving ABCDE

    Though it seems F was not particularly difficult this time

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I tried a DP solution for D. Somehow getting TLE in 6 TCs. I tried to optimize it as much as I could. Am I missing something?

Link to Submission

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    4 years ago, # ^ |
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    post qns only after contest is over.

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    4 years ago, # ^ |
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    i used fenwick tree to optimize it, note that k <= 10, your solution fails because ranges can be very big, so your solution is n^2

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      4 years ago, # ^ |
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      But the sum of ranges won't be more than O(n) I suppose. Because the segments are nonintersecting.

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        4 years ago, # ^ |
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        Yes, the sum of ranges is $$$O(n)$$$, but you also have $$$O(n)$$$ states and for each, you iterate through all of them. In total it is $$$O(n^2)$$$.

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    4 years ago, # ^ |
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    you can try it in O(n*k). Refer to below link for reference.

    https://www.geeksforgeeks.org/constant-time-range-add-operation-array/

    My Submission

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    4 years ago, # ^ |
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    your code runs in O(K*N) where K is the number of possible jumps, theoretically in worst case it would reach O(n^2) time complexity which could defenitely get TLE. In this problem i used dynamic programming with fenwick tree, you can look at my code here

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    4 years ago, # ^ |
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    For d, you have to combine the dp you are talking about, with the range covering problem (it is easier than segment tree I leave you this video in which that guy explains that problem https://youtu.be/Zze-O2oxoEo?t=219 ) and then you just have to be careful about modulo operations (you might be doing them with negatives)

    I hope this was useful

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      4 years ago, # ^ |
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      Thanks for this. I checked out the video. Isn't what he is talking about, called the Fenwick tree?

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        4 years ago, # ^ |
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        No, Fenwick tree is a little more complex than that, it is about having sum until last significant bit and when doing a query going up from last significant bit to most significant bit (with a complexity of logn) and when updating is almost the same, but instead of getting an answer, updating; that is why it is also called Binary Indexed Tree

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    4 years ago, # ^ |
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    My code is similar with yours and I wonder how to optimize it too. [submission:#17148205]

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4 years ago, # |
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First time in ABC I was able to solve all problems. Here are short explenations.

F
E
D
C

Funfact, I had no WA, all AC on first submission.

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    4 years ago, # ^ |
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    For d instead of segment tree + dp you can use prefix array + dp.

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      4 years ago, # ^ |
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      please share your code

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        4 years ago, # ^ |
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          4 years ago, # ^ |
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          plz give a brief explanation also. thnx

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            4 years ago, # ^ |
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            Basic
            Solution
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    4 years ago, # ^ |
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    You don't really need a segment tree for D
    I just used prefix sums

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    4 years ago, # ^ |
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    D and E should have been swapped ,I didn't even attempt E after continuous TLE's in D .

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      4 years ago, # ^ |
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      D is more difficult in terms of the idea I guess

      E is heavier in terms of implementation

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      4 years ago, # ^ |
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      even i think so

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    4 years ago, # ^ |
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    You don't need segment tree for F either.

    Code
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      4 years ago, # ^ |
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      can you explain your approach for F

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        4 years ago, # ^ |
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        I'm more than a month late, you may have got the answer till now, but here is what I think the approach is:

        Here two sets are used to maintain position of lines of white stones. There are two kinds of lines length wise :

        1. Partial lines(going from one end and ending halfway)

        2. Complete lines (going from one end to other end)

        There are two kinds of lines possible orientation wise :

        a) Horizontal Line - Can be stored as $$$y$$$ co-ordinate it starts on and $$$x$$$ co-ordinate it ends on.

        b) Vertical Line - Can be stored as $$$x$$$ co-ordinate it starts on and $$$y$$$ co-ordinate it ends on.

        Now what exactly those two sets do, set[0] stores position for horizontal lines, and set[1] stores position of vertical lines.

        Formally a pair stored in respective sets looks like:

        • s[0]= (Ending abscissa, originating ordinate)
        • s[1]=(Ending ordinate, originating abscissa)

        Note that at the start we only have two complete lines, one horizontal and one vertical that's why a pair $$$(n,n)$$$ is stored in both the sets.

        • Query 1 We search for nearest horizontal line we can find that ends after $$$x$$$, and thus add remove black stones in that interval from the answer accordingly and now one vertical line is created in the process that starts at $$$x$$$ and ends just before the horizontal line we found.

        • Query 2 Search for nearest vertical line, remove black stones in that range from the answer and add a horizontal line in the corresponding set.

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    4 years ago, # ^ |
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    F: you can use std::map for the same query (lower_bound to find closest) Submission

    D: you can use only partial sums of your dp and fill dp[i] with sum of previous values Submission

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    4 years ago, # ^ |
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    I also thought of using segtree with lazy prop on F, but I had no segment tree template with lazy propagation :P and I had only 35mins left. So I quit. Missed my chance of solving all problems in ABC for the first time.

    I'm taking your template now XD.

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      4 years ago, # ^ |
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      I also got idea of two lazy segment tree immediately but I haven't implemented even a basic lazy segment tree before so I gave up. Now I am seeing there are other solutions too

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      4 years ago, # ^ |
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      I had to fix a bug in that template in function rangeInc(). Not sure if everything works fine.

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        4 years ago, # ^ |
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        Oh,thanks for informing. I'll test it properly after customizing it for myself. :)

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          4 years ago, # ^ |
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          In AtCoders ACL library there is a segment tree, too. I would like to switch to that one, but need time to get used to it.

          ACL lib

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      4 years ago, # ^ |
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      Question: I thought that we should need to be able to get the historically minimum of some certain index? Won't that be more than just a plain old segtree?

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    4 years ago, # ^ |
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    I got tle in this, could you plz tell me why?

    int ans = 0;
    for(int i = 1 ; i <n; i++){
    	int x = n-i;
    	for(int j = 1; j*j<=x; j++){
    		if(x%j == 0){
    			ans++;
    		if(x/j != j){
    			ans++;
    		}
    		}
    	}
    }
    // ans += ans;
    cout << ans ;
    
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      4 years ago, # ^ |
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      You have asymptotic of $$$O(n \sqrt{n})$$$ which is too big. You can enumerate only the smallest number of $$$A, B$$$ which would be $$$\sqrt{n}$$$ and calculate how many multipliers bigger than your number you can have such that $$$A \times B$$$ is smaller than $$$N$$$.

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      4 years ago, # ^ |
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      n is 1e6, so the inner loop runs up to 1e3 times... that are aprox 5e8 iterations.

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    4 years ago, # ^ |
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    Me too for the first time solved set in any contest lol

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    4 years ago, # ^ |
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    Can you please further explain how did segment tree help with this problem? I know segment tree but I cannot utilize it to solve this problem.

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      4 years ago, # ^ |
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      I assume problem D.

      If we would do standard knapsack we would need to loop over all the values in the K ranges, which is O(n).

      With the segment tree we are able to do these updates in O(log n).

      Note that the solution with the partial sums do these updates in virtually O(1). see here

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        4 years ago, # ^ |
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        Thanks a lot:) you always provide useful insights

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    4 years ago, # ^ |
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    correct me if I am wrong, the time complexity of your D solution is N*k*log(n) and not N*log(n) which the editorial says is the optimum

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      4 years ago, # ^ |
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      Yes, you are right. Because of this my submission needs 800+ms instead of possible 10ms.

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    21 month(s) ago, # ^ |
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    In problem E , How can I know that the series will repeat itself ?

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      21 month(s) ago, # ^ |
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      It is the pidgeonhole principle. Since all values of the series are in range 0..M-1, there is a repitition after at most M values.

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How to solve problem D?

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    4 years ago, # ^ |
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    Spoiler
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    4 years ago, # ^ |
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    I solved it without segment tree

    I used difference array. Time complexity of $$$O(N*K)$$$

    basically let dp[i] be no of ways to each i.

    Then you increment, $$$i$$$+$$$L_j$$$ to $$$i$$$+$$$R_j$$$ with dp[i], increment using difference array method, and keep summing as you move toward right.

    The answer would be dp[n]

    Submission

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    4 years ago, # ^ |
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    I solved it maintaining a prefix sum array and using dp.

    dp[i] = sum of dp[j] for all j, i is reachable. As the range was contiguous it was easy to get the sum of dp[j] using prefix sum.

    My Sumbission

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      4 years ago, # ^ |
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      could u plz explain more

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        4 years ago, # ^ |
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        Here dp[i] is the number of ways to reach i.

        dp[i] was calculated by the sum of dp[j] for all j from where i is reachable. Assume i = 4 and i is reachable from 2 and 3. If the number of ways to reach 2 is 1 aka dp[2] = 1 and the number of ways to reach 3 is 2, dp[3] = 2 then dp[4] = dp[2] + dp[3] which is dp[4] = 3 (rule or sum)

        And for any i corresponding js were calculated from the ranges and their sum was calculated from the the prefix sum array. For any position iand a range L, R i is reachable from all valid i-R, i-L.

        Finally the answer is dp[n].

        Complexity O(N*K)

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I got runtime error in sometest cases in prob D with segtree can somebody help me? https://atcoder.jp/contests/abc179/submissions/16889464

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void test_case() {
	int n , x;
	cin >> n >> x;
	for(int i = 0 ; i < 2*x ; i++){
		int p;
		cin >> p;
		c.insert(p);
	}
	dp[1] = 1;
	for(int i = 2 ; i <= n ; i++){
		for(auto j : c){
			if(j <= i){
				dp[i] = (dp[i]%M+dp[i-j]%M)%M;
			}
		}
	}
	cout<<dp[n]%M;
}
 

what WA in my D https://atcoder.jp/contests/abc179/submissions/16888327[my sol link](http://https://atcoder.jp/contests/abc179/submissions/16888327)

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    4 years ago, # ^ |
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    they want you to union the segments [Li,Ri] i mean {Li,1+Li,2+Li,3+Li,....,Ri} not only values {Li,Ri}

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It`s the first time I AK abc! LOL

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I tried to solve D by repeating knapsack dp approach with time complexity O(N*m).But TLE destroyed my today's contest.Can anyone help me with that?

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    4 years ago, # ^ |
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    D can be done in O(n*k).

    Use range update operation on array, it will take O(n) for each range.

    My submission

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      4 years ago, # ^ |
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      Can you elaborate this part?

      if(i+range[j].F<=n) temp[i+range[j].F]=(temp[i+range[j].F]+dp[i])%M; if(i+range[j].S+1<=n) temp[i+range[j].S+1]=(temp[i+range[j].S+1]-dp[i]+M)%M;

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        4 years ago, # ^ |
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        Suppose you have array 1 2 3 4 5 You want to add 2 from position 2 to 4 [1 based-indexing) You can take a temp array and make temp[2]=2 and temp[5]=-2

        Now you can iterate over array and take sum+=temp[i]

        And add sum to a[i] . You can get the required array after update in O(n) this way.

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Eagerly waiting for geothermal's editorial.

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Hello in D no I used the similar pattern of dp like CSES-Coin Combinations I. but get TLE from that.Isn't it the similar pattern problem.

My solution is here

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    4 years ago, # ^ |
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    Thats' so because the time complexity of your solution is O(n*n).

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      4 years ago, # ^ |
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      I thought it's like Coin Combinations I problem.My bad.How can I solve it using dp ?

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        4 years ago, # ^ |
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        Well, usually i solve 4-5 problems in ABC contest but today i was able to solve only the first three!

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Could anyone please explain the 1 testcase this is failing on? I have tried to find a cycle and then adding to the sum the sum of that cycle and then adding the rest seperately.Submission

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how to solve D? someone with good explanation?

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    4 years ago, # ^ |
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    You can use this dp: dp[i] means how many ways do you have to go to the i-th cell.

    Let's get an array of long long dp[2n]. First we fill it with zeros. Denote the sums[2n] as prefix sums array of dp.

    1. dp[n] = 1, sums[n] = 1

    2. for each cell from n+1 to 2*n and each segment (l[j], r[j]) we calculate: dp[i] = (mod + dp[i] + sums[i - l[j]] - sums[i - r[j] - 1]) % mod

    You can understand that part as: to get how many ways I have to go to the i-th cell, I should sum the ways from all dp[the previous one, from which you can get into this].

    The value sums[i - l[j]] - sums[i - r[j] - 1] gets us sum: dp[i - r] + dp[i - r + 1] + ... + dp[i - l]

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      4 years ago, # ^ |
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      Why are you taking dp[2*n] and not dp[n] or dp[n+1]?

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        4 years ago, # ^ |
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        Just because I don't want to care about bounds

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how do you solve c? I got TLE my solution is only O(n^2)

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    4 years ago, # ^ |
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    $$$O(N^2)$$$ is too large for $$$N=10^6$$$.

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      4 years ago, # ^ |
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      How can i know that it's large or not?

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        4 years ago, # ^ |
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        You just put N into the time complexity expression and evaluate it. Generally speaking, $$$10^7$$$ is totally acceptable, while $$$10^8$$$ can be a bit dangerous (on some OJs it cannot pass), and $$$10^9$$$ is almost impossible to pass the time limits.

        However, sometimes we also need to consider the constant, which is omitted in the time complexity expression.

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    4 years ago, # ^ |
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    can somebody please explain the logic for c.

    n = int(input())
    cnt = 0
    for i in range(1, n):
        cnt += (n - 1) // i
    print(cnt)
    
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      4 years ago, # ^ |
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      Since N is fixed, we don't need to enumerate all possible triplet of (A, B, C), we only need to enumerate (A, B) and compute C accordingly. Such method can be improved by enumerating only A and find the upperbound and lowerbound of B. All the values between lowerbound and upperbound are valid B. Note that B should be integer.

      $$$0 \lt A \cdot B = N - C \lt N \implies 0 < B < N/A$$$

      N = int(input())
      ans = 0
      for A in range(1, N):
          lb = 1
          ub = math.ceil(N / A) - 1
          ans += ub - lb + 1
      print(ans)
      
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I believe F can be solved using monotonic set

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I have written an unofficial English editorial.you can find it here.

UPD: Added editorial for problem F.

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    4 years ago, # ^ |
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    In the explanation for C : Could you please explain how the number of ways of choosing is N/A ?

    I have tried to do it on paper but couldnot come up with the intution.

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      4 years ago, # ^ |
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      It should be $$$\frac{N - 1}{A}$$$ fixed it now, thanks for notifying me.

      The reason for that is for every $$$A$$$ there can be $$$\frac{N}{A}$$$ numbers for be such that $$$ A \times B \le N$$$, however we should subtract $$$1$$$ because $$$C$$$ can't be equal to zero.

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Can E be solved with Matrix exponentiation??

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I tried to solve E,but for some unkown reason,I got two RE.

Can you help me?

Code
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4 years ago, # |
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Editorial D

How to solve in O(nlogn) when the transitions cannot be written as a sum of small number of segments?

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    4 years ago, # ^ |
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    [ignore this, too much complex]

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    4 years ago, # ^ |
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    let $$$J$$$ be the set of possible jump lengths, now $$$ J = \cup[l_i,r_i] $$$

    Now we can use generating function $$$ G $$$ to define that set.
    $$$G = \sum_{i=0}^N c_i x^i $$$ where $$$ c_i = 1$$$ if $$${i \in J}$$$ else $$$c_i = 0$$$.

    Now, number of ways to reach from $$$1$$$ to $$$N$$$ using exactly $$$k$$$ jumps = $$$[x^{N-1}] G^k$$$.

    As there is no restriction of jumps,so we sum it over all possible $$$k$$$.
    so, finally what we need is $$$[x^{N-1}]\sum_{k>=0} G^k = [x^{N-1}]\frac{1}{1-G}$$$

    now coefficient of $$$ x^{N-1} $$$ in the polynomial $$$\frac{1}{1-G}$$$ can be easily calculated by calculating inverse of polynomial $$$ 1 - G $$$ restricted to max degree $$$N$$$.

    Overall complexity is $$$O(N\log{}N)$$$

    For calculating inverse of polynomial you can refer Operations on polynomials and series

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In problem E , How can I know that the series will repeat itself?

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    21 month(s) ago, # ^ |
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    Pigeon hole principle

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      21 month(s) ago, # ^ |
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      I know Pigeon hole principle but i can't understand how can we know that the series will repeat itself?

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        21 month(s) ago, # ^ |
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        There can only be $$$M$$$ different values of $$$A_i$$$. Therefore, it's inevitable that the sequence will repeat in at most $$$M$$$ iterations.

        Also this topic is 2 years old

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        21 month(s) ago, # ^ |
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        A[i] is only dependent on A[i-1]. Therefore, given the last value, there is only one and only one possible current value. It also means there can only be one and only one possible next value A[i+1].