chokudai's blog

By chokudai, history, 5 days ago, In English

We will hold AtCoder Beginner Contest 180.

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

We are looking forward to your participation!

 
 
 
 
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5 days ago, # |
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Wow that's an early announcement

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4 days ago, # |
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How to solve D?

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    4 days ago, # ^ |
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    The idea is that it will only be optimal to use Kakomon Gym (multiply by A) a few times, after which it will always be better to add instead of multiply. You can simulate the first part while x*a <= x+b, then use division to find out the number of operations of the second type to do.

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      4 days ago, # ^ |
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      Can you share your code?

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        4 days ago, # ^ |
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          4 days ago, # ^ |
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          Any Idea why this submission gets WA?

          https://atcoder.jp/contests/abc180/submissions/17481021

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            4 days ago, # ^ |
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            Because x*a overflow when x = 10^18 && a>10 something like that

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          4 days ago, # ^ |
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          can you please tell me why you subtract 1 from the answer(int ans) at the end??

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            4 days ago, # ^ |
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            The way the code is implemented has the 2 following cases: 1. x stopped right at y, i.e, x == y; 2. x exceeds y after the last increase by b operation, i.e, x > y.This is reflected in the else block inside the while loop. We add 1 to take the remaining y — (y — x) / b * b into consideration. So this add 1 makes x exceeding y.

            In both case, we must have 1 fewer operation to make x strictly smaller than y. So we subtract 1.

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          25 hours ago, # ^ |
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          can you explain LLONG_MAX/x >= a why this was done in if condition?

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            19 hours ago, # ^ |
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            To prevent overflow. If that condition is not my, then x*a will overflow, which can lead to wrong answers and infinite loops.

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      4 days ago, # ^ |
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      idk why i thought of dp here may be coz i thought inorder to maintain maximum exp you gotta maintain minmimum str

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      32 hours ago, # ^ |
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      Can you prove why that strategy is optimal?

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    4 days ago, # ^ |
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    Just brute force over all the steps,and calculate the maximum. By steps I mean, steps until you reach target by visiting 1st place some k times and visiting 2nd place to complete the remaining EXP. Check for overflows while dong so.

    Solution

    Sorry for my English.

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4 days ago, # |
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How to solve B(only Euclidian distance part) and D ??

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    4 days ago, # ^ |
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    Sum up the squares of elements in a long long, then cast to long double and take square root.

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    4 days ago, # ^ |
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    The easier thing would be to use Python for questions like D, or Java's BigInteger.

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4 days ago, # |
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Solution for F, anyone?

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    4 days ago, # ^ |
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    Let DP[n][m][b] be the number of ways to build a graph with n labeled nodes and m unlabeled edges with b being a flag for restricting the size of the maximal connected component.

    From there at each iteration and to avoid double counting you only look at the components covering the first node.

    For example if I want a component of size 5 and n = 10 first I find the number of ways I can choose nodes to form this component which is 9 choose 4 (node #1 is already included). Then I find the number of way to build a path (5! / 2) or a cycle (4! / 2) and call the appropriate state.

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      4 days ago, # ^ |
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      I didn't get how are you avoiding double counting?

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        4 days ago, # ^ |
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        cause if you reverse a chain, the result will be the same as the origin, so there are k!/2 ways to form a chain with size k, or you will count a chain twice.

        as to circle, you should consider symmetry and shift, so there are k!/2/k = (k-1)!/2 ways to form a circle with size k.

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          4 days ago, # ^ |
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          He is referring to the over counting of the total number of ways to make the connected components using DP, not the over counting of a particular component itself.

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        3 days ago, # ^ |
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        sorry, i misunderstood it.

        for example, with node 1, 2, 3, 4 and 5,

        • choose 1,2 to form a component, and then choose 3,4,5 to form a component.
        • choose 3,4,5 to form a component, and then choose 1,2 to form a component.

        these 2 ways generate same graph, but if you count it more than once, then double counting happen.

        if you only look at the components covering the first node, you will only count the first way.

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4 days ago, # |
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How was E intended to be solved? I just copy pasted a DP code from stack overflow and calculated distances of graph[i][j] using the formula in the question and got AC.

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    4 days ago, # ^ |
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    Travelling Salesman Problem using dynamic programming

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      4 days ago, # ^ |
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      That's what I did, I am saying if it was meant to be solved in a simpler way.

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        4 days ago, # ^ |
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        Given that $$$n \le 17$$$, I guess the intended solution $$$O(n^{2} \times 2^{n})$$$ bitmask dp.

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        4 days ago, # ^ |
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        Well I guess we'll find out when(if) the editorial comes out

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    4 days ago, # ^ |
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    It is a standard question of Dp with bitmasking.

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    4 days ago, # ^ |
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    Well, I just honestly implemented dp for traveling salesman problem... I guess it was the intended solution

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    4 days ago, # ^ |
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    I guess the main difference in the problem was to visit at least once, while in standard TSP it is exactly once.

    So consider if the problem was a general directed graph, then the standard DP solution would not have worked, but since the shortest path between two vertices, in this case, is the same as their direct edge between them, the problem had the exact same solution.

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      4 days ago, # ^ |
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      Even for the general graph, we can run floyd warshall first, then the classic TSP. It's a shame I took 50 minutes in this problem.

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        4 days ago, # ^ |
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        I never thought they would straight away give classic tsp for a problem at "E". Man .. I took so long to implement floyd warshall version of it .. . Should have tested with the classic version first

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        4 days ago, # ^ |
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        Can you elaborate on the Floyd-Warshall + TSP Solution for general graphs, or share any resource to learn from.

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          4 days ago, # ^ |
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          This is what I did in my submission in contest,now you would have run through comments that direct edge is always optimal in this problem which i never observed so instead what i did was to first apply floyd warshall on the graph with saving a "next" Matrix in order to trace the path when i move from a node to other because when I would do so unintentionally I would also visit some unvisited cities(maybe or maybe not). Now after doing this i just do normal tsp and updating the mask. However i did see printing some paths and saw that direct edges are the shortest ones but thought they have intentionally given such samples.

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    4 days ago, # ^ |
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    if n<=20 its 95% of the time bitmask dp

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    4 days ago, # ^ |
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    Can you share the link? I searched for such one, but did not found any, and was not able to implement a bugfree version.

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    4 days ago, # ^ |
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    Can you share the link please?

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4 days ago, # |
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Was solution to E was intended to be solved using floyd warshall and then tsp, or did I do some overkill ?

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    4 days ago, # ^ |
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    I don't think you need floyd warshall. Since the distance metric used always gives shortest path between two vertices.

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    4 days ago, # ^ |
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    I solved it using dp

    hint1
    hint2

    my code :

    Code
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4 days ago, # |
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This is my solution to B problem, can anyone tell me why it's WA My code

Edit: After Changing everything to long long, it is working now, even long double didn't work, can anyone explain to me why long long is working and remaining are not.

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    4 days ago, # ^ |
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    Try long double instead of double, and sqrt instead of pow.

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      4 days ago, # ^ |
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      Tried now, but still shows WA, can you please help me out and point the mistake in this

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        4 days ago, # ^ |
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        You have to take max of absolute of arr[i] instead of just arr[i] to calculate 3rd distance. This is the same which I did and llost 700 ranks bcoz of that

        EDIT: sorry I overlooked that part.

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          4 days ago, # ^ |
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          Hey, but I did that right, as you can see, while taking the input of the numbers, if the number is negative, I have written -arr[i].

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4 days ago, # |
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How to solve Problem F? DP?

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4 days ago, # |
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how to solve C? UPDATE:got it.

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    4 days ago, # ^ |
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    find divisors of number in root(N) time, add it to a list ,sort the list and print it.

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    4 days ago, # ^ |
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    So, first, you have to find the numbers which are divisible by the given number N.

    For this, we can run a for loop from 1 to N, but you will notice that we shorten up this process, when you get a number i, which is divisible by N, you will also get a number N / i, which is also a divisor of N.

    when you need to essentially go up to sqrt(n) numbers, as you will cover everything divisor by that time. (Note that a square number wont 2 divisor at sqrt(n))

    for(int i = 1; i * i <= n; i++)  {
       if(n % i == 0) {
             cout << i << '\n';
             if(i * i != n)
                cout << n / i << '\n';
         }
    }
    

    But, now you need to print it in the ascending order, just take a vector, and push_back all the (N / i) numbers, print it in the reverse order.

    vector<int> div;
    for(int i = 1; i * i <= n; i++)  {
       if(n % i == 0) {
             cout << i << '\n';
             if(i * i != n)
                div.push_back(n / i);
         }
    }
    
    for(int i = div.size()- 1; i >= 0; i--)
        cout << div[i] << '\n';
    
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4 days ago, # |
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Hi, could someone point out my mistake in B

Edit: But x[i] is long long right?

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4 days ago, # |
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Could someone help me.. i dont know why my dp runs incorrectly :< https://paste.ubuntu.com/p/nrYYHqvxw7/

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4 days ago, # |
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  1. TLE
  2. AC Solution after contest 4th problem why anyone?
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4 days ago, # |
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How to solve C

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How to solve d please explain anyone?? i was thinking it was dp but other's solution just amazed me

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    4 days ago, # ^ |
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    Solution Just think about a short case and try to work on it. Keep a check on overflows.

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Why this solution gets TLE while this gets AC? Using int instead of long long should result in wrong answer instead of TLE, then what other reasons behind it?

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    4 days ago, # ^ |
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    For bigger values of n, i * i will overflow and will result in i * i <= n being true.

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Hi, can you explain to me why i got WA in problem B? Both sample cases work;

n = int(input()) x = list(map(int, input().split())) y = list(map(abs, x)) print(sum(y)) #1 res = [a ** 2 for a in x] j = sum(res) #2 print(math.sqrt(j)) print(max(x)) #3

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Please can somebody explain why using even unsigned long long is giving WA on D? I tried different values of x,y,a,b and checked but they were all running fine, but during contest ((ull)x*a<2e18) dosen't work but ((double)x*a) worked; why??

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How to solve E?

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    3 days ago, # ^ |
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    Use dynamic programming with bitmask, dp[i][j] is the min cost of visting all cities specified by state i with the last visiting city as j. For a given dp[i][j], if we already know its result, we use it to make the next hop. For all valid hops from a city that has been included in i, to a city j from 0 to n — 1, update dp[next][j] where next is i | (1 << j).

    The final answer is dp[(1 << n) — 1][0].

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3 days ago, # |
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t = int(input()) li = list(map(int, input().split()))

manhattan_distance = 0 for i in range(t): manhattan_distance = manhattan_distance + abs(li[i])

print(manhattan_distance)

import math euclidian_distance = 0 for i in range(t): euclidian_distance = euclidian_distance + (li[i]*li[i])

print(math.sqrt(euclidian_distance))

chebyshev_distance = max(li) print(chebyshev_distance)

Would anyone debug it please?

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    2 days ago, # ^ |
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    li[i]*li[i] may overflow .type cast it into long.

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39 hours ago, # |
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I have a problem with B. I am using long long and long double wherever needed and I am still getting compilation error. I think I am getting an error while using "abs(x)". Here's my code, plz help : problem B

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    39 hours ago, # ^ |
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    There are a few problems:

    The error with abs function is due to ll being defined as unsigned.

    Then after this, you will be getting a WA because you have your cd as cd=max(cd,x) instead of cd=max(cd,abs(x))

    After that, you still are getting a WA because you are submitting problem B's solution for problem A :)