AtCoder | Learning library

by MathProgrammer C++ Version: Current License: No License

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kandi X-RAY | AtCoder Summary

AtCoder is a C++ library typically used in Tutorial, Learning applications. AtCoder has no bugs, it has no vulnerabilities and it has low support. You can download it from GitHub.

AtCoder

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Support

AtCoder has a low active ecosystem.

It has 25 star(s) with 16 fork(s). There are 1 watchers for this library.

It had no major release in the last 6 months.

There are 1 open issues and 0 have been closed. There are no pull requests.

It has a neutral sentiment in the developer community.

The latest version of AtCoder is current.

Quality

AtCoder has 0 bugs and 0 code smells.

Security

AtCoder has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.

AtCoder code analysis shows 0 unresolved vulnerabilities.

There are 0 security hotspots that need review.

License

AtCoder does not have a standard license declared.

Check the repository for any license declaration and review the terms closely.

Without a license, all rights are reserved, and you cannot use the library in your applications.

Reuse

AtCoder releases are not available. You will need to build from source code and install.

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AtCoder Key Features

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AtCoder Examples and Code Snippets

No Code Snippets are available at this moment for AtCoder.

Community Discussions

Trending Discussions on AtCoder

AtCoder DP Contest Q -I Coins (getting wrong answer may be using double)

atcoder educational dp problem a runtime error problem

Getting Overflow and TLE issues in this Atcoder question

[C++]The value has changed even though a variable has never been assigned since the first declaration

C++ push function in priority queue library won't work on my environment

How to save memory when using a trie?

Why this Knapsack problem's solution is not working?

How do I optimise the code to return the number closest to a given integer, not present in the given list?

Problem in code to divide N conditionally

QUESTION

AtCoder DP Contest Q -I Coins (getting wrong answer may be using double)

Asked 2022-Jan-13 at 08:38

Link to Problem -> https://atcoder.jp/contests/dp/tasks/dp_i

question Coins-I
Code I wrote gave wrong answer to TEST CASE 3

...

ANSWER

Answered 2022-Jan-13 at 08:38

The number of digits to be printed via output streams is controlled by their precision parameter. The default is 6 digits, if you want more you need to adjust it accordingly by std::setprecision :

Source https://stackoverflow.com/questions/70693002

QUESTION

atcoder educational dp problem a runtime error problem

Asked 2021-Nov-25 at 07:46

I am getting a runtime error in some test cases when I try to submit my code. Problem Link: https://atcoder.jp/contests/dp/tasks/dp_a

My code:

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ANSWER

Answered 2021-Nov-25 at 07:46

I'm not sure this is the only problem, in minCost() but you're passing dp by value, not reference. That means the compiler will make a copy of dp, not the actual dp.

Change your code to:

Source https://stackoverflow.com/questions/70107366

QUESTION

Getting Overflow and TLE issues in this Atcoder question

Asked 2021-Mar-06 at 09:29

I am solving this question: D-Base n (https://atcoder.jp/contests/abc192/tasks/abc192_d)

It states that:

Given are a string X consisting of 0 through 9, and an integer M . Let d be the greatest digit in X . How many different integers not greater than M can be obtained by choosing an integer n not less than d + 1 and seeing X as a base- n number?

I solved this question iteratively (which gave me correct answers for small inputs) but it gave me TLE and overflow issues (see my submission at https://atcoder.jp/contests/abc192/submissions/20651499). The editorial said that I must use Binary Search and I have implemented it as follows. But i still dont get the correct outputs. Any advice is welcomed.

...

ANSWER

Answered 2021-Mar-06 at 09:29

ll convert(string x, ll base, ll m){
    ll ans=0;
    ll p=0;
    ll cur=1;
    for(int i=x.size()-1;i>=0;i--){
        if(cur<0 || ans<0)
            return m+1;
        ans+=(ll)(x[i]-'0')*cur;
        cur*=base;
        if(ans>m)
            return m+1;
    }
    return ans;
}
void solve(){
    ll i, j, m;
    string x;
    cin>>x>>m;
    int n=x.size();
    ll mx=0;
    for(auto i:x){
        mx=max(mx, (ll)(i-'0'));
    }
    if(n==1){
        cout<<(((ll)stoi(x))<=m)<

Source https://stackoverflow.com/questions/66481482

QUESTION

[C++]The value has changed even though a variable has never been assigned since the first declaration

Asked 2020-Oct-17 at 02:02

Envs

g++ (Ubuntu 7.5.0-3ubuntu1~18.04) 7.5.0

Background

I practice implementing DFS.
I tried to solve this problem.

Code

sorry for dirty code

...

ANSWER

Answered 2020-Oct-17 at 02:02

The array arr has only 3 elements, so arr[3] is out-of-range and no read and write to there are allowed. It seems params is happened to be placed just after arr and writing to arr[3] is breaking data in params.

To easily fix this, allocate one more element for arr.

Better fix is to fix range of i not to cause this out-of-range access.

Source https://stackoverflow.com/questions/64398353

QUESTION

C++ push function in priority queue library won't work on my environment

Asked 2020-Sep-21 at 21:15

i was trying to solve a competitive programming problem (on AtCoder) and i wanted to use priority queue, but then i got some problem.

i wrote the code below but when i tried to compile that, i got error messages that said "no matching member function for call to 'push' " (the entire error messages are shown below)

i have no idea why this is happening, and i tried it on AtCoder Code Test online judge.

https://atcoder.jp/contests/abc176/custom_test

it worked completely fine on the online judge.

adding to that, i talked about this problem with my friend and he said the code worked just fine on his environment (he uses Windows).

Here is the code.

...

ANSWER

Answered 2020-Sep-21 at 20:58

Do you have an ancient version of g++? I tried compiling your code as far back as 4.8.1 (from 2013) and it worked in c++11 mode. In older versions than that, it didn't know some of the headers.
(Current version is 10.2.0)

I suggest you try playing with your code on Compiler Explorer. It gives you access to online versions of g++, clang, msvc++, and others, going back for many years over many versions, without any hassle of setup or installing anything. Here's your code with g++ 4.8:

https://godbolt.org/z/5TTjqd

Source https://stackoverflow.com/questions/63999551

QUESTION

How to save memory when using a trie?

Asked 2020-Aug-10 at 10:03

I came across this quesiton: https://atcoder.jp/contests/agc047/tasks/agc047_b and my code is as follows. My solution is very similar to the editorial.

I'm confused why this has memory issues. I make 10^6 * 26 nodes in the worst case scenario. The memory limit is 1024MB which seems like much more than enough. How is memory measured for a self referential class?

Additionally, how can I salvage my code?

...

ANSWER

Answered 2020-Aug-10 at 10:03

With 26 million nodes and 1024 megabytes, each node will need to be less than 40 bytes. But sizeof(node) is going to be at least 112 bytes, just for its members. The "self-referential" part comes from 26 pointers, each of which needs to be 4 bytes to support 1024MB.

You'll need another data structure. One solution is to have just a single firstChild per node, but also a nextSibling and a char c.

Source https://stackoverflow.com/questions/63335336

QUESTION

Why this Knapsack problem's solution is not working?

Asked 2020-Aug-01 at 07:12

I was recently solving a knapsack problem. In this version of knapsack, instead of weight, value is taken into the state.

The problem is the same as general Knapsack: There are n items, Item i has a weight of wi and a value of vi. The capacity of the knapsack is W. Find the maximum possible sum of the values of items that can be filled in the knapsack.

Constraints: 1<= n <=100 1<= W <=10^9 1<= wi <=W 1<= vi <=1000

Input: n W
w1 v1 w2 v2 w3 v3 ..... wn vn

Due to large value of weight, We have to take value in the state and the result will be the minimum weight. Although I have explained the problem but in case you need more details this is the problem link.

The below code is my attempt to solve the problem. I have used 1-based indexing. I'm not able to find out the error. I have tried debugging the code but that didn't help. I'm stuck on this from 2 days. Please help.

...

ANSWER

Answered 2020-Aug-01 at 07:12

There's a few problems with your code:

Max value can be 0
weight needed to reach some value can be > INT_MAX
dp[1][v[1]]=w[1]; uneeded line
for (int j = 1; j < 100001; ++j) j of 0 is valid weight though
if(j-v[i]>=1){ same as above

Here's a fixed version:

Source https://stackoverflow.com/questions/63202595

QUESTION

How do I optimise the code to return the number closest to a given integer, not present in the given list?

Asked 2020-Jun-16 at 15:22

I solved this problem statement (Yeah Yeah, I know, I am putting the problem statement below).

Given are an integer X and an integer sequence of length N: p1, …, pN. Among the integers not contained in the sequence p1, …, pN (not necessarily positive), find the integer nearest to X, i.e. the integer whose absolute difference with X is the minimum. If there are multiple such integers, report the smallest such integer

This is the code I used:

...

ANSWER

Answered 2020-Jun-16 at 14:39

According to your link, the total number of integers is at most 100. So 100 bits are enough to hold the flags, which numbers appear in the sequence. Those can be held in the processor registers.

The following code shows only the storage, afterwords you would have to chose suitable bit scan operations:

Source https://stackoverflow.com/questions/62409554

QUESTION

Problem in code to divide N conditionally

Asked 2020-Jun-01 at 13:55

So, I'm trying to solve the following Problem Statement

For an integer N, find the number of times the following operation can be performed:

...

ANSWER

Answered 2020-Jun-01 at 13:55

The number N can be factored into (p1 ^ e1)(p2 ^ e2) .... For each of these factors p^e, the most that you can divide N by is (p)(p ^ 2)(p ^ 3).... Of course, all those powers need to sum to less than or equal to e.

So you only need to compute all the factors once, and then for each e, find the largest triangular number that is less than or equal to e. The repeated factor calculations are wasted work, and is unnecessary.

Source https://stackoverflow.com/questions/62128390

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