hii everyone..i am new here...wer can i ask doubts regarding any problems??i am not seeing any active dicussion about problems.
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You can ask them just in these posts — it is the most usual way. However try to respect English language to increase chance of being answered, instead of being severely downvoted ;-)
You are not only new, you are also very stupid.
whether or not he is stupid, you are rude!
You watch your football.
i will, thank you very much! :)
Each problem here was a part of contest. When you solve a problem , see the tutorial , probably you will find some contestants who was discussing these problems after each contest. And as RodionGork said , try to respect English language to increase chance of being answered. Welcome to Codeforces =)
As Druid mentioned you can always look up the tutorial for a problem that was in a competition here, but you can also always ask any questions about any problems you encounter at all, in just these posts (blogs). Codeforces community is friendly and helpful, I have asked questions myself and someone always had the right answer, so if there is any task you are struggling with you can always ask here for help. As previously mentioned, try describing the problem you have in proper English so people can understand and help you (it also irritates a lot of people when they see posts with bad English), and I'm sure you will get the answers you are looking for.
Welcome to Codeforces! :)
What is the means of this in any program #define mod 998244353
if you want to use the value 998244353 anywhere in your code you can write mod instead of writing 998244353.
Like helping someone is a crime here.Why down-votes?
You will have to define this value as mod you cannot use it otherwise as per my knowledge
No, "mod" is just a name given to that number, you can name it anything you like. As #define is a preprocessor command, so every instance of "mod" will be replaced by 998244353 before compiling.
That's what I was saying that you will have to define it using #define mod 998244353 or you can use it like int mod=998244353
What is the different between 1. v.push_back({v[i].first, v[i].second + 1}); Vs 2. v.push_back((v[i].first, v[i].second + 1));
Edit- I know 1 is not wrong, 2 is wrong, but somehow idk how to fix it, see my reply below for more context
1 isn't wrong. The following works fine.
Yeah, ur right, I know 1 works which is why I thought i had mistyped it but infact i had written it correctly and cf somehow changed it, idk This link will explain better . Am i missing something obvious here?
that's funny, I think it got confused with markdown for a numbered list and those forcibly start at 1
I played around with it a bit. Looks like it's treating 2. as the start of a list and substituting the html with a list wrapper. Seems to cause the same result whenever you put a number followed by a . on a new line followed by some text. 255. a becomes.
Or you could also use:
v.emplace_back(v[i].first, v[i].second + 1)
20C][SUBMISSION:239959904 - Dijkstra? please see fails on large test case test case 31 failed. Dijkstra 20c <-problem
Use long long or unsigned long long types instead of int type. ALSO USE ....king spoilers.
i tried both ll and ull but doesnt work
the graph is undirected. (check graph theory) So, use this too:
vector < long long > dist(n + 1, 1e18)
in 31 tests the minimum path is 1e9
thank you appreciate!
Here is the link with the corrected solution :"Link"
[submission:239953552][submission:Codeforces Round 473 (Div. 2)] Can you please debug this snippet.test case 3 is failing. i have attached the image also.
Your error was in the UnionBySize function .Here is the corrected solution:"Link. "
You mixed up the lines: "cost[ulp_u] = min(cost[ulp_u], cost[ulp_v]) with ; cost[ulp_v] = min(cost[ulp_u], cost[ulp_v]);"
Appreciate that!
please guide me which of the following topics are important for Competitive programming. also if you can share some of the problems/links related to the following topics in mathematics:
Root finding for non linear equations and systems of equation
1 Introduction to Numerical computations and floating point arithmetic
2 Types of errors, significant digits and error propagation.
3 Introduction to iterative methods, initial approximation and intermediate value theorem
4 Iterative Methods based on first degree equation
5 Iterative Methods based on second degree equation: Muller Method
6 Solved Problems using Muller Method
7 Chebyshev Method
8 Solution of the system of non-linear equations using the Newton Raphson method.
** Interpolation**
11 Finite Difference Operators
12 Interpolating polynomials using Finite Difference(Gauss Forward)
13 Interpolating polynomials using Finite Difference(Gauss Backward)
14 Hermite Interpolation
15 Problem solving
16 Tutorial(Problem Solving)
17 Lagrange Bivariate Interpolation
18 Newton’s Bivariate Interpolation for Equispaced Points
19 Inverse Interpolation
Numerical Integration
21 Introduction to Numerical Integration
22 Numerical Integration with known nodes
23 Rombergh Integration
24 Gauss-Legendre Integration
25 Gauss-Chebyshev
26 Gauss-Laguerre and Gauss-Hermite Integration Methods.
27 Problem solving
28 Double Integration by Trapizoidal methods.
29 Double Integration by Simpsons’ methods.
30 Problem solving Numerical Solution of ODEs and PDEs
32 Introduction to the numerical solution of ODEs
33 Initial Value Problems and it’s numerical solutions
34 Milne’s method
35 Boundary Value Problems using Shooting Method
36 Problem Solving
37 Finite Difference approximation of derivatives
38 Using Finite difference methods to solve 1D heat equation using explicitn method(Schmidt Method)
39 Finite difference methods to solve 1Dheat equation using implicit scheme(Crank Nicholson Method)
i don't get it how do i do it without using arithmetic operations please help!!
Problem Statement:
Given 'd' you have to calculate f(d) where , f(d)=2*d(d+1)+1 using bit manipulation. Dont't use +,-,/,* arithmetic operations either. f(d) is only for reference for checking for given d whether f(d) is valid or not.
constraints: 0<=d<=100, 1<=f(d)<=20201
test case1 d=0 op=1; test case2 d=1 op=5; test case3 d=2 op=13 test case4 d=3 op=25 test case5 d=4 op=41
int function(int d){
//don't do -> return 2*d(d+1)+1 instead try to do it using bit manipulation on given 'd' parameter.
}