Is doing div2 A<B<C regularly fast is enough for reaching expert?Do I ever need to solve D?

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Is doing div2 A<B<C regularly fast is enough for reaching expert?Do I ever need to solve D?

Is it not a simple and correct assumption that if anyone solving a particular rated problem like all under and inclusive 1400 in contest time regulary like for 4-5 contests,he/she will become 1400/specialist?

Am I wrong here ?

For this Problem

The solution is written in O(n*m)

But How it is possible if there is recursion calls inside the nested solution.

Solution in tutorial is:

https://codeforces.com/blog/entry/116108 (Problem E)

Please someone help me understand the time complexity?Shouldn't it be O(n*m*n*m).I know we are not checking same grid twice.

Please someone help me.

Can anybody give me idea about leetcode contests problem difficulty compared to codeforces level?

I have a string which is empty initially.Then I added characters to it.

```
string s="";
s=s+'a';
```

it works fine in my pc but getting error in codeforces submission because it prints garbage at the end of hathe printed string.

Same things happen I use vector of char.In that case I use push_back method to add chars.But it still prints garbage in codeforces submission.

```
vector<char>vc;
vc.push_back('a')
```

Why is that happening? /predownloaded/42/08/4208fdc2b1a8d618adc8327184077c6b4baa0b60.png

Codeforces problems are always entitled with many tags assuming there are diverse solution approach for problems.

If one solved a problem with greedy approach which is in the tags,then if he want to know the data structure approach to solve a problem how can he find the other tags approach?

Here is the link of problem at atcoder:

I tried it with a naive recursion approach. It's running good with the testcases in my pc but getting RE in atcoder compiler. I have no clue how it is getting RE.

Here is the solution

#include<bits/stdc++.h>
using namespace std;
int N,sum=0,mn=INT_MAX,h[100006];
int recur(int i,int sum)
{
if(N>=4)
{
return mn=min(sum,mn);
}
else
{
recur(i+1,sum+abs(h[i+1]-h[i]));
recur(i+2,sum+abs(h[i+2]-h[i]));
}
}
int main()
{
int n,i,j,k;
cin>>N;
for(i=1;i<=N;i++)
{
cin>>h[i];
}
cout<<recur(1,0)<<endl;
return 0;
}

We have an array which is produced from a sorted array by right shifting array elements some number of times.

Like [4,5,6,1,2,3] is produced from [1,2,3,4,5,6] by right shifting the sorted array 3 times.

The sorted array is always increasing.

If the given array cannot be produced from any increasingly sorted array the answer is no, otherwise answer is yes.

https://onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=565

does it allow multiple solution?

like 43 2 is a correct answer for the fifth test case?

I am getting wrong answer and the problem description is not clearing it.

Here is my submission Submission

It seems to be both output and answer is same.Though it is giving wa. Please can anybody help me out?

In this problem,I have come up with a solution that is only equal bit length numbers meet the condition of & operation >= XOR operation. That means number 2-3,4-7,8-15.... meet the conditions given in this problem.

Based on this idea, I write a solution, But this is giving always wrong answer. I am so frustrated and don't have a clue what is missing.

Can anybody please help me,what is wrong with my approach?

We all know that, in recent years codeforces popularity rose very high. Thus participants in contests have grown to a large number. I want to know that, does increased participant number somehow devalued the ranking of those who participated with fewer number of participants?

I mean ,is 1400+ rating in 6-7 years back is devalued compare with today's 1400.

Why second loop of sieve of Eratosthenes starting from square of current prime number?

```
vector<bool>vc(100006,1);
void seive(int n)
{
vc[0]=vc[1]=0;
int i,j;
for(i=2;i*i<=n;i++)
{
if(vc[i]==1)
{
for(j=i*i;j<=n;j=j+i)
{
vc[j]=0;
}
}
}
}
```

My question is why the second loop starting from square of i. How the multiples before the square of current prime(i) getting marked ?Actually how are we so sure of it?

A famous company in our country recruits people by asking them to solve problems without using C++ STL. Their problemset having problems on Searching Techniques, BFS, DFS, Tree Traversal, Prefix Tree or Trie , Backtracking and A glimpse of Dynamic Programming and many others.

Now I wonder how one can solve problems without STL. Like when using BFS how one can not using queue, vector, etc.

Can anybody please help me how to solve problems without STL?

Problem:

A man just come to codeforces with basic knowledge of CP. After looking the contests and taking part in some of it,he decided to do a strange thing that is he will never solve a problem none but the DIV2-D.He decides that with all the 2/2.5/3 hours he will get in a live contest, he will use those time to solve his favourite problem DIV2-D.

But as his very early days in CP he does not know any DSA or anything. He become frightened and helpless that he might not be able to gain the capability to solve just his favourite problem DIV2-D.Now he ask for help?

What path he should follow now to fulfill his dream.Let help help him .But there is a condition you cannot say him practice more D.He need to know the learning and practicing roadmap for achiving his dream.

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