Problem Link Anyone please help me to understand the problem . For this example :

3 3 3 1 2 1 2 3

If we set `label[ 2 ] = 1 , label[ 3 ] = 2 , label[ 1 ] = 3 `

; Answer should be : 2 3 1 but answer is 3 2 1 , but why ?

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Problem Link Anyone please help me to understand the problem . For this example :

3 3 3 1 2 1 2 3

If we set `label[ 2 ] = 1 , label[ 3 ] = 2 , label[ 1 ] = 3 `

; Answer should be : 2 3 1 but answer is 3 2 1 , but why ?

How to solve this problem ? Thanks!

In Vanya and Exams problem ,minimum of min(avg * n - sum, r - a_{i}) will be added to the answer . But why do we subtract ( r — a_{i} ) ? Anyone else please help me .

Thanks in advance

Need a list of problems on Bitmask from novice to medium hard..I am a beginner. Thanks in advance :)

How could i solve this problem.Please help me to solve this problem. Thanks in advance :)

Here is two implementation of this problem

Using Sparse Table: Submission Using Segment Tree: Submission

Time took by Sparse Table is 0.336 s ,where Segment Tree took 0.324 s. Memory took by Sparse Table is 8732 KB ,where Segment Tree took 3640 KB.

Problem is static,thats why Sparse Table should be fast!But here i couldn't see this..Where is the problem?Could anyone please find that out ?

Thanks in advance :)

Could anyone please explain me clearly why we have to do

`low[vertex]=min(low[vertex],dis[i]);`

Yeah!I know that is for lowest discovery time determination.But my question is that all the time we put minimum time in the current node comparing with its adjacent nodes,but not increase the time.My question is two or more nodes cant visit at a time.But lowest time show that,it can do that.But how? Could anyone please clear me out?Actually,i read more than 5/6 resources,but i couldn't get that.There all says about subtree,ancestor,blah blah,but my memory couldn't make a sense for that,what they actually want to say.

Thanks in advance.

Following is the psuedocode of finding out articulation point.

```
time = 0
function DFS(adj[][], disc[], low[], visited[], parent[], AP[], vertex, V)
visited[vertex] = true
disc[vertex] = low[vertex] = time+1
child = 0
for i = 0 to V
if adj[vertex][i] == true
if visited[i] == false
child = child + 1
parent[i] = vertex
DFS(adj, disc, low, visited, parent, AP, i, n, time+1)
low[vertex] = minimum(low[vertex], low[i])
if parent[vertex] == nil and child > 1
AP[vertex] = true
if parent[vertex] != nil and low[i] >= disc[vertex]
AP[vertex] = true
else if parent[vertex] != i
low[vertex] = minimum(low[vertex], disc[i])
```

I have tried my level best to understand this problem.Could anyone else please explain me how i solve this problem?

Note: I know Binary Indexed tree and want to solve this problem by this algorithm.Google search couldn't make a sense for myself to solve this problem.Everywhere,i saw,sort query according to it's value 'k'.And suggestion was that sort your given array also..But,i failed to get that!Pure explanation needed!

Thanks in advance!

Z[k] would be longest substring starting at k which is also prefix of the string. Now R-i + 1 gives the length of remaining characters in the Z box. Since we copy the value of Z[k] to optimize the calculation of Z values we need to check whether these values are inside the boundary of the Z box

Why not we calculate the remaining part including k index?K index is in the substring,so why skip this index?

To practise more on Number Theory realted problem,is there any oj except spoj,uva where i'll get a catagorized problem? And want to know that which topics on Number theory should i konw for competitive programming?Let's help me by giving helpfull tutorial link.

Thanks!Thanks in advance ! Happy coding!

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