Блог пользователя MagicSpark

Автор MagicSpark, история, 5 лет назад, По-английски

When I am solving 1197D,I didn't see m<=10 and then I find a solution that can solve m<=n<=3e5

Here is my solution:

The range [l,r] is equal to [l,l+m-1],[l+m,l+2m-1]....[l+xm,r]

then we will enumeration the value of l+xm.

We consider every reminder(from 0 to m-1) of position module m and solve them independently.

When we are solving each of the reminders,we have v[i]-->the value of the ith range(from the ith position fits the reminder to the next),then we will consider the value of two parts

First:[l,l+m-1],[l+m,l+2m-1].....[l+(x-1)m,l+xm-1]

Second:[l+xm,r]

We can use segment tree to get the max prefix sum of [l+mx,l+(m+1)x-1],then it's the second part.

When we are solveing the first part,we need to find the min value of the prefix sum of v[i].

This can also be done using segment tree.

Then the answer for each l+xm is First+Second-k.

Here is my submission https://codeforces.com/contest/1197/submission/57841454

The final answer is the maximum of them.

Overall complexy O(n log n) with big constants

I wanna know if there are better solutions.

If you have,please share under this blog.

  • Проголосовать: нравится
  • +24
  • Проголосовать: не нравится

»
5 лет назад, # |
  Проголосовать: нравится +8 Проголосовать: не нравится

I don't have one, but i found one

link

The first comment below tutorial QwQ

»
5 лет назад, # |
  Проголосовать: нравится +5 Проголосовать: не нравится

Knee new bee.

»
4 года назад, # |
  Проголосовать: нравится +15 Проголосовать: не нравится

Please use $$$\LaTeX$$$.