We invite you to participate in CodeChef’s Starters113, this Wednesday, 20th December, rated for All.
Time: 8:00 PM — 10:30 PM IST
Joining us on the problem setting panel are:
Setters: Satyam satyam343 Kumar Roy.
Tester: Takuki tabr Kurokawa.
Statement Verifier: Kanhaiya notsoloud1 Mohan
Text Editorialists: Nishank IceKnight1093 Suresh.
Contest Admin : Satyam satyam343 Kumar Roy.
Written editorials will be available for all on discuss.codechef.com. Pro users can find the editorials directly on the problem pages after the contest. The video editorials of the problems will be available only to Pro users.
Also, if you have some original and engaging problem ideas, and you’re interested in them being used in CodeChef's contests, you can share them here.
Hope to see you participating.
Good Luck!
Omg satyam343 round,all problems of all divs!
Omg satyam343 round,all problems of all divs!
Omg satyam343 round,all problems of all divs!
Omg satyam343 round, all problems of all divs!
Omg satyam343 round, all problems of all divs!
As someone who has seen the problems, i regret missing the chance to participate
rejected a party invitation to give this contest , hopefully it goes well
Friendly reminder: The contest starts in less than 60 minutes. I hope you like the problems.
Here, we go again. Post contest solution discussion stream
unable to understand this counting in fun problem statement. It's too complicated.
You can use the comment section on Codechef to ask questions. Also 8 participants in Div 1 have ACed the problem. So, probably, the statement isn't that bad.
Loved the problemset, SCORESUM7 is such a cute problem.
Scores and ranks (for me at least) are not up-to-date until now ...
Same here, even after solving the problem i didnt got the points and its showing tick mark
Counting is Fun was amazing!
Glad that you liked it. It is one of my best problems.
I hope you liked the problems! I would love to hear your feedback.
By the way, do check out the intended solution for COUNTIFUN7. It is one of my favourite problems that I have authored.
Once you establish a nice bijection, the solution is just 4-5 lines of code.
Yes, a math problem is the best CP problems you've authored :clown:
Damn, nice solution. I solved it in a quite complicated way. This was my solution:
Let us fix the value of prefix minimum, call it $$$x$$$. Thus $$$1$$$, $$$2$$$,..., $$$x-1$$$ all occur after $$$x$$$. If the nearest smaller element occuring after $$$x$$$ is $$$y$$$ positions ahead, then contribution to answer is $$${y+k-1 \choose k}$$$. So if we denote by $$$ans(x,l)$$$ as the sum of $$${(t_2-1)+(k-1) \choose k}$$$ over all $$$1=t_1<t_2<t_3<...<t_x<t_{x+1}=l+1$$$ ($$$t_i$$$ are the relative positions of elements $$$<x$$$ wrt $$$x$$$) , Then contribution to the final answer for $$$x<V$$$ is $$$\sum_{l=1}^{n-1} ans(x,l)(x-1)!(n-1-x)!$$$ and for $$$x=V$$$ is $$$ans(V,n)(V-1)!(n-V)!$$$.
To determine $$$ans(x,l)$$$, we observe that $$$ans(x,l)=ans(x,l-1)+ans(x-1,l-1)$$$ for $$$x>1$$$ and $$$ans(1,l)={l+k-1 \choose k}$$$. So if we represent a generating function $$$G_x = \sum ans(x,l)z^l$$$, then $$$G_x = z(G_x+G_{x-1})\Rightarrow G_x=\frac{zG_{x-1}}{1-z}$$$ and $$$G_1$$$ is $$$z(1-z)^{-k-1}$$$. Thus $$$G_x=z^{x}(1-z)^{-k-x}\Rightarrow ans(x,l)={l+k-1 \choose l-x}={l+k-1\choose k+x-1}$$$
Plugging this in summation gives a telescopic sum, thus contribution for each $$$x$$$ can be found in $$$O(1)$$$.
What is mistake in this https://www.codechef.com/viewsolution/1036038527 can any one give any counter test
Can anybody give me a counter testcase for MAKE ALL EQUAL problem, here is the code https://www.codechef.com/viewsolution/1036034543
What is the solution to Make All Equal?
Consider the case $$$N = 8, M = 2$$$ and $$$A = [3, 4, 5, 8, 8, 8, 8, 8]$$$. How does the official solution make the array all equal in 6 steps?
Do operations on index (1,2) 3 times, (1,3) 2 times and (2,3) once.
i think your rating is bugged, it shows -279 in last edu round
I loved this problemset and liked MAKEALLEQUAL and SCORESUM7 problems
Maximise the Min. problem can be solved without dp too , using set.
it's just case works
no it isn't. It is more than that.
my solution only includes case works and little math
I was wondering what would be the solution to the MAKE ALL EQUAL problem , if we allowed exactly m operations?
Find the smallest $$$x$$$ which is greater than or equal to maximum element of initial $$$a$$$ and $$$(x \cdot n - \sum_{i=1}^{n}a_i) \mod m=0$$$. Finding $$$x$$$ is not that hard.
For $$$(x \cdot n - \sum_{i=1}^{n}a_i) \mod m=0$$$ condition, only $$$x\mod m$$$ matters and we have $$$m \leq n$$$.
It might be the case that such $$$x$$$ doesn't exist, in which case it is impossible to make all the elements same.
Now it might be the case that $$$\frac{(x \cdot n - \sum_{i=1}^{n}a_i)}{m} > x - a_{min}$$$. So you need to do some algebraic manipulations to find the optimal $$$x$$$.
Alternate method to Minimise max , using prefix sums and sets.