Omg satyam343 round,all problems of all divs!

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.

Glad that you liked it. It is one of my best problems.

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 $$$O(1)$$$.

Do operations on index (1,2) 3 times, (1,3) 2 times and (2,3) once.

it's just case works

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.

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 $$$x$$$.