Ok. It's $$$\mathcal O(\sqrt N \log N)$$$.

you are manually performing the operations, ofcourse it will give TLE

For some reason I still can't understand the idea behind B? I don't get why you can take a, b and perform a//g and b//g, when g != 1 for starters. How can that be the same thing?

First we iterate through all $$$s$$$, and check whether we can make all elements equal using exactly $$$s$$$ steps. We can easily get rid of the case where the sum doesn't fit and some other trivial cases, so now we try to find a way to make every element $$$V$$$ in $$$s$$$ steps.

Note that if we construct a $$$n\times n$$$ matrix $$$M$$$ where $$$0\le m_{i,j}\le s$$$ denotes the number of permutations where $$$p_i=j$$$, if the matrix satisfy the constraint that every row and every column sum up to $$$s$$$, we can always find $$$s$$$ permutations that corresponds to this matrix. This can be done using bipartite matching in $$$O(sn\sqrt n)$$$ time, which is obviously enough since $$$s\le 10^4$$$. Thus we try to find this matrix with the additional condition that for all $$$i$$$ , $$$\sum_{j=1}^n j\times m_{i,j}=V-a_i$$$.

Instead of the original matrix, we try to construct the suffix sum matrix $$$A$$$, such that $$$A_{i,j}=\sum_{k=j}^n m_{i,j}$$$, now we re-write the constraints on $$$A$$$: row sum equals $$$V-a_i$$$, column sum equals $$$(n-i+1)s$$$, $$$A_{i,1}=s$$$, and $$$A_{i,j}\ge A_{i,j+1}$$$.

Since the constraint on row sum is annoying, we first construct any such matrix without this condition. This can be done easily by greedy or whatever method. Then we try to adjust this matrix so that every row satisfy the constraint without breaking the other ones. Let $$$b_i=V-a_i-\sum_{j=1}^n A_{i,j}$$$, then if all $$$b_i=0$$$, we are finished. Otherwise, WLOG, suppose $$$b_1$$$ is the smallest one, then we need to move some other elements to this row to make this row good. To do this, we simply iterate through all $$$i$$$ such that $$$b_i\ge 0$$$, and for every $$$j$$$, we can calculate the maximum value that can be subtracted from $$$A_{i,j}$$$ and can be added to $$$A_{1,j}$$$ without violating other constraints (we can only care about whether $$$A_{i,j}\ge A_{i,j+1}$$$ here because the other conditions are always ok). We keep doing this until all $$$b_i=0$$$, in which case we have constructed the matrix succesfully, or we can't make the smallest value bigger, which results in a failure.

Now we analyse the complexity. Let $$$v$$$ be the number of $$$i,j$$$ such that $$$A_{i,j}\ne A_{i,j+1}$$$. Since when we move every time, we make at least one $$$A_{i,j}=A_{i,j+1}$$$ or we make at least one $$$b_i=0$$$. Though it is possible that previously $$$A_{i,j}=A_{i,j+1}$$$, and later we make $$$A_{i,j+1}=A_{i,j+2}$$$, when we iterate through all $$$j$$$, $$$v$$$ decreases by at least $$$1$$$ every time we move from an $$$i$$$, and $$$v$$$ never increases. Since $$$v$$$ is at most $$$O(n^2)$$$, and it takes $$$O(n)$$$ time to move from $$$i$$$, the complexity for checking is $$$O(n^3)$$$. Since the answer is always less than $$$O(n^2)$$$, the total complexity is $$$O(n^5)$$$. However, we can never actually reach this upper bound since 1. the minimum number of operations is far fewer than $$$n^2$$$, and 2. the number of operations needed to adjust the matrix is far fewer than $$$n^3$$$ since it is just a theoretical bound.