Please speak English

何意味

E in my opinion is implementation-only problem, D requires some thought, but a lot easier implementation, so they are probably same difficulty

you don't need 2D prefix sums to solve F. the title tells you how to solve (Chebyshev)

Editorial mentions an inequality which I didn't know, I have another solution with dp.

$$$hbit(mask)$$$ is highest bit in $$$mask$$$ in 0-indexation ($$$-1$$$ if $$$mask = 0$$$)

We search for the longest non-increasing subsequence in $$$a_L, \ldots, a_R$$$, where $$$a_i = popcnt(i)$$$. Let's go from higher bits to lower. While $$$hbit(L) = hbit(R)$$$ we can just remove the bit. After that let $$$h = hbit(R)$$$. We split the interval into parts $$$[L, 2^h - 1]$$$ and $$$[2^h, R]$$$ and try to merge optimal subsequence from subsequences in these intervals. From the left interval we're only interested in the value of the last element, and from the right interval — only in the first value. Notice also, that the first $$$h$$$ bits in $$$2^h - 1$$$ are on and are off in $$$2^h$$$.

This leads to following dynamics: $$$suf[k][e]$$$ — length of the longest non-increasing subsequence on the interval $$$[L_k, 2^k - 1]$$$ with the last value being at least $$$e$$$. ($$$L_k$$$ means the first $$$k$$$ bits of $$$L$$$), and $$$pref[k][s]$$$ — length on the interval $$$[0, R_k]$$$ with the first value at most $$$s$$$. Answer is $$$\max_{mid} suf[h][mid] + pref[h][mid - 1]$$$. This $$$-1$$$ comes from the bit $$$h$$$ that is on in the second interval.

To find $$$suf[k][e]$$$ we will further split the interval by the highest bit. If $$$k$$$-th bit is on in $$$L$$$, then $$$suf[k][e] = suf[k - 1][e - 1]$$$, otherwise the two intervals are $$$[L, 2^{k-1} - 1]$$$ and $$$[2^{k-1}, 2^k - 1]$$$. If we switch bit $$$k-1$$$ off, then the interval becomes $$$[0, 2^{k-1} - 1]$$$, which is good since it doesn't depend on $$$L$$$ and $$$R$$$ at all, which leads to one more dynamic that will be precalced — $$$dp[k][s][e]$$$ — length on the interval $$$[0, 2^k]$$$ with start at most $$$s$$$ and end at least $$$e$$$.

All of these dps have similar transitions that look as follows:

$$$dp[k][s][e] = \max_{mid} dp[k - 1][s][mid] + dp[k - 1][mid - 1][\max(0, e - 1)]$$$

$$$suf[k][e] = \max_{mid} suf[k - 1][mid] + dp[k - 1][mid - 1][\max(0, e - 1)]$$$

$$$pref[k][s] = \max_{mid} dp[k - 1][s][mid] + pref[k - 1][mid - 1]$$$.

Let $$$K = 60$$$. $$$dp$$$ is calculated in $$$O(K^4)$$$, $$$suf$$$ and $$$pref$$$ are calculated in $$$O(K^3)$$$, so we get $$$O(\log^3 R)$$$ per testcase, that passes because of good constant factor.

I did too, but I implemented it from scratch. How did you just use

segtree<int, mn, e1> s1(n + 1); and segtree<int, mx, e2> s2(n + 1);?

Yes, I agree with you.