can you please explain your solution?

I solved it with divide and conquer.

I used sqrt-decomposition with time complexity $$$O(N^{1.5} \log N)$$$.

First, for a fixed $$$R$$$, we define $$$v_L = \min(A_L, A_{L+1}, A_{L+2},..., A_R) - \min(B_L, B_{L+1}, B_{L+2},..., B_R)$$$. When $$$R$$$ increases by $$$1$$$, some value of $$$v$$$ changes, but the overall change is mostly composed with few range-add. Actually, the overall number of range-adds is $$$O(N)$$$.

So, we can transform into the following problem:
You are given $$$N$$$, and $$$Q$$$ queries. The array $$$c_1, c_2, ..., c_N$$$ is initially all-zero, and there are two types of queries.
Type 1: You are given $$$l, r, x$$$. Increase $$$c_l, c_{l+1}, ..., c_r$$$ by $$$x$$$.
Type 2: You are given $$$a, b, r$$$. Calculate the number of $$$i \ (i \leq r)$$$ which is $$$a \leq c_i \leq b$$$.
This query can be answered in $$$O(\sqrt{N} \log N)$$$ when we do sqrt-decomposition of array.
In this problem, $$$a = -K$$$ and $$$b=K$$$. And also, as stated above, $$$Q = O(N)$$$. Hence, the overall time complexity is $$$O(N^{1.5} \log N)$$$.

My solution counts the number of intervals $$$[L, R]$$$ s.t. $$$max(a[L, R]) < max(b[L, R]) - K$$$. For each $$$i$$$ let $$$[u, v]$$$ be the interval in which $$$b[i]$$$ is maximal. We'll take the smaller interval between $$$[u, i]$$$ and $$$[i, v]$$$. Assuming it's $$$[u, i]$$$. Let's iterate $$$j$$$ over $$$[u, i]$$$ and find the largest $$$k$$$ s.t. $$$max(a[j, k]) < b[i] - K$$$. Subtracting $$$k - i + 1$$$ from the answer.

See here https://mirror.codeforces.com/blog/entry/66765?#comment-508361

Find R1 and R2 using some value which gives strictly increasing value (I use 50 as 2^50*R1 > 2^25*100(as 100 can be the max value of any R)). So, we get R1 and R2.

Now, use some value which give increasing sequence for R3 and R4 as well (I use 168). After that we have two eqn and two variables. So, just solve for R5 and R6!

Here's my solution.

First, write down a table of $$$[n/i]$$$ for various $$$n$$$ and $$$i$$$ from $$$1$$$ to $$$6$$$. I picked two lines:

  n   n/1 n/2 n/3 n/4 n/5 n/6
 55:   55  27  18  13  11   9
185:  185  92  61  46  37  30

Now, when we ask for day 185, we get the value $$$v_{185} = (r_1 \cdot 2^{185} + r_2 \cdot 2^{192} + r_3 \cdot 2^{61} + r_4 \cdot 2^{46} + r_5 \cdot 2^{37} + r_6 \cdot 2^{30}) \bmod 2^{63}$$$. Everything after power $$$63$$$ is lost. Note that $$$r_i$$$ are from $$$0$$$ to $$$100$$$, so when we have at least $$$7$$$ powers of two between consecutive values, we can just restore them directly. Thus

r6 = (v185 >> 30) & 127;
r5 = (v185 >> 37) & 127;
r4 = (v185 >> 46) & 127;

The second question will be about $$$v_{55} = (r_1 \cdot 2^{55} + r_2 \cdot 2^{27} + r_3 \cdot 2^{18} + r_4 \cdot 2^{13} + r_5 \cdot 2^{11} + r_6 \cdot 2^{9}) \bmod 2^{63}$$$. Here, note that $$$r_4$$$ influences the bits of $$$r_3$$$, $$$r_5$$$ can influence $$$r_4$$$, and $$$r_6$$$ can influence $$$r_5$$$. But as we know them already, calculate $$$v_{55}' = v_{55} - r_4 \cdot 2^{13} - r_5 \cdot 2^{11} - r_6 \cdot 2^{9}$$$. Now the restoration is straightforward:

r3 = (v55p >> 18) & 127;
r2 = (v55p >> 27) & 127;
r1 = (v55p >> 55) & 127;

Number of rings after day $$$d$$$ is $$$\sum_{i=1..6}R_i2^{x_i}$$$ where $$$x_i=\lfloor{d/i}\rfloor$$$. If $$$2^{x_5}\cdot maxR < 2^{x_6}$$$ then no numbers overlap in the binary representation of the sum and we can read all $$$R$$$s that fit in $$$2^{63}$$$. So at $$$d=210$$$ (lowest solution of above inequality) we can read $$$R_4$$$, $$$R_5$$$, $$$R_6$$$ directly. Once we know them, we ask the same condition $$$2^{x_2}\cdot maxR < 2^{x_3}$$$ for the next $$$R$$$s and the solution $$$d=42$$$ gives $$$R_1$$$, $$$R_2$$$ and $$$R_3$$$ directly after subtracting the known sum of $$$R_42^{x_4}+R_52^{x_5}+R_62^{x_6}$$$.

i'm old man. 54 years old.

when i read it i thought wtf how second problem can be so hard. i looked at the scoreboard and my place was 220. that's why my idea was the following: print 220 and 54.

When you print 220, you get $$$2^{55} \cdot r_4 + 2^{44} \cdot r_5 + 2^{36} \cdot r_6$$$. $$$r_i <= 100$$$, so you need maximum $$$7$$$ bits to store such a number. That's why representations of $$$r_4$$$, $$$r_5$$$ and $$$r_6$$$ are non-intersecting.

Printing 54 use the idea above for $$$r_1,r_2,r_3$$$, knowing $$$r_4,r_5,r_6$$$ and subtracting them from input given.

There's an easier solution to B. Take the first query to be 56, you can get r1 and r2 as you said. Take the next query to be 170. Since 2^170 and 2^85 are greater than 2^63, they'll get wiped out after mod. Using the above technique like for 56, you can find r3 and r4. Now for r5 and r6 you can simply use the 2 above queries and solve the system of 2 linear equations using Cramer's rule.

If I understand your approach correctly, multiple pairs of values for r3,4... can satisfy the equation

You should read $$$w$$$ after $$$T$$$ and judge's response after writing your answer

You forgot to input w :/

Compress the coordinates. There will be up to P + 2 different coordinates for each axis. Then you can select the best one among all intersection in the compressed grid.
Code

The problem is independent for each coordinate.

One set, no binsearch, no inclusion-exclusion, code.

We subtract numbers of bad intervals by passing through all possible skill values. Now it works for $$$O(\max(c_i,d_i)\cdot\log{n})$$$, but can be easily reduced to $$$O(n\cdot\log{n})$$$ by compressing the skill values (but why should one do it).

Is there a way to see scores by language?

I'm surprised not finding Greenland in the list. :(

Ofcourse a sparse table on array D would reduce the extra logN for range max queries to get this down to O(N*logN)

I've implemented my O(n) idea and it works — although again, care needs to be taken to handle ties correctly.