The subtasks are meant to reward all suboptimal solutions and are not tied to any specific methods. Here are some common constructions:

Method 1: $\sim {\left(N/2\right)}^2$ pairs

Set all $a_i=1$.

Then, set $a_{\lfloor N/2\rfloor }={10}^{10}$.

Any pair $(l,r)$ satisfying $1\le l\le \lfloor N/2\rfloor \le r\le N$ is valid.

Thus, there are approximately ${\left(N/2\right)}^2$ such pairs.

Method 2: $\to N(N+1)/2$ pairs

Recurse Method 1 on both halves of the current range. The values you set for the numbers in the middle should decrease a sufficient amount (e.g. various functions work, but floor dividing by four is a simple example).

This method approaches $N(N+1)/2$ pairs, but may run into issues for larger $N$. Fine tuning the decreasing function may result in full points.

Intended Solution: $N(N+1)/2$ pairs

Instead of building the array top down, we can build it bottom up. Define a recursive function that constructs a range of the array, $(L,R)$, ensuring that all pairs $(l,r)$ satisfying $L\le l\le r\le R$ are valid. The intended solution uses the following method:

1) If L > R, return.

2) If L == R, set $a_L=1$

• $(L,L)$ is a valid pair now. $✓$

3) Otherwise, let M = (L + R) / 2. Recurse on (L, M-1) and (M+1, R).

• All pairs $(l,r)$ satisfying $L\le l\le r\le M-1$ or $M+1\le l\le r\le R$ are valid now. $✓$

4) Set $a_M$ to $(a_L+a_{L+1}+\cdots +a_{M-1})+(a_{M+1}+a_{M+2}+\cdots +a_R)+1$

• $a_M$ is the minimum amount required to make all pairs $(l,r)$ satisfying $L\le l\le M\le r\le R$ valid. $✓$

Now, all pairs satisfying $L\le l\le r\le R$ are valid! $✓$

We can call this recursive function on (1, N) to receive full points. The maximum value that this construction uses when $N=2\times {10}^5$ is 7380642475, around $7\times {10}^9$. This fits comfortably under the ${10}^{10}$ limit.

Bonus

Python golf solution (87 bytes): https://dmoj.ca/src/6848763 (Credits to Riolku for the bulk of the logic + Josh for optimizing it with walrus operator)