Well I don't think you should do this. Well my approach is to use the quadratic formula since you have this nice inequality here

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And since it is always rounding down so it can be guaranteed that the inequality is satisfied. So when coding it should looks something like this

    ll x = (-1 + sqrt((double)(8*r-8*l+1)))/2;
    if (l + x*(x+1)/2 > r) x--; // Just for sure (it can be omitted)
    cout<<x+1<<"\n";

You can view my solution here 279086455

When you build the array the difference between abjacent elements is like this:

1 2 3 4 5 6 ......

so if l = 10 and r = 20 the longest good array is 10 11 13 16 20

If you keep adding 1, 2, 3, 4, 5, 6 .... to build the next element in the array you will realize that after 44730 iterations you are already above 10^9 which is the max of r, this is why your n = 44730 works and without pre-computing and bruce forcing should work too

In this case, brute force works because of a mathematical concept known as the harmonic series, which states that 1 + 1/2 + 1/3 + 1/4 + ... + 1/n ≈ log n. For this problem, we would be adding 1 + 2 + 3 + 4 + ... + n such that the sum of all elements ≤ r-l, where the upper limit for r is 1e9 and the lower limit for l is 0. Therefore, we can assume that for each testcase, it runs in roughly log 1e9(note that log = log_2 and not log_10) operations. Multiply that by t, and we get roughly 300000(assuming my math isn't off) operations per test.

Hope this helps!

You dont need to precompute values, Just use the quadratic formula for the difference (r-l). you just need to equate n^2+n = 2*(r-l)

then we can use the quadratic formula to find the no of elements (n+1) in the array.

#include<bits/stdc++.h>
#include<cmath>
using namespace std;
#define int long long
signed main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin>>t;
    while(t--){
        int l,r;
        cin>>l>>r;
        int diff = r-l;
        int x = (8*diff)+1;
        int sq = sqrt(x);
        int ans = (sq-1)/2;
        cout<<ans+1<<endl;
    }
}

This code works fine if you want to try it for debugging purposes.

The maximum length of array $$$a$$$ is obtained when the difference of two consecutive differences is minimized ; Thus we set the difference of two consecutive differences is only 1 , For example consider $$$l=1,r=10$$$ thus we can consider the following array as the optimal choice

thus the maximum length is $4$ , Notice we can form an $$$\textbf{Optimal Model}$$$ , $$$a$$$ can be considered as :

since each time we've an element $l$ + Sum of First $$$n$$$ integers called Triangular Numbers

We want to determine the value of $x$ above i.e.

First approach Simulate the process with brute force in

for _ in range(int(input())):
    l,r=map(int,input().split())
    x=1
    m=1
    while(l<=r):
        l+=x
        x+=1
        m+=1
    if(l==r):
        print(m)
    else:
        print(m-1)

Second approach : $T(n)$ is increasing , thus we can use binary search in $$$\mathcal{O}(\log(r-l))$$$

def f(x):
    l,r=0,x
    while(l<=r):
        m=(l+r)//2
        v=(m*(m+1))//2
        if(v>x): r=m-1
        else: l=m+1
    return r+1
 
 
for _ in range(int(input())):
    l,r=map(int,input().split())
    print(f(r-l))

$$$\textbf{Third Approach}$$$ : We can reverse this with Quadratic Formula in $$$\mathcal{O}(1)$$$ :

Consider

Since we may have rational part We've to perform ceiling , thus

from math import *
def f(n):
    return ceil((sqrt(1+8*n)-1)/2)
for _ in range(int(input())):
    l,r=map(int,input().split())
    print(f(r-l+1))