This is my Own Signature Solution!!! I spent 4 days just for this, and it surpassed the editorial's speed. The implementatoin is straightforward and highly intuitive. Hope you like it.

include

using namespace std;

int main(){ int t; scanf("%d", &t); long long a, b, dts, mod, ret;

while(t--) {
    scanf("%lld %lld", &a, &b); 

    ret = 0;

    //leading partial block 
    mod = a % 4;
    if(mod == 1) { ret = a + 1; a += 3; }//ret = a + 1 is key here, which reduces computational steps 
    else if(mod == 2) { ret = a + 1; a += 2; } 
    else if(mod == 3) { ++a; }

    //Full 4-blocks 
    dts = b - a + 1; 
    mod = dts / 4; 
    if(mod % 2 > 0) ret ^= 2; //xOr results in 2. 

    //Trailing partial blocks 
    mod = dts % 4; 
    if(mod == 1 || mod == 2) ret ^= b;
    else if(mod == 3) ret ^= 2; 

    printf("%lld\n", ret); 
}

return 0; 

}

Here is my Python code!

def xorSequence(l, r):
    if l == 0:
        n = r % 8
        if n == 0 or n == 1:
            return r
        elif n == 2 or n == 3:
            return 2
        elif n == 4 or n == 5:
            return r + 2
        return 0
    return xorSequence(0, r) ^ xorSequence(0, l - 1)

The same modulus 8 solution in JavaScript/Node.js:

function getXor(x) { let n = x % 8; if (n === 0 || n === 1) return x; if (n === 2 || n === 3) return 2; if (n === 4 || n === 5) return x + 2; return 0; }

function xorSequence(l, r) { return BigInt(getXor(r)) ^ BigInt(getXor(l - 1)); }

There is a pattern of 4 consecutive elements in the array A, for example: A[0] -> A[3], and so on. The first number of the sequence = it's index, the 2nd number is 1, the third = index of the fourth and the fourth = 0. And, when you xor all 4 together, you always get 2. When you xor 2 sequences, you get 0, so it doesn't affect the remaining xor. Base on that, you can just find the effective xor, which are from left to the next start sequence index and from right back to the index after the last 8-element sequence.

function getValue (idx) {
    let dp = 0n;
    let remain = idx % 4n;
    if(remain === 0n) dp = idx;
    if(remain === 1n) dp = 1n;
    if(remain === 2n) dp = idx+1n;
    if(remain === 3n) dp = 0n;   
    return dp;
}
function xorSequence(l, r) {
    [l, r] = [BigInt(l),BigInt(r)];
    if(l === r) return getValue(l);
    let output = getValue(l);
    let nextSequenceIdx = 4n*((l - l%4n)/4n + 1n);
    let leftoverIdxs  = (r-nextSequenceIdx+1n) % 8n;
    for(let i=l+1n; i<nextSequenceIdx;i++) {
        output = output^getValue(i);
    }
    for(let i=r; i>r-leftoverIdxs;i--) {
        output = output^getValue(i);
    }
    return output;
}

Haskell

So, if you look at the sequence of running XORs over the XORs, there's a pattern. I didn't give it much thought after that -- just implement the lookup function.

module Main where

import Control.Monad (replicateM_)
import Data.Bits (xor)

lu :: Int -> Integer
lu 0 = 1
lu 1 = 2
lu 2 = 2
lu x = do
    let (q, r) = (x - 3) `quotRem` 8
        b = 8 * (fromIntegral q) + 6 :: Integer
     in [b, b + 1, 0, 0, b + 2, b + 3, 2, 2] !! r

solve :: Int -> Int -> Integer
solve 1 b = lu (b - 1)
solve a b = xor (lu (a - 2)) (lu (b - 1))

main :: IO ()
main = do
    cases <- readLn
    replicateM_ cases $ do
        [a, b] <- map read . words <$> getLine
        print $ solve a b