• + 5 comments

    We know that A[x] = 1 ^ 2 ^ 3 ^ ... ^ x.

    now, A[L] ^ A[L+1] ^ A[L+2] ^ ... ^ A[R] = A[1] ^ A[2] ^ ... A[L-1] ^ (A[L] ^ A[L+1] ^ ... ^ A[R]) ^ A[1] ^ A[2] ^ ... ^ A[L-1],

    because a ^ b = b ^ a, a ^ a = 0, and 0 ^ a = a.

    construct a function G(X) = A[1] ^ A[2] ^ ... ^ A[X].

    the answer is G(R) ^ G(L-1). search the patern for G(X).