There is an operation program, can make: a ⊕ B = n (n is a constant), get (a + 1) ⊕ B = n + 1, a ⊕ (B + 1) = n-2, now known, 1 ⊕ 1 = 2, then 2013 ⊕ 2013=______ ;2014⊕2014=________

There is an operation program, can make: a ⊕ B = n (n is a constant), get (a + 1) ⊕ B = n + 1, a ⊕ (B + 1) = n-2, now known, 1 ⊕ 1 = 2, then 2013 ⊕ 2013=______ ;2014⊕2014=________

a♁b=n （a+1）♁b=n+1 （a+2）♁b=(a+1+1)♁b=n+1+1=n+2
By analogy, (a + k) ♁ B = n + K
By the same method, we can get a ♁ B = n a ♁ (B + 1) = n-2 a ♁ (B + 2) = n-2-2 = n-4
So we can get a ♁ (B + m) = n-2m
So (a + k) ♁ (B + m) = a ♁ (B + m) + k = a ♁ b-2m + K
2013♁2013=1♁1 -2×2012+2012=2-2012=-2010
2014♁2014=1♁1 -2×2013+2013=2-2013=-2011