dp
λ°°μ΄ : μ§κΈκ» μ νν κ²λ€ μ€ μ΅λμ κ° (μ°μ 3κ° μ ν x)
dp[1] = drink[1]
dp[2] = drink[1] + drink[2]
...
dp[n] = Math.max(dp[n-2] + drink[n], drink[n-3] + drink[n-1] + drink[n])
dp[n] = Math.max(dp[n-1], dp[n]) /* μ°μ 2κ° μ ν x */