毒酒桶 II

10 名仆人对应 10 位二进制信息，可区分 $2^{10}$ 种死亡模式。

把桶编号为 0..$n-1$ 的二进制；第 $k$ 位为 1 的桶由第 $k$ 名仆人去尝。

1 个月后死亡模式唯一对应桶编号。

因此最大 $n=2^{10}=1024$。

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Original Explanation

The kegs can be labeled in binary from $0$ to $n-1$, and the servants are labeled from $0$ to $9$. With $10$ servants, we can represent up to $10$ binary digits. Each keg label $i$ can be written as $i = a_{0i} + a_{1i} \cdot 2^1 + \dots + a_{9i} \cdot 2^9$, where $a_{0i}, a_{1i}, \dots, a_{9i} = 0$ or $1$. We give wine from keg $i$ to all servants whose indices match the binary sequence:

$$S_i = \{0 \leq k \leq 9 : a_{ki} = 1\}$$

For example, if $i = 17 = 0000010001_2$, servants $0$ and $4$ receive wine from keg $17$. After a month, the pattern of servants who die reveals which keg is poisoned. If servants $0$ and $4$ die, then keg $17$ is poisoned. This creates a one-to-one mapping between the kegs and binary sequences of length $10$. Thus, the maximum $n$ is $2^{10} = 1024$ kegs. We can't test more than this, as there are only $2^{10}$ subsets of servants, which wouldn't allow unique identification if $n > 1024$.