The following is a series of monomer molecules $A_i$ that can undergo polymerization reactions with each other:

<MultiModal>![The monomer molecules are as follows: 1. CCC(O)=O 2. CCC=O 3. NCCN 4. O=CCCC=O 5. [C-]#[N+]CC[N+]#[C-] 6. O=C(CCC(O)=O)O 7. CC[N+]#[C-] 8. NCC](/media/uploads/8993e8ce-eaa2-406a-9062-1dc4e341cb28.png)</MultiModal>

The following is a series of repeating units $P_j$ of polymer, which are obtained from the above monomer molecules (in the figure, `*` indicates the connection site of the repeating unit fragment in the polymer):

<MultiModal>![The repeating units of the polymer are as follows: 1. O=C([*])CCC(N(CC)C(CCC(N(CC)[*])C(NCC)=O)C(NCC)=O)=O 2. [*]NCCNC(C(CC)N(C(CC)=O)CCN(C(C([*])=O)C)C(CC)=O)=O 3. O=C([*])CCC(N(CCN(C(CC)C(NCC)=O)[*])C(C(NCC)=O)CC)=O 4. O=C(NCC)C([*])N(C(CC)=O)CCN(C(C(NCC)=O)CC[*])C(CC)=O 5. O=C([*])C(CC)N(C(CCC(N(CC)C(C(NCCN[*])=O)CC)=O)=O)CC 6. CCN(C(CC)=O)C(CCC(N(CC)C(CC)=O)C(NCCNC([*])=O)=O)[*]，where [*] indicates the connection site of the repeating unit in the polymer](/media/uploads/27566190-e759-48c6-8536-cce21165f742.png)</MultiModal>

For each polymer, calculate the value of $z_j$:

$$
z_j = \prod_{i \in p_j} A_i\times \prod_{i \notin p_j} \frac{1}{A_i}
$$

Where:
- $p_j$: The set of monomers present in the repeating unit $P_j$
- $A_i$: The serial number of monomer $A_i$

Finally, calculate the sum of $z_j$:

$$
S = \sum_jz_j
$$

Select the correct option, requiring the final result to retain 3 significant figures, and select the option with an error within $1\%$ of the result; otherwise, select option **A**: All other options are incorrect.
以下一系列单体分子 $A_i$，其彼此之间可以发生聚合反应：

<MultiModal>![单体分子有以下几种：1. CCC(O)=O 2. CCC=O 3. NCCN 4. O=CCCC=O 5. [C-]#[N+]CC[N+]#[C-] 6. O=C(CCC(O)=O)O 7. CC[N+]#[C-] 8. NCC](/media/uploads/8993e8ce-eaa2-406a-9062-1dc4e341cb28.png)</MultiModal>

以下一系列高分子聚合物的重复单元 $P_j$，是由以上单体分子反应得到的（图中 `*` 表示重复单元片段在聚合物中的连接位点）：

<MultiModal>![聚合物的重复单元有以下几种：1. O=C([*])CCC(N(CC)C(CCC(N(CC)[*])C(NCC)=O)C(NCC)=O)=O 2. [*]NCCNC(C(CC)N(C(CC)=O)CCN(C(C([*])=O)C)C(CC)=O)=O 3. O=C([*])CCC(N(CCN(C(CC)C(NCC)=O)[*])C(C(NCC)=O)CC)=O 4. O=C(NCC)C([*])N(C(CC)=O)CCN(C(C(NCC)=O)CC[*])C(CC)=O 5. O=C([*])C(CC)N(C(CCC(N(CC)C(C(NCCN[*])=O)CC)=O)=O)CC 6. CCN(C(CC)=O)C(CCC(N(CC)C(CC)=O)C(NCCNC([*])=O)=O)[*]，其中 [*] 表示重复单元在聚合物中的连接位点](/media/uploads/27566190-e759-48c6-8536-cce21165f742.png)</MultiModal>

对于每种聚合物，计算 $z_j$ 的值：

$$
z_j = \prod_{i \in p_j} A_i\times \prod_{i \notin p_j} \frac{1}{A_i}
$$

其中：
- $p_j$：在重复单元 $P_j$ 中存在的单体集合
- $A_i$：单体 $A_i$ 的序号

最终，计算 $z_j$ 的总和：

$$
S = \sum_jz_j
$$

选择正确的选项，要求最终结果保留 3 位有效数字，选择与结果的误差在 $1\%$ 以内的选项，否则选择选项 **A**: 其他选项均不正确。
All other options are incorrect
52.9
44.7
10.1
5.31
2.18
0.454
55.8
70.5
91.5
150
其他选项均不正确
52.9
44.7
10.1
5.31
2.18
0.454
55.8
70.5
91.5
150
Observing the structure of the polymers, we find that each complex polymer repeating unit is generated through its corresponding four different monomers via the Ugi multi-component polymerization reaction.

<Checkpoint>[Each polymer is generated from four monomer molecules through Ugi multi-component polymerization](2)</Checkpoint>

| Polymer Repeating Unit | Corresponding Monomer Set ($p_j$) | Monomer Number |
|:--------------:|:----------------------:|:--------:|
| $P_1$          | $A_4, A_6, A_7, A_8$   | {4, 6, 7, 8} |
| $P_2$          | $A_1, A_2, A_3, A_5$   | {1, 2, 3, 5} |
| $P_3$          | $A_2, A_3, A_6, A_7$   | {2, 3, 6, 7} |
| $P_4$          | $A_1, A_3, A_4, A_7$   | {1, 3, 4, 7} |
| $P_5$          | $A_2, A_5, A_6, A_8$   | {2, 5, 6, 8} |
| $P_6$          | $A_1, A_4, A_5, A_8$   | {1, 4, 5, 8} |

<Checkpoint>[The corresponding monomer numbers for $P_1$ are 4, 6, 7, 8](1)</Checkpoint>

<Checkpoint>[The corresponding monomer numbers for $P_2$ are 1, 2, 3, 5](1)</Checkpoint>

<Checkpoint>[The corresponding monomer numbers for $P_3$ are 2, 3, 6, 7](1)</Checkpoint>

<Checkpoint>[The corresponding monomer numbers for $P_4$ are 1, 3, 4, 7](1)</Checkpoint>

<Checkpoint>[The corresponding monomer numbers for $P_5$ are 2, 5, 6, 8](1)</Checkpoint>

<Checkpoint>[The corresponding monomer numbers for $P_6$ are 1, 4, 5, 8](1)</Checkpoint>

We will use the following formula to calculate the value of each $z_j$. In the formula, $A_i$ represents the monomer number, $p_j$ is the set of monomer numbers that constitute the polymer $P_j$, and we assume that each monomer appears only once in the reaction (i.e., $n_{i,j}=1$).

$$
z_j = \frac{\prod_{i \in p_j} A_i}{\prod_{i \notin p_j} A_i}
$$

$$
z_1 = \frac{4 \times 6 \times 7 \times 8}{1 \times 2 \times 3 \times 5} = \frac{1344}{30} \approx 44.8000
$$

<Checkpoint>[$z_1 = 44.8000$](0.5)</Checkpoint>

$$
z_2 = \frac{1 \times 2 \times 3 \times 5}{4 \times 6 \times 7 \times 8} = \frac{30}{1344} \approx 0.0223
$$

<Checkpoint>[$z_2 = 0.0223$](0.5)</Checkpoint>

$$
z_3 = \frac{2 \times 3 \times 6 \times 7}{1 \times 4 \times 5 \times 8} = \frac{252}{160} = 1.5750
$$

<Checkpoint>[$z_3 = 1.5750$](0.5)</Checkpoint>

$$
z_4 = \frac{1 \times 3 \times 4 \times 7}{2 \times 5 \times 6 \times 8} = \frac{84}{480} = 0.1750
$$

<Checkpoint>[$z_4 = 0.1750$](0.5)</Checkpoint>

$$
z_5 = \frac{2 \times 5 \times 6 \times 8}{1 \times 3 \times 4 \times 7} = \frac{480}{84} \approx 5.7143
$$

<Checkpoint>[$z_5 = 5.7143$](0.5)</Checkpoint>

$$
z_6 = \frac{1 \times 4 \times 5 \times 8}{2 \times 3 \times 6 \times 7} = \frac{160}{252} \approx 0.6349
$$

<Checkpoint>[$z_6 = 0.6349$](0.5)</Checkpoint>

Finally, calculate:
$$
S = \sum_{j=1}^{6} z_j = z_1 + z_2 + z_3 + z_4 + z_5 + z_6
$$

$$
S \approx 44.8000 + 0.0223 + 1.5750 + 0.1750 + 5.7143 + 0.6349
$$

$$
S \approx 52.9
$$

<Checkpoint>[$S \approx 52.9$](0.5)</Checkpoint>
观察聚合物的结构我们发现，每个复杂的聚合物重复单元是通过其对应的四种不同单体，经由Ugi多组分聚合反应生成的。

<Checkpoint>[每种聚合物是由四个单体分子经过Ugi多组分聚合生成的](2)</Checkpoint>

| 聚合物重复单元 | 对应的单体集合 ($p_j$) | 单体序号 |
|:--------------:|:----------------------:|:--------:|
| $P_1$          | $A_4, A_6, A_7, A_8$   | {4, 6, 7, 8} |
| $P_2$          | $A_1, A_2, A_3, A_5$   | {1, 2, 3, 5} |
| $P_3$          | $A_2, A_3, A_6, A_7$   | {2, 3, 6, 7} |
| $P_4$          | $A_1, A_3, A_4, A_7$   | {1, 3, 4, 7} |
| $P_5$          | $A_2, A_5, A_6, A_8$   | {2, 5, 6, 8} |
| $P_6$          | $A_1, A_4, A_5, A_8$   | {1, 4, 5, 8} |

<Checkpoint>[$P_1$ 的对应单体序号为 4, 6, 7, 8](1)</Checkpoint>

<Checkpoint>[$P_2$ 的对应单体序号为 1, 2, 3, 5](1)</Checkpoint>

<Checkpoint>[$P_3$ 的对应单体序号为 2, 3, 6, 7](1)</Checkpoint>

<Checkpoint>[$P_4$ 的对应单体序号为 1, 3, 4, 7](1)</Checkpoint>

<Checkpoint>[$P_5$ 的对应单体序号为 2, 5, 6, 8](1)</Checkpoint>

<Checkpoint>[$P_6$ 的对应单体序号为 1, 4, 5, 8](1)</Checkpoint>

我们将使用以下公式计算每个 $z_j$ 的值。公式中，$A_i$ 代表单体的序号，$p_j$ 是构成聚合物 $P_j$ 的单体序号集合，我们假设每个单体在反应中只出现一次（即 $n_{i,j}=1$）。

$$
z_j = \frac{\prod_{i \in p_j} A_i}{\prod_{i \notin p_j} A_i}
$$

$$
z_1 = \frac{4 \times 6 \times 7 \times 8}{1 \times 2 \times 3 \times 5} = \frac{1344}{30} \approx 44.8000
$$

<Checkpoint>[$z_1 = 44.8000$](0.5)</Checkpoint>

$$
z_2 = \frac{1 \times 2 \times 3 \times 5}{4 \times 6 \times 7 \times 8} = \frac{30}{1344} \approx 0.0223
$$

<Checkpoint>[$z_2 = 0.0223$](0.5)</Checkpoint>

$$
z_3 = \frac{2 \times 3 \times 6 \times 7}{1 \times 4 \times 5 \times 8} = \frac{252}{160} = 1.5750
$$

<Checkpoint>[$z_3 = 1.5750$](0.5)</Checkpoint>

$$
z_4 = \frac{1 \times 3 \times 4 \times 7}{2 \times 5 \times 6 \times 8} = \frac{84}{480} = 0.1750
$$

<Checkpoint>[$z_4 = 0.1750$](0.5)</Checkpoint>

$$
z_5 = \frac{2 \times 5 \times 6 \times 8}{1 \times 3 \times 4 \times 7} = \frac{480}{84} \approx 5.7143
$$

<Checkpoint>[$z_5 = 5.7143$](0.5)</Checkpoint>

$$
z_6 = \frac{1 \times 4 \times 5 \times 8}{2 \times 3 \times 6 \times 7} = \frac{160}{252} \approx 0.6349
$$

<Checkpoint>[$z_6 = 0.6349$](0.5)</Checkpoint>

最终，计算：
$$
S = \sum_{j=1}^{6} z_j = z_1 + z_2 + z_3 + z_4 + z_5 + z_6
$$

$$
S \approx 44.8000 + 0.0223 + 1.5750 + 0.1750 + 5.7143 + 0.6349
$$

$$
S \approx 52.9
$$

<Checkpoint>[$S \approx 52.9$](0.5)</Checkpoint>