diff --git a/农作物种植策略/solve/solve.py b/农作物种植策略/solve/solve.py
new file mode 100644
index 0000000..3409239
--- /dev/null
+++ b/农作物种植策略/solve/solve.py
@@ -0,0 +1,104 @@
+import pulp as pl
+import pandas as pd
+from itertools import product
+
+df1 = pd.read_excel (io='附件1.xlsx', sheet_name = "乡村的现有耕地")
+df2 = pd.read_excel (io='附件2.xlsx', sheet_name = "2023年的农作物种植情况").fillna(method='ffill')
+df3 = pd.read_excel (io='附件2.xlsx', sheet_name = "2023年统计的相关数据")
+
+trans_dict = {"A": "平原地", "B": "林地", "C": "山坡地", "D": "水浇地", "E": "普通大棚", "F": "智能大棚"}
+# 地块定义
+A = ["A"+str(i) for i in range(1,7)]
+B = ["B"+str(i) for i in range(1,15)]
+C = ["C"+str(i) for i in range(1,7)]
+D = ["D"+str(i) for i in range(1,9)]
+E = ["E"+str(i) for i in range(1,17)]
+F = ["F"+str(i) for i in range(1,5)]
+
+# 作物定义
+# 粮食（豆类）
+LDN = ["黄豆", "黑豆", "红豆", "绿豆", "爬豆"]  
+LDI = list(range(1, 6))
+
+# 粮食
+LN = ["小麦", "玉米", "谷子", "高粱", "黍子", "荞麦", "南瓜", "红薯", "莜麦", "大麦"]  
+LI = list(range(6, 16))
+
+# 水稻类
+LSN = ["水稻"]  
+LSI = list(range(16, 17))
+
+# 蔬菜（豆类）
+SDN = ["豇豆", "刀豆", "芸豆"]  
+SDI = list(range(17, 20))
+
+# 蔬菜水浇地第一季
+SN = ["土豆", "西红柿", "茄子", "菠菜", "青椒", "菜花", "包菜", "油麦菜", "小青菜",
+        "黄瓜", "生菜", "辣椒", "空心菜", "黄心菜", "芹菜"]
+SI = list(range(20, 35))
+
+# 蔬菜水浇地第二季
+SAN = ["大白菜", "白萝卜", "红萝卜"]
+SAI = list(range(35, 38))
+
+# 食用菌普通大棚第二季
+SYJN = ["榆黄菇", "香菇", "白灵菇", "羊肚菌"]
+SYJI = list(range(38, 42))
+
+K = [1,2] # 季
+Y = [y for y in range(2024,2031)] # 年
+I = A + B + C + D + E + F # 全部地块
+J = LDN + LN + LSN + SDN + SN + SAN + SYJN # 所有作物
+
+V_1 = list(product(A + B + C, LDI + LI, [K[0]], Y))
+V_2 = list(product(D, LSI, [K[0]], Y))
+V_3 = list(product(D + E + F, SDI + SI, [K[0]], Y))
+V_4 = list(product(F, SDI + SI, [K[1]], Y))
+V_5 = list(product(D, SAI, [K[1]], Y))
+V_6 = list(product(E, SYJI, [K[1]], Y))
+
+V = V_1 + V_2 + V_3 + V_4 + V_5 + V_6
+
+W1 = product(A + B + C + D + E + F, [K[0]], Y) # 第一季
+W2 = product(D + E + F, [K[1]], Y) # 第二季
+W = list(W1) + list(W2)
+
+# 数据
+# 从地块名称转到序号
+I2N = {"A": "平旱地", "B": "梯田", "C": "山坡地", "D": "水浇地", "E": "普通大棚", "F": "智慧大棚"}
+N2I = {"平旱地": A, "梯田": B, "山坡地": C, "水浇地": D, "普通大棚": E, "智慧大棚": F}
+q = {(I2N[df2["种植地块"][i]], df2["作物编号"][i],
+        K[1] if df2["种植季次"][i] == "第二季" else K[0]):
+            df3["种植成本/(元/亩)"][i]
+        for i in range(len(df2))}
+
+def get_data(name:str):
+    return {(df3["地块类型"][i].strip(), df3["作物编号"][i], 
+           K[1] if df3["种植季次"][i] == "第二季" else K[0]): 
+           df3[name][i]
+            for i in range(len(df3))}
+
+r = get_data("亩产量/斤")
+c = get_data("种植成本/(元/亩)")
+p = get_data("销售单价/(元/斤)")
+
+
+# 问题求解
+problem = pl.LpProblem("Agricultural Planting",sense=pl.LpMinimize)
+
+x = pl.Dict([V,Y])
+y = pl.LpVariable("y")
+z = y > x
+
+problem += min([x,y,z]) 
+for var in vars:
+    problem += (x+y > 8)
+    problem += (x+y+z > 13)
+    problem += (x+z > 7)
+
+problem.solve()
+
+
+for v in problem.variables():
+    print(v.name,"=",v.varValue)
+print(pl.value(problem.objective))
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diff --git a/随机模型/main.md b/随机模型/main.md
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--- /dev/null
+++ b/随机模型/main.md
@@ -0,0 +1,101 @@
+# 随机模型
+
+## 第一题
+
+### 1.1
+
+旧赛制进球数概率
+
+|        | 0         | 1         | 2     |
+| ------ | --------- | --------- | ----- |
+| 先 $X$ | $(1-p)^2$ | $2p(1-p)$ | $p^2$ |
+| 后 $Y$ | $(1-q)^2$ | $2q(1-q)$ | $q^2$ |
+
+$P\{先手胜\}=P\{X>Y\}=2p^2q(1-q)+2(1-q)^2p(1-p)+p^2(1-q)^2=(1-q)p[2pq+2(1-p)(1-q)+p(1-q)]$
+
+$P\{后手胜\}=P\{X<Y\}=2q^2p(1-q)+2(1-p)^2q(1-q)+q^2(1-p)^2=2(1-p)q[2pq+2(1-p)(1-q)+q(1-p)]$
+
+$P\{相同\}=P\{X=Y\}=[pq+(1-p)(1-q)]^2$
+
+新赛制进球数概率
+
+
+|        | 0            | 1               | 2    |
+| ------ | ------------ | --------------- | ---- |
+| 先 $X$ | $(1-p)(1-q)$ | $p(1-q)+q(1-p)$ | $pq$ |
+| 后 $Y$ | $(1-p)(1-q)$ | $p(1-q)+q(1-p)$ | $pq$ |
+
+ $P\{先手胜\}=P\{X>Y\}=p(1-p)[q^2+(1-q)^2]+q(1-q)[p^2+(1-p)^2]+pq(1-p)(1-q)$
+
+$P\{后手胜\}=P\{X<Y\}=p(1-p)[q^2+(1-q)^2]+q(1-q)[p^2+(1-p)^2]+pq(1-p)(1-q)$
+
+$P\{相同\}=P\{X=Y\}=[(1-p)(1-q)]^2+[p(1-q)+q(1-p)]^2+p^2q^2$
+
+在 $p=\frac{3}{4},q=\frac{2}{3}$ 时
+
+|      | 先胜             | 后胜             | 平               |
+| ---- | ---------------- | ---------------- | ---------------- |
+| 旧   | $\frac{17}{48}$  | $\frac{2}{9}$    | $\frac{61}{144}$ |
+| 新   | $\frac{41}{144}$ | $\frac{41}{144}$ | $\frac{62}{144}$ |
+
+### 1.2
+
+无论在新旧规则下，加赛第一轮都是先罚球队先罚球，后罚球队后罚球。
+
+所以，先罚球队获胜概率为: $p(1-q)$，后罚球队获胜概率为: $q(1-p)$
+
+## 第二题
+
+### 2.1
+
+若 A 第一轮得分，则获胜，若不得分，若对手也不得分，则进入下一轮循环
+
+所以概率为
+$$
+P\{A获胜\}=(\alpha+\beta)\sum\limits_{i=0}\gamma^i=\frac{\alpha+\beta}{1-\gamma^2}
+$$
+
+### 2.2
+
+$$
+P\{A获胜\}=\alpha+a\beta+b\gamma
+$$
+
+考虑 $a$ 的情况
+
+1. B 不得分，A胜利 
+2. B射门，A在突然死亡中胜利
+
+所以 $a=\gamma+\beta\frac{\alpha+\beta}{1-\gamma^2}$
+
+考虑 $b$ 的情况
+
+B 在突然死亡中获胜的概率为$\frac{\alpha+\beta}{1-\gamma^2}$ ，则 A 获胜的概率 $b=P\{\bar{B}\}=1-P\{B\}=1-\frac{\alpha+\beta}{1-\gamma^2}$
+
+### 2.3
+
+由 **2.2** 的结论可知 
+$$
+P\{A获胜\}=\alpha+(\gamma+\beta\frac{\alpha+\beta}{1-\gamma^2})\beta+(1-\frac{\alpha+\beta}{1-\gamma^2})\gamma
+$$
+修改以前，代入 $\alpha+\beta=1-\gamma$ 我们可以知道
+$$
+P\{A获胜\}=\frac{\alpha+\beta}{1-\gamma^2}=\frac{1}{1+\gamma}
+$$
+所以，先手获胜的概率是更大的，尤其是不得分的概率很小时
+
+修改之后， 
+
+1. $\gamma$ 很大, 即 $\gamma\to1$ 时
+
+   $P=\frac{\gamma^2}{1+\gamma}=\frac{1}{2}$
+
+2. $\gamma$ 很小, 即 $\gamma\to0$ 时
+
+   $P=\alpha+\beta^2$ ，此时我们令 $\alpha=1-\beta$, 可知 $P=1-\beta+\beta^2$
+
+   ![](.\res\figure1.png)
+
+   先手仍更有可能获胜，但是在 $\beta$ 没有很大或很小的情况时，比修改之前小。
+
+   注：$\beta$ 很大导致先手获胜概率大是因为第三轮突然死亡先手得分概率很大，$\beta$ 很小则先手在第一轮就容易获胜。如果要使得先手后手更加公平，需要继续推迟突然死亡的机制
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