diff --git a/工具/修改题目数据库.py b/工具/修改题目数据库.py
index ba02fda5..307d7ffd 100644
--- a/工具/修改题目数据库.py
+++ b/工具/修改题目数据库.py
@@ -1,6 +1,6 @@
 import os,re,json
 """这里编辑题号(列表)后将在vscode中打开窗口, 编辑后保存关闭, 随后运行第二个代码块"""
-problems = "14725   ,14931"
+problems = "14995"
 
 def generate_number_set(string,dict):
     string = re.sub(r"[\n\s]","",string)
diff --git a/题库0.3/Problems.json b/题库0.3/Problems.json
index 97be7fce..c7606c88 100644
--- a/题库0.3/Problems.json
+++ b/题库0.3/Problems.json
@@ -392638,7 +392638,7 @@
         "objs": [],
         "tags": [],
         "genre": "解答题",
-        "ans": "(1) $2,5,8,11,8,5,2$;\\\\\n(2) $k=13$时$S_{2k-1}$取到最大, 最大值为$626$;\\\\\n(3) 共有四种满足要求的数列:\\\\\n第一种: $1,2,\\cdots,2^{m-2},2^{m-1},2^{m-1},2^{m-2},\\cdots,2,1$($2m$项), $S_{2022}=\\begin{cases}2^{m+1}-2^{2m-2022}-1, & 1500<m\\le 2022,\\\\2^{2022}-1, & m \\ge 2022;\\end{cases}$\\\\\n第二种: $1,2,\\cdots,2^{m-2},2^{m-1},2^{m-2},\\cdots,2,1$($2m-1$项), $S_{2022}=\\begin{cases}3\\cdot 2^{m+1}-2^{2m-2023}-1, & 1500<m\\le 2022,\\\\2^{2022}-1, & m \\ge 2022;\\end{cases}$\\\\\n第三种: $2^{m-1},2^{m-2},\\cdots,2,1,1,2,\\cdots,2^{m-2},2^{m-1}$($2m$项), $S_{2022}=\\begin{cases}2^m+2^{2022-m}-2, & 1500<m\\le 2022,\\\\2^m-2^{m-2022}, & m \\ge 2022;\\end{cases}$\\\\\n第四种: $2^{m-1},2^{m-2},\\cdots,2,1,2,\\cdots,2^{m-2},2^{m-1}$($2m$项), $S_{2022}=\\begin{cases}2^m+2^{2023-m}-3, & 1500<m\\le 2022,\\\\2^m-2^{m-2022}, & m \\ge 2022;\\end{cases}$",
+        "ans": "(1) $2,5,8,11,8,5,2$;\\\\\n(2) $k=13$时$S_{2k-1}$取到最大, 最大值为$626$;\\\\\n(3) 共有四种满足要求的数列:\\\\\n第一种: $1,2,\\cdots,2^{m-2},2^{m-1},2^{m-1},2^{m-2},\\cdots,2,1$($2m$项), $S_{2022}=\\begin{cases}2^{m+1}-2^{2m-2022}-1, & 1500<m\\le 2022,\\\\2^{2022}-1, & m \\ge 2022;\\end{cases}$\\\\\n第二种: $1,2,\\cdots,2^{m-2},2^{m-1},2^{m-2},\\cdots,2,1$($2m-1$项), $S_{2022}=\\begin{cases}3\\cdot 2^{m-1}-2^{2m-2023}-1, & 1500<m\\le 2022,\\\\2^{2022}-1, & m \\ge 2022;\\end{cases}$\\\\\n第三种: $2^{m-1},2^{m-2},\\cdots,2,1,1,2,\\cdots,2^{m-2},2^{m-1}$($2m$项), $S_{2022}=\\begin{cases}2^m+2^{2022-m}-2, & 1500<m\\le 2022,\\\\2^m-2^{m-2022}, & m \\ge 2022;\\end{cases}$\\\\\n第四种: $2^{m-1},2^{m-2},\\cdots,2,1,2,\\cdots,2^{m-2},2^{m-1}$($2m$项), $S_{2022}=\\begin{cases}2^m+2^{2023-m}-3, & 1500<m\\le 2022,\\\\2^m-2^{m-2022}, & m \\ge 2022;\\end{cases}$",
         "solution": "",
         "duration": -1,
         "usages": [],