Merge branch 'dev'
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commit
e69be89dc1
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import os,re,json
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import os,re,json
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"""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"""
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"""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"""
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problems = "14725 ,14931"
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problems = "14995"
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def generate_number_set(string,dict):
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def generate_number_set(string,dict):
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string = re.sub(r"[\n\s]","",string)
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string = re.sub(r"[\n\s]","",string)
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@ -392711,7 +392711,7 @@
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"objs": [],
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"objs": [],
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"tags": [],
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"tags": [],
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"genre": "解答题",
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"genre": "解答题",
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"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), $S_{2022}=\\begin{cases}2^m+2^{2023-m}-3, & 1500<m\\le 2022,\\\\2^m-2^{m-2022}, & m \\ge 2022;\\end{cases}$",
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"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), $S_{2022}=\\begin{cases}2^m+2^{2023-m}-3, & 1500<m\\le 2022,\\\\2^m-2^{m-2022}, & m \\ge 2022;\\end{cases}$",
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"solution": "",
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"solution": "",
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"duration": -1,
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"duration": -1,
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"usages": [],
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"usages": [],
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