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        <title>CVBB ACM Team 2020-2021:teams:the_great_wave_off_kanagawa</title>
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       <dc:date>2026-04-29T23:35:17+0800</dc:date>
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        <dc:date>2020-07-31T17:40:53+0800</dc:date>
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        <title>2020-2021:teams:the_great_wave_off_kanagawa:front_page</title>
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        <description>The Great Wave off Kanagawa

周报

2020/05/02 -- 2020/05/08 周报

2020/05/09 -- 2020/05/15 周报

2020/05/16 -- 2020/05/22 周报

2020/05/23 -- 2020/05/29 周报

2020/05/30 -- 2020/06/05 周报

2020/07/12 -- 2020/07/18 周报

2020/07/19 -- 2020/07/25 周报

2020/07/26 -- 2020/07/31 周报

个人

airbust

kazamori

Ket98

Wiki

隔板法（Stars and bars）</description>
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        <dc:format>text/html</dc:format>
        <dc:date>2020-06-01T19:34:55+0800</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>2020-2021:teams:the_great_wave_off_kanagawa:stars_and_bars</title>
        <link>https://wiki.cvbbacm.com/doku.php?id=2020-2021:teams:the_great_wave_off_kanagawa:stars_and_bars&amp;rev=1591011295&amp;do=diff</link>
        <description>格式：英文/公式两边接汉字时请空格。

内容：建议找几个例题。

隔板法（Stars and bars）

隔板法是解决某些组合问题的一种数学技术。每当要计算对相同对象进行分组的方式时，就可以使用隔板法。$n$$ k $$$\binom {n+k-1} {n}。$$$\bigstar | \bigstar \bigstar |~| \bigstar \bigstar$$n$$k-1$$n$$k-1$$n$$k$$n$$k-1$$x_1 + x_2 + \dots + x_k = n,x_i \ge 0$$n = 4$$k = 3$$1 + 3 + 0 = 4$$\bigstar | \bigstar \bigstar \bigstar |$$\binom {n + k-1} {n}$$x_i \ge a_i$$$x_1 + x_2 + \dots + x_k = n$$$x_i':= x_i-a_i$$$(x_1' + a_i) + (x_2' + a_i) + \dots + (x_k' + a_k) = n$$$$\Leftrightarrow ~ ~ x_1' + x_2' + …</description>
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        <dc:date>2020-05-25T20:45:33+0800</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>2020-2021:teams:the_great_wave_off_kanagawa:week_summary_1</title>
        <link>https://wiki.cvbbacm.com/doku.php?id=2020-2021:teams:the_great_wave_off_kanagawa:week_summary_1&amp;rev=1590410733&amp;do=diff</link>
        <description>2020/05/02 -- 2020/05/08 周报

本周推荐

airbust

CF 1342D Mulitple Cases

先求出一共要多少个case，假设大于等于$i$的$m_i$个数是$b_i$，根据抽屉原理，case的个数要大于等于$\lceil \frac {b_i}{c_i} \rceil$，所以一共需要$ans=max(\lceil \frac {b_i}{c_i} \rceil)$个case。然后是构造方案，将$m_i$$m_i$$(i\ mod\ ans)$$n$$a_i$$b_i$$k$$n$$i$$i$$j$$0\leq i&lt;n$</description>
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        <dc:date>2020-05-16T23:31:31+0800</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>2020-2021:teams:the_great_wave_off_kanagawa:week_summary_2</title>
        <link>https://wiki.cvbbacm.com/doku.php?id=2020-2021:teams:the_great_wave_off_kanagawa:week_summary_2&amp;rev=1589643091&amp;do=diff</link>
        <description>2020/05/09 -- 2020/05/15 周报

本周推荐

airbust

CF 1350C Orac and LCM

	*  分类：数论
	*  简要题意： 给定一个长度为$n$的数组，求$gcd\{lcm(a_i,a_j)|i&lt;j\}$
	*  解法： 对于$a_i$，产生的lcm有$lcm(a_i,a_1),\ldots,lcm(a_i,a_{i-1}),lcm(a_i,a_{i+1}),\ldots,lcm(a_i,a_n)$，则它们的gcd为$gcd_i=gcd(lcm(a_i,a_1),\ldots,lcm(a_i,a_{i-1}),lcm(a_i,a_{i+1}),\ldots,lcm(a_i,a_n))$，由于它们中的每一项都含有$a_i$$a_i$$gcd_i$$gcd_i=lcm(a_i,gcd(a_1,\ldots,a_{i-1},a_{i+1},\ldots,a_n))$$gcd(gcd_1,\ldots,gcd_n)$$n$$a_i$$b_i$$k$$n$$i$$i$$j$$M$$N$$K$$998244353$$i$$N-1$$N-1-i$$N…</description>
    </item>
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        <dc:format>text/html</dc:format>
        <dc:date>2020-05-25T21:56:00+0800</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>2020-2021:teams:the_great_wave_off_kanagawa:week_summary_3</title>
        <link>https://wiki.cvbbacm.com/doku.php?id=2020-2021:teams:the_great_wave_off_kanagawa:week_summary_3&amp;rev=1590414960&amp;do=diff</link>
        <description>2020/05/16 -- 2020/05/22 周报

本周推荐

airbust

CF 1354D Multiset

	*  分类：二分，数据结构
	*  简要题意： 给定一个长度为$n(n \leq 1e6)$的数组$a_1,\ldots,a_n$，$q$次询问，每次插入一个数或删除第$k$小数，保证每次操作有$1 \leq a_i \leq n$，输出最后结果$1$$n$$k$$k$$$ans=\frac{cos(\frac{\pi}{4n})}{sin(\frac{\pi}{2n})}$$$M$$N$$K$$998244353$$i$$N-1$$N-1-i$$N-i$$M$$M-1$$0\le i \le K$</description>
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        <dc:format>text/html</dc:format>
        <dc:date>2020-05-29T23:06:57+0800</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>2020-2021:teams:the_great_wave_off_kanagawa:week_summary_4</title>
        <link>https://wiki.cvbbacm.com/doku.php?id=2020-2021:teams:the_great_wave_off_kanagawa:week_summary_4&amp;rev=1590764817&amp;do=diff</link>
        <description>2020/05/23 -- 2020/05/29 周报

本周推荐

airbust

CF 1354D Multiset

	*  分类：二分，数据结构
	*  简要题意： 给定一个长度为$n(n \leq 1e6)$的数组$a_1,\ldots,a_n$，$q$次询问，每次插入一个数或删除第$k$小数，保证每次操作有$1 \leq a_i \leq n$，输出最后结果$1$$n$$k$$k$$$ans=\frac{cos(\frac{\pi}{4n})}{sin(\frac{\pi}{2n})}$$$M$$N$$K$$998244353$$i$$N-1$$N-1-i$$N-i$$M$$M-1$$0\le i \le K$</description>
    </item>
    <item rdf:about="https://wiki.cvbbacm.com/doku.php?id=2020-2021:teams:the_great_wave_off_kanagawa:week_summary_5&amp;rev=1591534866&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2020-06-07T21:01:06+0800</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>2020-2021:teams:the_great_wave_off_kanagawa:week_summary_5</title>
        <link>https://wiki.cvbbacm.com/doku.php?id=2020-2021:teams:the_great_wave_off_kanagawa:week_summary_5&amp;rev=1591534866&amp;do=diff</link>
        <description>2020/05/30 -- 2020/06/05 周报

本周推荐

airbust

无

kazamori

CF 1354C2  Not So Simple Polygon Embedding 

	*  分类：计算几何
	*  简要题意： 给出奇数n，求覆盖边数为 2n 边长为 1 的正凸多边形的最小正方形的边长。 $$ans=\frac{cos(\frac{\pi}{4n})}{sin(\frac{\pi}{2n})}$$$M$$N$$K$$998244353$$i$$N-1$$N-1-i$$N-i$$M$$M-1$$0\le i \le K$</description>
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        <dc:format>text/html</dc:format>
        <dc:date>2020-07-24T14:38:32+0800</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>2020-2021:teams:the_great_wave_off_kanagawa:week_summary_6</title>
        <link>https://wiki.cvbbacm.com/doku.php?id=2020-2021:teams:the_great_wave_off_kanagawa:week_summary_6&amp;rev=1595572712&amp;do=diff</link>
        <description>2020/07/12 -- 2020/07/18 周报

本周推荐

airbust

AIsing Programming Contest 2020 E Camel Train

分类：贪心，数据结构

题意：有$n$只骆驼。对于第$i$只骆驼，给出三个正整数$K_i,L_i,R_i$​，表示如果把这只骆驼放在前$K_i$的位置，就会有收益$L_i$$R_i$$L$$R$$l&gt;r$$k$$k$$r$$L$$\{k,l-r\}$$l&lt;r$$n-k$$n-k$$l$$R$$\{n-k,r-l\}$$l=r$$L,R$$n$$A$$B$$ABO$$B$$AO$</description>
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    <item rdf:about="https://wiki.cvbbacm.com/doku.php?id=2020-2021:teams:the_great_wave_off_kanagawa:week_summary_7&amp;rev=1595583502&amp;do=diff">
        <dc:format>text/html</dc:format>
        <dc:date>2020-07-24T17:38:22+0800</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>2020-2021:teams:the_great_wave_off_kanagawa:week_summary_7</title>
        <link>https://wiki.cvbbacm.com/doku.php?id=2020-2021:teams:the_great_wave_off_kanagawa:week_summary_7&amp;rev=1595583502&amp;do=diff</link>
        <description>2020/07/19 -- 2020/07/25 周报

本周推荐

airbust

CF 1381A1 Prefix Flip (Easy Version)

	*  分类：思维
	*  题意：给定一个01字符串，每次可以选择一个$x$，然后操作。操作方法为把该字符串的$x$前缀全部反转（0变成1, 1变成0），然后翻转（首尾交换）。目标是使该字符串变成另外一个01字符串。要求给出翻转方案。字符串的长度为$n（n \leq 1000）$$3n$$i$$i$$i$$i$$n$</description>
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        <dc:format>text/html</dc:format>
        <dc:date>2020-07-31T18:01:03+0800</dc:date>
        <dc:creator>Anonymous (anonymous@undisclosed.example.com)</dc:creator>
        <title>2020-2021:teams:the_great_wave_off_kanagawa:week_summary_8</title>
        <link>https://wiki.cvbbacm.com/doku.php?id=2020-2021:teams:the_great_wave_off_kanagawa:week_summary_8&amp;rev=1596189663&amp;do=diff</link>
        <description>2020/07/26 -- 2020/07/31 周报

团队训练

2020.07.27 2020牛客暑期多校训练营（第六场）

本周推荐

airbust

无

kazamori

无

Ket98

Fraction Construction Problem

&lt;https://ac.nowcoder.com/acm/contest/5668/F&gt;

分类：数学

这题考察的数学知识很棒，证明也很精彩，所以作为推荐题目。

题目大意：给定两个整数 $a,b$$c,d,e,f$$\frac{c}{d}-\frac{e}{f}=\frac{a}{b}$$d&lt;b$$f&lt;b$$1\le c,e \le 4 \times 10^{12}$$a,b$$\frac{\frac{a}{g}+1}{\frac{b}{g}}-\frac{\frac{a}{g}}{\frac{b}{g}}=\frac{a}{b}$$b$$\frac{cf-de}{df}$$\frac{a}{b}$$a$$b$$b=g^{k_b}$$d=c_1g^{k_d}, f=c_2g^{k_f}$$g$$g$$min(k_d,…</description>
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