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Warning: Cannot modify header information - headers already sent by (output started at /data/wiki/inc/init.php:239) in /data/wiki/inc/httputils.php on line 29 CVBB ACM Team 2020-2021:teams:farmer_john:2sozx:数学
https://wiki.cvbbacm.com/
2026-06-18T01:02:12+0800CVBB ACM Team
https://wiki.cvbbacm.com/
https://wiki.cvbbacm.com/lib/exe/fetch.php?media=favicon.icotext/html2020-05-09T20:37:37+0800Anonymous (anonymous@undisclosed.example.com)2020-2021:teams:farmer_john:2sozx:数学:一道没有来源的题目
https://wiki.cvbbacm.com/doku.php?id=2020-2021:teams:farmer_john:2sozx:%E6%95%B0%E5%AD%A6:%E4%B8%80%E9%81%93%E6%B2%A1%E6%9C%89%E6%9D%A5%E6%BA%90%E7%9A%84%E9%A2%98%E7%9B%AE&rev=1589027857&do=diff
题意
* 平面上有$n(n{\le}8)$个点,告诉你每个点距离原点的距离,求这$n$个点所围成的凸包的最大面积
题解
* 枚举哪些点在凸包上,并且这些点极角排序后的顺序。假设极径依次为$r_1,r_2,⋯,r_n$。
面积$S={\frac{1}{2}}(r_1r_2sinθ_1+r_2r_3sinθ_2+⋯+r_nr_1sinθ_n)$并且${\sum_{i=1}^n}{\theta}_i=2\pi$。
令$F(θ_1,θ_2,⋯,θ_n)=S+{\lambda}g(θ_1,θ_2,⋯,θ_n)$,其中$g(θ_1,θ_2,⋯,θ_n)={\sum_{i=1}^n}{\theta}_i-2\pi$$-λ=r_1r_2cosθ_1=r_2r_3cosθ_2=⋯=r_nr_1cosθ_n$$λ$$g=0$$\theta$…text/html2020-06-12T22:13:12+0800Anonymous (anonymous@undisclosed.example.com)2020-2021:teams:farmer_john:2sozx:数学:知识点
https://wiki.cvbbacm.com/doku.php?id=2020-2021:teams:farmer_john:2sozx:%E6%95%B0%E5%AD%A6:%E7%9F%A5%E8%AF%86%E7%82%B9&rev=1591971192&do=diff
格式:
* 向量建议写成 $\boldsymbol{x}_{0}$
内容:
* 没有例题吗
知识点
前言
对于一元函数的极值问题相信大家都十分熟悉,但是对于多元函数的极值问题可能就会比较陌生。大家都学过淑芬怎么可能陌生呢$f(\boldsymbol{x})$${\boldsymbol{\varphi}}(\boldsymbol{x})=({\varphi}_1(\boldsymbol{x}),{\varphi}_2(\boldsymbol{x}),\cdots,{\varphi}_m(\boldsymbol{x}))$$D\subset \mathbb{R}^n (m<n)$${\boldsymbol{x_0}}=({x_1}^0,{x_2}^0,\cdots,{x_n}^0)\in D$$f(\boldsymbol{x})$$$\begin{cases}{\varphi}_1(\boldsymbol{x})=0 \\{\varphi}_2(\boldsymbol{x})=0 \\ \vdots\\{\varphi}_m(\boldsymbol{x})=0\end{…text/html2020-05-20T14:44:09+0800Anonymous (anonymous@undisclosed.example.com)2020-2021:teams:farmer_john:2sozx:数学:类欧几里得
https://wiki.cvbbacm.com/doku.php?id=2020-2021:teams:farmer_john:2sozx:%E6%95%B0%E5%AD%A6:%E7%B1%BB%E6%AC%A7%E5%87%A0%E9%87%8C%E5%BE%97&rev=1589957049&do=diff
占坑text/html2020-05-17T21:29:22+0800Anonymous (anonymous@undisclosed.example.com)2020-2021:teams:farmer_john:2sozx:数学:遗留的知识点
https://wiki.cvbbacm.com/doku.php?id=2020-2021:teams:farmer_john:2sozx:%E6%95%B0%E5%AD%A6:%E9%81%97%E7%95%99%E7%9A%84%E7%9F%A5%E8%AF%86%E7%82%B9&rev=1589722162&do=diff
知识点
前言
对于一元函数的极值问题相信大家都十分熟悉,但是对于多元函数的极值问题可能就会比较陌生。大家都学过淑芬怎么可能陌生呢
对于没有限制条件的多元函数来说,只需要对函数求导即可,但是若有了限制条件,即函数的值要在一定条件下才能取到,则需要用到拉格朗日乘子法。$f(\vec{x})$${\vec{\varphi}}(\vec{x})=({\varphi}_1(\vec{x}),{\varphi}_2(\vec{x}),…,{\varphi}_m(\vec{x}))$$D\subset \mathbb{R}^n (m<n)$${\vec{x_0}}=({x_1}^0,{x_2}^0,…,{x_n}^0)\in D$$f(\vec{x})$$$\begin{cases}{\varphi}_1(\vec{x})=0 \\{\varphi}_2(\vec{x})=0 \\…\\{\varphi}_m(\vec{x})=0\end{cases}$$${\varphi}'(x_0)$$m$${\lambda}_1,{\lambda}_2,…,{\lambda}_3{\in}\mathbb…text/html2020-05-20T14:41:22+0800Anonymous (anonymous@undisclosed.example.com)2020-2021:teams:farmer_john:2sozx:数学:exgcd
https://wiki.cvbbacm.com/doku.php?id=2020-2021:teams:farmer_john:2sozx:%E6%95%B0%E5%AD%A6:exgcd&rev=1589956882&do=diff
扩展欧几里得
用途
解裴蜀方程
写法
通过普通的欧几里得算法来求解
int exgcd(int a, int b, int& x, int& y){
if(!b) {y=0,x=1;return a;}
int gcd=exgcd(b,a%b,y,x);
y-=a/b*x;
return gcd;
}