https://wiki.cvbbacm.com/ 2026-04-30T03:29:22+0800 https://wiki.cvbbacm.com/ https://wiki.cvbbacm.com/lib/exe/fetch.php?media=favicon.ico text/html 2020-07-31T11:20:47+0800 Anonymous (anonymous@undisclosed.example.com) https://wiki.cvbbacm.com/doku.php?id=2020-2021:teams:alchemist:hardict:haversine_formula&rev=1596165647&do=diff Haversine formula $h(\theta)=sin^2(\frac{\theta}{2})=\frac{1-cos(\theta)}{2}$ 则$h(\theta)=h(\frac{d}{R})=h(\Delta \beta)+cos(\beta_1)cos(\beta_2)h(\Delta \alpha)$ - $R表示球面半径,d表示球面距离,\theta表示两点与圆心夹角弧度$ - $\alpha_i分别表示两点经度,\beta_i表示两点维度,\Delta表示差值$ - 公式全称应该为$half-versine$,即$versine: 1-cos(\theta)的一半$ - 计算时可进一步化解：$cos(\theta)=sin(\beta_1)sin(\beta_2)+cos(\beta_1)cos(\beta_2)cos(\Delta \alpha)$ 这里求$\overset{\frown}{AB}$，显然求得$|AB|$即可 以$OEF$为例，$\angle OEF=\Delta \alpha,|EF|=2sin(\frac{\Delta \a… text/html 2020-07-31T11:20:07+0800 Anonymous (anonymous@undisclosed.example.com) https://wiki.cvbbacm.com/doku.php?id=2020-2021:teams:alchemist:hardict:math&rev=1596165607&do=diff 自然数幂和以及伯努利数 二次剩余与三次剩余 经纬度求解两点球面距离(Haversine formula) text/html 2020-05-07T17:54:33+0800 Anonymous (anonymous@undisclosed.example.com) https://wiki.cvbbacm.com/doku.php?id=2020-2021:teams:alchemist:hardict:myself&rev=1588845273&do=diff 摸鱼万岁 text/html 2020-05-09T10:56:07+0800 Anonymous (anonymous@undisclosed.example.com) https://wiki.cvbbacm.com/doku.php?id=2020-2021:teams:alchemist:hardict:powersum&rev=1588992967&do=diff 递推 $令f_{k}(n)=\sum_{i=1}^{n}i^{k}$ $$ \begin{align} f_{k+1}(n+1) & =f_{k+1}(n)+(n+1)^{k+1} &\\ & =1+\sum_{i=1}^{n}(i+1)^{k+1} &\\ & =1+\sum_{i=1}^{n}\sum_{j=0}^{k+1} \binom{k+1}{j} i^{j} &\\ & =1+\sum_{j=0}^{k+1}\binom{k+1}{j} \sum_{i=1}^{n} &\\ & =1+\sum_{i=0}^{k+1}\binom{k+1}{i} f_{i}(n) &\\ & =1+f_{k+1}(n)+(k+1)f_{k}(n)+\sum_{i=0}^{k-1}\binom{k+1}{i} f_{i}(n) &\\ \end{align} $$ $得到了:(k+1)f_{k}(n)=(n+1)^{k+1}-1 - \sum_{i=0}^{k-1}\binom{k+1}{i} f_{i}(n)$ $可以写成: f_{k-1}(n)=\frac{(n+1)^{k}-1… text/html 2020-05-16T16:13:30+0800 Anonymous (anonymous@undisclosed.example.com) https://wiki.cvbbacm.com/doku.php?id=2020-2021:teams:alchemist:hardict:qrp&rev=1589616810&do=diff 二次剩余(QRP) Cipolla算法(素数情况下) wiki百科 对于 $x^{2} \equiv a \pmod{p}$ 可以随机找一个数$s,\quad s.t:(\frac{s^{2}-a}{p})=-1,即s不是p的二次剩余$,可以知道找到$s$的期望次数为2 考虑$\mathbb{Z}(w=\sqrt{s^{2}-a})=\{j+kw\}$以及如下定理 * $w^{p} = w*(w^{2})^{\frac{p-1}{2}}=w*(s^{2}-a)^{\frac{p-1}{2}} \equiv -w \pmod{p}$ * $(a+b)^{p} \equiv a^{p}+b^{b} \pmod p$ 解为$x \equiv (s+w)^{\frac{p+1}{2}}: (s+w)^{p+1} = (s+w)^{p}(s+w) \equiv (s^{p}+w^{p})(s+w) \equiv (s-w)(s+w) = (s^{2}-w^{2}) = a \pmod{p}$$x^{3} \equiv a \pmod{p}$$a=0,p \leq 3$$…