类欧几里得算法乱搞记

这三个f,g,h让我的脑子快要爆炸，还是终于推了出来，记录一下。

记得初一的时候就无意间在ZJY的PPT翻到了这个东西，当时和WYT推了一波，到现在连个印象都没有。

据说有几何推法，我这么渣肯定是不会的了。

参考博客：Xdl.

定义：

f(a,b,c,n)=∑ni=0⌊ai+bc⌋ f(a,b, c, n) =\sum_{i=0}^n{\lfloor {ai + b \over c} \rfloor}
g(a,b,c,n)=∑ni=0i⌊ai+bc⌋ g(a,b, c, n) =\sum_{i=0}^n i{\lfloor {ai + b \over c} \rfloor}
h(a,b,c,n)=∑ni=0⌊ai+bc⌋2 h(a,b, c, n) =\sum_{i=0}^n {\lfloor {ai + b \over c} \rfloor}^2

f的推导：

设 a′=a mod c,b′=b mod c a'=a~mod ~c,b'=b~mod~c

f(a,b,c,n)=∑ni=0⌊a′i+b′c⌋+⌊ac⌋∗i+⌊bc⌋ f(a,b, c, n) =\sum_{i=0}^n{\lfloor {a'i + b' \over c} \rfloor}+{\lfloor {a\over c} \rfloor}*i+{\lfloor {b\over c} \rfloor}
=f(a′,b′,c,n)+n(n+1)/2∗⌊ac⌋+(n+1)∗⌊bc⌋ = f(a',b',c, n) +n(n+1)/2*{\lfloor {a \over c}\rfloor}+(n+1)*{\lfloor {b \over c}\rfloor}

进过上面这个调整后可以使 a,b<c a,b,这也是为什么它叫类欧几里得算法的原因，从最终的式子可以看出来这点。

设 m=⌊an+bc⌋ m={\lfloor {an + b \over c} \rfloor}。

f(a,b,c,n) f(a,b, c,n)
=∑ni=0⌊ai+bc⌋ =\sum_{i=0}^n{\lfloor {ai + b \over c} \rfloor}
=∑mj=1∑ni=0[⌊ai+bc⌋>=j] =\sum_{j=1}^{m}\sum_{i=0}^n[{\lfloor {ai + b \over c} \rfloor}>=j]
=∑m−1j=0∑ni=0[⌊ai+bc⌋>=j+1] =\sum_{j=0}^{m-1}\sum_{i=0}^n[{\lfloor {ai + b \over c} \rfloor}>=j+1]
=∑m−1j=0∑ni=0[ai>=jc+c−b] =\sum_{j=0}^{m-1}\sum_{i=0}^n[ai>=jc+c-b]
=∑m−1j=0∑ni=0[ai>jc+c−b−1] =\sum_{j=0}^{m-1}\sum_{i=0}^n[ai>jc+c-b-1]
=∑m−1j=0∑ni=0[i>⌊jc+c−b−1a⌋] =\sum_{j=0}^{m-1}\sum_{i=0}^n[i>{\lfloor {jc+c-b-1 \over a}}\rfloor]
=∑m−1j=0n−⌊jc+c−b−1a⌋ =\sum_{j=0}^{m-1}n-{\lfloor {jc+c-b-1 \over a}}\rfloor
=nm−∑m−1j=0⌊jc+c−b−1a⌋ =nm-\sum_{j=0}^{m-1}{\lfloor {jc+c-b-1 \over a}}\rfloor
=nm−f(c,c−b−1,a,m−1) =nm-f(c,c-b-1,a,m-1)

这个可以直接递归去求。我设的边界是m=0。

发现a,c的位置互换了。因为a是已经mod过c的了，所以这和欧几里得算法的复杂度是一模一样的。

g的推导：

g(a,b,c,n)=∑ni=0∑ni=0i∗⌊a′i+b′c⌋+⌊ac⌋∗i2+⌊bc⌋∗i g(a,b,c,n)=\sum_{i=0}^n\sum_{i=0}^ni*{\lfloor {a'i + b' \over c} \rfloor}+{\lfloor {a\over c} \rfloor}*i^2+{\lfloor {b\over c} \rfloor}*i
=g(a′,b′,c,n)+n(n+1)(2n+1)/6∗⌊ac⌋+n(n+1)/2∗⌊bc⌋ =g(a',b',c, n) +n(n+1)(2n+1)/6*{\lfloor {a \over c}\rfloor}+n(n+1)/2*{\lfloor {b \over c}\rfloor}
接下来的变换和f类似，跳一下步：
g(a,b,c,n) g(a,b,c,n)
=∑m−1j=0∑ni=0i∗[i>⌊jc+c−b−1a⌋] =\sum_{j=0}^{m-1}\sum_{i=0}^ni*[i>{\lfloor {jc+c-b-1 \over a}}\rfloor]
到这里，实际上可以用个等差数列搞搞。
=∑m−1j=0(n−⌊jc+c−b−1a⌋)(n+⌊jc+c−b−1a⌋+1)/2 =\sum_{j=0}^{m-1} (n-{\lfloor {jc+c-b-1 \over a}}\rfloor)(n+{\lfloor {jc+c-b-1 \over a}}\rfloor+1)/2
=12∑m−1j=0n(n+1)−⌊jc+c−b−1a⌋−⌊jc+c−b−1a⌋2 ={1 \over 2}\sum_{j=0}^{m-1} n(n+1)-\lfloor {jc+c-b-1 \over a}\rfloor-{\lfloor {jc+c-b-1 \over a}\rfloor}^2
=12(mn(n+1)−f(c,c−b−1,a,m−1)−h(c,c−b−1,a,m−1)) ={1 \over 2}(mn(n+1)-f(c,c-b-1,a,m-1)-h(c,c-b-1,a,m-1))

h的推导：

h(a,b,c,n)=(∑ni=0⌊a′i+b′c⌋+⌊ac⌋∗i+⌊bc⌋)2 h(a,b,c,n)=({\sum_{i=0}^n{\lfloor {a'i + b' \over c} \rfloor}+{\lfloor {a\over c} \rfloor}*i+{\lfloor {b\over c} \rfloor}})^2
=h(a′,b′,c,n) =h(a',b',c, n)
+n(n+1)(2n+1)/6∗⌊ac⌋+(n+1)∗⌊bc⌋ +n(n+1)(2n+1)/6*{\lfloor {a \over c}\rfloor}+(n+1)*{\lfloor {b \over c}\rfloor}
+2⌊bc⌋∗f(a′,b′,c,n)+2⌊ac⌋∗g(a′,b′,c,n) +2{\lfloor {b \over c}\rfloor}*f(a',b',c, n)+2{\lfloor {a \over c}\rfloor}*g(a',b',c, n)
+n(n+1)⌊ac⌋⌊bc⌋ +n(n+1){\lfloor {a \over c}\rfloor}{\lfloor {b \over c}\rfloor}

一种有用的写法： x2=(2∑xi=0i)−x x^2=(2\sum_{i=0}^x i) - x

h(a,b,c,n)=∑ni=0(2(∑⌊ai+bc⌋j=1j)−⌊ai+bc⌋) h(a,b,c,n)=\sum_{i=0}^n(2(\sum_{j=1}^{{\lfloor {ai + b \over c} \rfloor}}j)-{\lfloor {ai + b \over c} \rfloor})
=2(∑m−1j=0(j+1)∑ni=0[⌊ai+bc⌋>=j+1])−f(a,b,c,n) =2(\sum_{j=0}^{m-1}(j+1)\sum_{i=0}^n[{\lfloor {ai + b \over c}\rfloor }>=j+1])-f(a,b,c,n)
=2(∑m−1j=0j∑ni=0[⌊ai+bc⌋>=j+1])+2(∑m−1j=0∑ni=0[⌊ai+bc⌋>=j+1])−f(a,b,c,n) =2(\sum_{j=0}^{m-1}j\sum_{i=0}^n[{\lfloor {ai + b \over c}\rfloor }>=j+1])+2(\sum_{j=0}^{m-1}\sum_{i=0}^n[{\lfloor {ai + b \over c}\rfloor }>=j+1])-f(a,b,c,n)
=2(∑m−1j=0j∑ni=0[i>⌊jc−c−b−1a⌋])+2(∑m−1j=0∑ni=0[i>⌊jc−c−b−1a⌋])−f(a,b,c,n) =2(\sum_{j=0}^{m-1}j\sum_{i=0}^n[i>\lfloor{jc-c-b-1\over a}\rfloor])+2(\sum_{j=0}^{m-1}\sum_{i=0}^n[i>\lfloor{jc-c-b-1\over a}\rfloor])-f(a,b,c,n)
=2(∑m−1j=0j∗(n−⌊jc−c−b−1a⌋)+2(∑m−1j=0n−⌊jc−c−b−1a⌋)−f(a,b,c,n) =2(\sum_{j=0}^{m-1}j*(n-\lfloor{jc-c-b-1\over a}\rfloor)+2(\sum_{j=0}^{m-1}n-\lfloor{jc-c-b-1\over a}\rfloor)-f(a,b,c,n)
=m(m−1)n−2g(c,c−b−1,a,m−1)+2mn−2f(c,c−b−1,a,m−1)−f(a,b,c,n) =m(m-1)n-2g(c,c-b-1,a,m-1)+2mn-2f(c,c-b-1,a,m-1)-f(a,b,c,n)
=m(m+1)n−2g(c,c−b−1,a,m−1)−2f(c,c−b−1,a,m−1)−f(a,b,c,n) =m(m+1)n-2g(c,c-b-1,a,m-1)-2f(c,c-b-1,a,m-1)-f(a,b,c,n)

至此，三种常见的形式已经弄完。

实现的话只求f还好，由于g和h相互调用，可能要用结构体来存。

代码的话暂时没有。