【类欧几里得学习小记】

类欧几里得算法因为形式上像欧几里得算法而得名，其时间复杂度也和欧几里得算法一致，类欧几里得算法有很多应用，下面只举部分例子。

目标

求出以下三个函数值

$f(a,b,c,n)=\sum_{i=0}^n\lfloor{ai+b\over c}\rfloor$

$g(a,b,c,n)=\sum_{i=0}^ni\lfloor{ai+b\over c}\rfloor$

$h(a,b,c,n)=\sum_{i=0}^n{\lfloor{ai+b\over c}\rfloor}^2$

设$m=\lfloor{an+b\over c}\rfloor$

f

当a==0时

$f(a,b,c,n)=\sum_{i=0}^n\lfloor{ai+b\over c}\rfloor$

$={b\over c}(n+1)$

当(a>=c)||(b>=c)时

$f(a,b,c,n)=\sum_{i=0}^n\lfloor{ai+b\over c}\rfloor$

$=\sum_{i=0}^n({a\over c}i+{b\over c}+{(a\%c)i+b\%c\over c})$

$={a\over c}n(n+1)/2+{b\over c}(n+1)+f(a\%c,b\%c,c,n)$

当$(a<c)and(b<c)$时

$f(a,b,c,n)=\sum_{i=0}^n\lfloor{ai+b\over c}\rfloor$

$=\sum_{i=0}^n\sum_{j=1}^m[{ai+b\over c}>=j]$

$=\sum_{i=0}^n\sum_{j=0}^{m-1}[{ai+b\over c}>=j+1]$

$=\sum_{i=0}^n\sum_{j=0}^{m-1}[ai+b >=cj+c]$

$=\sum_{i=0}^n\sum_{j=0}^{m-1}[ai+b >cj+c-1]$

$=\sum_{i=0}^n\sum_{j=0}^{m-1}[i >{cj+c-b-1\over a}]$

$=\sum_{j=0}^{m-1}\sum_{i=0}^n[i >{cj+c-b-1\over a}]$

$=\sum_{j=0}^{m-1}(n-{cj+c-b-1\over a})$

$=nm-\sum_{j=0}^{m-1}{cj+c-b-1\over a}$

$=nm-f(c,c-b-1,a,m-1)$

g

当a==0时

$g(a,b,c,n)=\sum_{i=0}^ni\lfloor{ai+b\over c}\rfloor$

$=\lfloor{b\over c}\rfloor n*(n+1)/2$

当(a>=c)||(b>=c)时

$g(a,b,c,n)=\sum_{i=0}^ni\lfloor{ai+b\over c}\rfloor$

$=\sum_{i=0}^ni({a\over c}i+{b\over c}+{(a\%c)i+b\%c\over c})$

$={a\over c}n(n+1)(2n+1)/6+{b\over c}n(n+1)/2+g(a\%c,b\%c,c,n)$

当$(a<c)and(b<c)$

$g(a,b,c,n)=\sum_{i=0}^ni\lfloor{ai+b\over c}\rfloor$

$=\sum_{j=0}^{m-1}\sum_{i=0}^ni[i>{cj+c-b-1\over a}]$

$=\sum_{j=0}^{m-1}(n+{cj+c-b-1\over a}-1)(n-{cj+c-b-1\over a})/2$

$={1\over2}mn(n+1)-{1\over2}f(c,c-b-1,a,m-1)-{1\over2}h(c,c-b-1,a,m-1)$

h

当a==0时

$h(a,b,c,n)=\sum_{i=0}^n{\lfloor{ai+b\over c}\rfloor}^2$

$=({b\over c})^2(n+1)$

当(a>=c)||(b>=c)时

$h(a,b,c,n)=\sum_{i=0}^n{\lfloor{ai+b\over c}\rfloor}^2$

$x^2=2{x(x+1)\over 2}-x$

$=2\sum_{i=1}^xi-x$

$h(a,b,c,n)=\sum_{i=0}^n{\lfloor{ai+b\over c}\rfloor}^2$

$=2\sum_{i=0}^n\sum_{j=1}^{{ai+b\over c}}j-f(a,b,c,n)$

$=2\sum_{i=1}^mi\sum_{j=0}^n[{aj+b\over c}>=i]-f(a,b,c,n)$

$=2\sum_{i=0}^{m-1}(i+1)\sum_{j=0}^n[j>{ci+c-b-1\over a}]-f(a,b,c,n)$

$=nm(m+1)-2f(c,c-b-1,a,m-1)-2g(c,c-b-1,a,m-1)-f(a,b,c,n)$

至此我们已经可以解决这一类类欧几里得问题。