类欧几里得算法

学习背景

之前做了一道类欧＋(Stern-Brocot Tree)二分的题目，就学一下。

问题模型

求

[egin{aligned} f(n,a,b,c) &= sum_{i=0}^{n} leftlfloor frac{ai+b}{c} ight floor \ g(n,a,b,c) &= sum_{i=0}^{n} i leftlfloor frac{ai+b}{c} ight floor \ h(n,a,b,c) &= sum_{i=0}^{n} left(leftlfloor frac{ai+b}{c} ight floor ight)^2 end{aligned} ]

解法推导

为了方便，我们设(m= leftlfloor frac{na+b}{c} ight floor)
注意，因为取模符号太宽了，所以用C++的"%"来表示取模。

（一）(f(n,a,b,c))

• 第一种情况：(a<c)且(b<c)
  这个可以直接推导，比较简单。

[egin{aligned} f(n,a,b,c) &= sum_{i=0}^{n} leftlfloor frac{ai+b}{c} ight floor \ &= sum_{i=0}^{n} sum_{j=0}^{n} left[j < leftlfloor frac{ai+b}{c} ight floor ight] \ &= sum_{j=0}^{m-1} sum_{i=0}^{n} left[j < leftlfloor frac{ai+b}{c} ight floor ight] \ &= sum_{j=0}^{m-1} sum_{i=0}^{n} left[j < leftlceil frac{ai+b-c+1}{c} ight ceil ight] \ &= sum_{j=0}^{m-1} sum_{i=0}^{n} left[j < frac{ai+b-c+1}{c} ight] \ &= sum_{j=0}^{m-1} sum_{i=0}^{n} left [cj < ai+b-c+1 ight] \ &= sum_{j=0}^{m-1} sum_{i=0}^{n} left[ i> frac{cj+c-b-1}{a} ight] \ &= sum_{j=0}^{m-1} n - leftlfloor frac{cj+c-b-1}{a} ight floor \ &= nm - sum_{j=0}^{m-1} leftlfloor frac{cj+c-b-1}{a} ight floor \ &= nm - f(m - 1, c, c - b - 1, a) end{aligned} ]

• 第二种情况：(ageq c)且(bgeq c)
  这时候我们只要将(leftlfloor frac{ai+b}{c} ight floor)拆开，就是：

[egin{aligned} f(n,a,b,c) &= sum_{i=0}^{n} leftlfloor frac{ai+b}{c} ight floor \ &= sum_{i=0}^{n} left( leftlfloor frac{(a \% c)i+(b \% c)}{c} ight floor + i leftlfloor frac{a}{c} ight floor + leftlfloor frac{b}{c} ight floor ight) \ &= frac{n(n+1)}{2} leftlfloor frac{a}{c} ight floor + n leftlfloor frac{b}{c} ight floor + sum_{i=0}^{n} leftlfloor frac{(a \% c)i+(b \% c)}{c} ight floor end{aligned} ]

（二）(g(n,a,b,c))

第二类和第一类比较类似，就写得简单一点。

• 第一种情况：(a<c)且(b<c)

[egin{aligned} g(n,a,b,c) &= sum_{i=0}^{n} i leftlfloor frac{ai+b}{c} ight floor \ &= sum_{j=0}^{m-1} sum_{i=0}^{n} i left[ leftlfloor frac{ai+b}{c} ight floor > j ight] \ &= sum_{j=0}^{m-1} sum_{i=0}^{n} i left[ i > leftlfloor frac{cj+c-b-1}{a} ight floor ight] \ &= frac{mn(n+1)}{2} - sum_{j=0}^{m-1} frac{leftlfloor frac{cj+c-b-1}{a} ight floor (leftlfloor frac{cj+c-b-1}{a} ight floor+1)}{2} \ &= frac{mn(n+1)}{2} - frac{1}{2} sum_{j=0}^{m-1} left( leftlfloor frac{cj+c-b-1}{a} ight floor ight)^2 - frac{1}{2} sum_{j=0}^{m-1} leftlfloor frac{cj+c-b-1}{a} ight floor \ &= frac{mn(n+1)}{2} - frac{1}{2}h(n,c,c-b-1,a) - frac{1}{2}f(n,c,c-b-1,a) end{aligned} ]

• 第二种情况：(ageq c)且(bgeq c)
  这个也和(f(n,a,b,c))类似，直接给出结论了：

[ g(n,a,b,c) = frac{n(n+1)(2n+1)}{6}leftlfloor frac{a}{c} ight floor + frac{n(n+1)}{2}leftlfloor frac{b}{c} ight floor + g(n,a \% c, b \% c, c) ]

（三）(h(n,a,b,c))

因为这里涉及到了平方，所以我们考虑用一种求和式表示平方。

[ n^2 = 2sum_{i=0}^{n} i - n ]

之后，有如下推导：

• 第一种情况：(a<c)且(b<c)

[ h(n,a,b,c) = sum_{i=0}^{n} left( 2sum_{j=0}^{left lfloorfrac{ai+b}{c} ight floor}j - leftlfloorfrac{ai+b}{c} ight floor ight) ]

之后，就用与求(f)和(g)类似的方法，得到：

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

• 第二种情况：(ageq c)且(bgeq c)
  对于这个，就直接将三项式暴力拆开，就有：

[egin{aligned} h(n,a,b,c) = leftlfloor frac{a}{c} ight floor^2 imes frac{n(n+1)(2n+1)}{6} + leftlfloor frac{b}{c} ight floor^2 imes (n+1) + leftlfloor frac{a}{c} ight floor imes leftlfloor frac{b}{c} ight floor imes n(n+1) + h(n,a \% c, b \% c, c) + 2leftlfloor frac{a}{c} ight floor g(n,a \% c, b \% c, c) + 2leftlfloor frac{b}{c} ight floor f(n,a \% c, b \% c, c) end{aligned} ]

（四）整理一下

[f(n,a,b,c) = egin{cases} frac{n(n+1)}{2} leftlfloor frac{a}{c} ight floor + n leftlfloor frac{b}{c} ight floor + f(m-1, a \% c, b \% c, c) & a < c And b < c \ nm - f(m - 1, c, c - b - 1, a) & otherwise end{cases} ]

[g(n,a,b,c) = egin{cases} frac{mn(n+1)}{2} - frac{1}{2}h(n,c,c-b-1,a) - frac{1}{2}f(n,c,c-b-1,a) & a < c And b < c \ frac{n(n+1)(2n+1)}{6}leftlfloor frac{a}{c} ight floor + frac{n(n+1)}{2}leftlfloor frac{b}{c} ight floor + g(n,a \% c, b \% c, c) & otherwise end{cases} ]

[h(n,a,b,c) = egin{cases} h(n,a,b,c) = nm(m+1) - 2g(m-1,c,c-b-1,a) - 2f(m-1,c,c-b-1,a) - f(n,a,b,c) & a < c And b < c \ leftlfloor frac{a}{c} ight floor^2 imes frac{n(n+1)(2n+1)}{6} + leftlfloor frac{b}{c} ight floor^2 imes (n+1) + leftlfloor frac{a}{c} ight floor imes leftlfloor frac{b}{c} ight floor imes n(n+1) + h(n,a \% c, b \% c, c) + 2leftlfloor frac{a}{c} ight floor g(n,a \% c, b \% c, c) + 2leftlfloor frac{b}{c} ight floor f(n,a \% c, b \% c, c) & otherwaise end{cases} ]