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(转载)有关反演和gcd

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积性函数 F (n) = Π F (piai )

若F (n), G (n)是积性函数则

F (n) * G (n)

Σd | n F (n)

是积性函数

n = Σd | n  φ (d)

1 = Σd | n  μ (d)

Σgcd (i, n) = 1 i = n * φ (n) / 2

Problem1

F (n) = Σ1<= i <= n gcd(i, n), n <= 1000000

Sol

枚举结果

F (n) = Σd | n d * Σgcd (i, n) = d 1

F (n) = Σd | n d * Σgcd (i / d, n / d) = 1 1

F (n) = Σd | n d * Σgcd (i / d, n / d) = 1 1

F (n) = Σd | n d * φ (n / d)

单次计算O (sqrt N)

筛法O (N)

Problem 2

F (n) = Σ1<= i <= n gcd (i, n), n <= 2147483647               (POJ longge‘s problem)

Sol

由P1可知F (n)是积性函数 因此考虑计算F (pk)

由P1 易知

F (pk) = p * F (pk - 1)  + (p - 1)pk - 1

单个F(n)可以O (sqrt N)时间有他的质因子分解计算得到

Problem 3

 F (n) = Σ1<= i <= n lcm(i, n), n <= 1000000             (SPOJ LCMSUM)

 

显而易见的变形

F (n) = Σ1<= i <= n i * n / gcd (i, n)

F (n) =Σd | n  Σgcd (i, n) = d i * n / d

F (n) =Σd | n n / d * Σgcd (i, n) = d i

F (n) =Σd | n n / d * Σgcd (i / d, n / d) = 1 i

F (n) =Σd | n n / d * d * Σgcd (i / d, n / d) = 1 i / d

令j = i / d

F (n) =Σd | n n * Σgcd (j, n / d) = 1 j

由Σgcd (i, n) = 1 i = n * φ (n) / 2

F (n) =Σd | n n * (n / d) * φ (n / d) / 2

F (n) =n * Σd | n (n / d) * φ (n / d) / 2

F (n) =n / 2 * Σd | n d * φ (d)

筛出 Σd | n d * φ (d)     O (N)-O(1)

Problem 4

F (n) = Σ1<= i <= n Σ1<= j <= n gcd (i, j)      n <= 1000000          (SPOJ GCDEX)

Sol

G (n) = Σd | n d * φ (n / d)

F (n) = Σ1<= i <= n G (i)

筛出G (i) 前缀和

Problem 5

求F (n, m) = [n / d] * [m / d]

Sol

研究退化情况 m = 1 F (n) = [n / d]

共有sqrt n种不同取值 

F (n, m) = [n / d] * [m / d]

共有sqrt n + sqrt m种不同取值  归并这两种取值

Problem 6

多组询问n, m 求F (n, m) = Σ1<= i <= n Σ1<= j <= m gcd (i, j)

Sol

F (n, m) = Σ1<= i <= n Σ1<= j <= m Σ d | gcd (i, j) φ (d)

F (n, m) = Σ d φ (d) Σ1<= i <= n d | i Σ1<= j <= m d | j 1

F (n, m) = Σ d φ (d) * [n / d] * [m / d]

可以经P5解决

Problem 6

多组询问n, m 求F (n, m) = Σ1<= i <= n Σ1<= j <= m gcd (i, j) = 1 1

Sol

F (n, m) = Σ1<= i <= n Σ1<= j <= m Σ d | gcd (i, j) μ (d)

F (n, m) = Σ d μ (d) Σ1<= i <= n d | i Σ1<= j <= m d | j 1

F (n, m) = Σ d μ (d) * [n / d] * [m / d]

可以经P5解决

 

Problem 7

F (n, m) = Σ1<= i <= n Σ1<= j <= m Σ gcd (i, j) <- prime 1            (BZOJ YY的GCD)

Sol

F (n, m) = Σ1<= i <= n Σ1<= j <= m Σ gcd (i, j) <- prime 1

F (n, m) = Σ1<= i <= n Σ1<= j <= m Σ p <- prime gcd (i, j) = p 1

F (n, m) = Σ p <- prime Σ1<= i <= n Σ1<= j <= m  gcd (i, j) = p 1

F (n, m) = Σ p <- prime Σ1<= i <= n / p Σ1<= j <= m / p gcd (n / p, m / p) = 1  1

可以经P6解决