（转载）有关反演和gcd

tips :

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


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


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


令j = i / d


由Σgcd (i, n) = 1 i = n * φ (n) / 2


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


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