51Nod1594 Gcd and Phi

题目看这里
一个简单的反演题目:

\sum_{i = 1}^{n} \sum_{j = 1}^{n} ϕ (g c d (ϕ (i), ϕ (j)))

$\sum_{i=1}^n\sum_{j=1}^n\phi\bigg(gcd(\phi(i),\phi(j))\bigg)$
首先做一下变换

\sum_{i = 1}^{n} \sum_{j = 1}^{n} ϕ (g c d (ϕ (i), ϕ (j)))

$\sum_{i=1}^n\sum_{j=1}^n\phi\big(gcd(\phi(i),\phi(j))\big)$

= \sum_{d = 1}^{n} ϕ (d) * f (d)

$=\sum_{d=1}^n\phi(d)* f(d)$
这里

f (d) = \sum_{i, j} [g c d (ϕ (i), ϕ (j)) = d]

$f(d)=\sum_{i,j}[gcd(\phi(i),\phi(j))=d]$
表示有多少对i,j满足它们的

ϕ

$\phi$ 值的最大公约数为d
我们令

F (d) = \sum_{i, j} [d | g c d (ϕ (i), ϕ (j))]

$F(d)=\sum_{i,j}[d|gcd(\phi(i),\phi(j))]$ ,就有

F (n) = \sum_{n | d} f (d)

$F(n)=\sum_{n|d}f(d)$
反演得到

f (n) = \sum_{d} F (n * d) * μ (d)

$f(n)=\sum_{d}F(n*d)*\mu(d)$
所以原式可以写成

\sum_{i * j <= n} ϕ (i) * F (i * j) * μ (j)

$\sum_{i*j<=n}\phi(i)*F(i*j)*\mu(j)$
预处理

μ, ϕ

$\mu,\phi$ 和

F

$F$ 的值就可以了

# p r a g m a G C C o p t i m i z e (" O 3 ")

$\#pragma\ GCC\ optimize("O3")$

# p r a g m a G + + o p t i m i z e (" O 3 ")

$\#pragma\ G++\ optimize("O3")$

# i n c l u d e < s t d i o . h >

$\#include<stdio.h>$

# i n c l u d e < s t r i n g . h >

$\#include<string.h>$

# i n c l u d e < a l g o r i t h m >

$\#include<algorithm>$

# d e f i n e N 2000010

$\#define\ N\ 2000010\$

# d e f i n e L L l o n g l o n g

$\#define\ LL\ long\ long$

u s i n g n a m e s p a c e s t d;

$using\ namespace\ std;$

i n t n, T; l o n g l o n g S;

$int\ n,T;\ long\ long\ S;$

i n t w [N >> 2], t, p h i [N], m u [N], v i s [N], c [N];

$int\ w[N>>2],t,phi[N],mu[N],vis[N],c[N];$

i n l i n e L L F (i n t x) {r e t u r n (L L) c [x] * c [x];}

$inline\ LL\ F(int\ x)\{\ return\ (LL)c[x]*c[x];\ \}$

i n l i n e v o i d c a l () {

$inline\ void\ cal()\{$

s c a n f (" % d ", & n); m e m s e t (c, S = 0, s i z e o f c);

$\quad scanf("\%d",\&n);\ memset(c,S=0,sizeof\ c);$

f o r (i n t i = 1; i <= n; + + i) + + c [p h i [i]];

$\quad for(int\ i=1;i<=n;++i)\ ++c[phi[i]];$

f o r (i n t i = 1; i <= n; + + i)

$\quad for(int\ i=1;i<=n;++i)$

f o r (i n t j = i + i; j <= n; j + = i) c [i] + = c [j];

$\quad \quad for(int\ j=i+i;j<=n;j+=i)\ c[i]+=c[j];$

f o r (i n t i = 1; i <= n; + + i) i f (m u [i])

$\quad for(int\ i=1;i<=n;++i)\ if(mu[i])$

f o r (i n t j = 1; i * j <= n; + + j)

$\quad \quad for(int\ j=1;i*j<=n;++j)$

S + = (L L) F (i * j) * m u [i] * p h i [j];

$\quad \quad \quad S+=(LL)F(i*j)*mu[i]*phi[j];$

p r i n t f (" % l l d ", S);

$\quad printf("\%lld",S);$

}

$\}$

i n t m a i n () {

$int\ main()\{$

p h i [1] = m u [1] = 1;

$\quad phi[1]=mu[1]=1;$

f o r (i n t i = 2; i <= 2000000; + + i) {

$\quad for(int\ i=2;i<=2000000;++i)\{$

i f (! v i s [i]) {m u [w [+ + t] = i] = - 1; p h i [i] = i - 1;}

$\quad \quad if(!vis[i])\{\ mu[w[++t]=i]=-1;\ phi[i]=i-1;\ \}$

f o r (i n t j = 1, k; (k = i * w [j]) <= 2000000; + + j) {

$\quad \quad for(int\ j=1,k;(k=i*w[j])<=2000000;++j)\{$

v i s [k] = 1;

$\quad \quad \quad vis[k]=1;$

i f (i % w [j] == 0) {p h i [k] = p h i [i] * w [j]; m u [k] = 0; b r e a k;}

$\quad \quad \quad if(i\%w[j]==0)\{\ phi[k]=phi[i]*w[j];\ mu[k]=0;\ break;\ \}$

p h i [k] = p h i [i] * (w [j] - 1); m u [k] = - m u [i];

$\quad \quad \quad phi[k]=phi[i]*(w[j]-1);\ mu[k]=-mu[i];$

}

$\quad \quad \}\$

}

$\quad \}$

f o r (s c a n f (" % d ", & T); T - -; c a l ());

$\quad for(scanf("\%d",\&T);T--;cal());$

}

$\}$