Spoj vlattice Visible Lattice Points first entry Möbius

Test instructions: The two-point (x, Y, z) connection does not go through a few other points

Solution: That is, for gcd (x, Y, Z) 1 points have several

Jie Yi: Because x, Y, Z are all within the 1~n, you can use Euler functions to solve

Solution Two: The inverse of the mo-Black

Set F[n] is the number of gcd (x, Y, z) =n

Set F[b] to b| The number of GCD (x, Y, z) is obvious f[b]= (n/i) * (n/i) * (n/i)

So F[n]=sigema (B|n,f[b]);

F[n]=sigema (N|b,mu[n],f[n])

#include <stdio.h> #include <string.h>const int maxn=1000005;int prime[maxn];int num[maxn];int MU[MAXN];    void Mobius () {memset (num,0,sizeof (num));    Mu[1]=1;    int all=0;            for (int i=2;i<maxn;i++) {if (!num[i]) {prime[all++]=i;        Mu[i]=-1;            } for (int j=0;j<all&&i*prime[j]<maxn;j++) {num[i*prime[j]]=1;            if (I%prime[j]) mu[i*prime[j]]=-mu[i];                else {mu[i*prime[j]]=0;            Break }}} return;}    int main () {int t;    int n;    a long long sum;    Mobius ();                       while (scanf ("%d", &t)!=-1) {while (t--) {sum=3;            X, y, Z axis scanf ("%d", &n);            for (int i=1;i<=n;i++) {sum+= (Long Long) mu[i]* (n/i) * (n/i) * ((n/i) +3);        The point on the face, so +3} printf ("%lld\n", sum); }} RETUrn 0;} 

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Spoj vlattice Visible Lattice Points first entry Möbius