记得以前是用容斥原理过的？好吧现在只记得奇加偶减了。。。
转化题目成求满足x/p和y/p互质的数对，那和上题就差不多了
先欧拉筛求出phi的前缀和a[i]，依次枚举每个素数p[i]，排除(1,1)答案就是sigma(a[n/p[i]]*2-1)
1 #include<bits/stdc++.h> 2 #define inc(i,l,r) for(int i=l;i<=r;i++) 3 #define dec(i,l,r) for(int i=l;i>=r;i--) 4 #define link(x) for(edge *j=h[x];j;j=j->next) 5 #define mem(a) memset(a,0,sizeof(a)) 6 #define inf 1e9 7 #define ll long long 8 #define succ(x) (1<<x) 9 #define NM 10000000+5 10 using namespace std; 11 int read(){ 12 int x=0,f=1;char ch=getchar(); 13 while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();} 14 while(isdigit(ch))x=x*10+ch-'0',ch=getchar(); 15 return x*f; 16 } 17 int n,p[NM],phi[NM],tot; 18 ll a[NM],s; 19 bool check[NM]; 20 int main(){ 21 freopen("data.in","r",stdin); 22 n=read(); 23 a[1]=phi[1]=1; 24 inc(i,2,n){ 25 if(!check[i]){ 26 p[++tot]=i; 27 phi[i]=i-1; 28 } 29 a[i]=a[i-1]+phi[i]; 30 inc(j,1,tot){ 31 if(p[j]*i>n)break; 32 check[p[j]*i]=true; 33 if(i%p[j])phi[i*p[j]]=phi[i]*phi[p[j]]; 34 else{ 35 phi[i*p[j]]=phi[i]*p[j]; 36 break; 37 } 38 } 39 } 40 // inc(i,1,n)printf("%d ",a[i]);printf(" "); 41 inc(i,1,tot)s+=a[(int)(0.999+n/p[i])]*2-1; 42 printf("%lld ",s); 43 return 0; 44 }