bzoj 2818

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)

#include<bits/stdc++.h>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define inf 1e9
#define ll long long
#define succ(x) (1<<x)
#define NM 10000000+5
using namespace std;
int read(){
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    return x*f;
}
int n,p[NM],phi[NM],tot;
ll a[NM],s;
bool check[NM];
int main(){
    freopen("data.in","r",stdin);
    n=read();
    a[1]=phi[1]=1;
    inc(i,2,n){
        if(!check[i]){
            p[++tot]=i;
            phi[i]=i-1;
        }
        a[i]=a[i-1]+phi[i];
        inc(j,1,tot){
            if(p[j]*i>n)break;
            check[p[j]*i]=true;
            if(i%p[j])phi[i*p[j]]=phi[i]*phi[p[j]];
            else{
                phi[i*p[j]]=phi[i]*p[j];
                break;
            }
        }
    }
//    inc(i,1,n)printf("%d ",a[i]);printf("\n");
    inc(i,1,tot)s+=a[(int)(0.999+n/p[i])]*2-1;
    printf("%lld\n",s);
    return 0;
}