bzoj 2818

本来想用容斥原理，但我居然不造还有欧拉函数这个神奇的东西，果然还是太弱orz

题解orzlsj：

要求gcd(x, y) = p (1 <= x, y <= n, p为质数 ) 的数对(x, y)个数.我们枚举素数p, 令x' = x / p, y' = y / p, 则只须求  f(p) = gcd(x', y') = 1的数对(x', y')个数(1 <= x', y' <= n / p), 显然f(p) = (∑ phi(x')) * 2 - 1(1 <= x' <= n / p). 所以最后答案为 ∑f(p)

#include<bits/stdc++.h>
#define inc(i,l,r) for(i=l;i<=r;i++)
#define dec(i,l,r) for(i=l;i>=r;i--)
#define inf 1e9
#define NM 10000000+5
#define mem(a) memset(a,0,siezof(a))
long long d[NM],ans;
int n,i,j,p[NM],m;
bool v[NM];
void init(){
    d[1]=1;
    inc(i,2,n){
        if(!v[i]){
            p[++m]=i;
            d[i]=i-1;
        }
        inc(j,1,m){
            if(i*p[j]>n)break;
            v[i*p[j]]++;
            if(i%p[j])d[i*p[j]]=d[i]*d[p[j]];
            else d[i*p[j]]=d[i]*p[j];
        }
    }
}
int main(){
    scanf("%d",&n);
    init();
    inc(i,1,n)d[i]+=d[i-1];
    inc(i,1,m)ans+=d[n/p[i]]*2-1;
    printf("%lld",ans);
    return 0;
}