【bzoj2818】: Gcd 数论-欧拉函数

【bzoj2818】: Gcd

考虑素数p<=n

gcd(xp,yp)=p 当 gcd(x,y)=1 xp,yp<=n满足条件

p对答案的贡献：

预处理前缀和就好了

/* http://www.cnblogs.com/karl07/ */
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

#define ll long long
const int N=1e7+5;
ll phi[N],prime[N];
int cnt=0,n;
ll ans;

void PHI(int n){
    phi[1]=1;
    for (int i=2;i<=n;i++){
        if (!phi[i]){
            prime[++cnt]=i;
            for (int j=i;j<=n;j+=i){
                if (phi[j]==0) phi[j]=j;
                phi[j]=phi[j]/i*(i-1);
            }
        }
    }
}

int main(){
    scanf("%d",&n);
    PHI(n);
    for (int i=1;i<=n;i++) phi[i]+=phi[i-1];
    for (int i=1;i<=cnt;i++){
        ans+=phi[n/prime[i]]*2-1;
    }
    printf("%lld
",ans);
    return 0;
}