bzoj 2818: Gcd

Description

给定整数N，求1<=x,y<=N且Gcd(x,y)为素数的
数对(x,y)有多少对.

Input

一个整数N

Output

如题

Sample Input

4

Sample Output

4

HINT

hint

对于样例(2,2),(2,4),(3,3),(4,2)

1<=N<=10^7

数论题心累。。总之就是各种变换。。

见http://wzimpha.sinaapp.com/archives/499#comment-41

#include<cstdio>#include<string>#include<cstring>using namespace std;int mu[10000001],prime[10000001];int g[10000001];int sum[10000001]; bool check[10000001]; int n;int tot;inline void findmu(){     memset(check,false,sizeof(check));     mu[1]=1;     int i,j;     for(i=2;i<=n;i++)     {          if(!check[i])          {               tot++;               prime[tot]=i;               mu[i]=-1;               g[i]=1;          }          for(j=1;j<=tot;j++)          {               if(i*prime[j]>n)                    break;               check[i*prime[j]]=true;               if(i%prime[j]==0)               {                    mu[i*prime[j]]=0;                    g[i*prime[j]]=mu[i];                    break;               }               else               {                    mu[i*prime[j]]=-mu[i];                    g[i*prime[j]]=mu[i]-g[i];               }          }     }     for(i=1;i<=n;i++)      sum[i]=sum[i-1]+g[i];}int main(){ scanf("%d",&n); findmu(); int d,j=0; long long ans=0; for(d=1;d<=n;d=j+1) {      j=n/(n/d);      ans+=(long long)(sum[j]-sum[d-1])*(n/d)*(n/d); } printf("%lld\n",ans);     return 0;}