BZOJ2818: 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

手推公式可得

#include<cstdio>
#include<cctype>
#include<queue>
#include<cmath>
#include<cstring>
#include<algorithm>
#define rep(i,s,t) for(int i=s;i<=t;i++)
#define dwn(i,s,t) for(int i=s;i>=t;i--)
#define ren for(int i=first[x];i;i=next[i])
using namespace std;
const int BufferSize=1<<16;
char buffer[BufferSize],*head,*tail;
inline char Getchar() {
if(head==tail) {
int l=fread(buffer,1,BufferSize,stdin);
tail=(head=buffer)+l;
}
return *head++;
}
inline int read() {
int x=0,f=1;char c=getchar();
for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
for(;isdigit(c);c=getchar()) x=x*10+c-'0';
return x*f;
}
typedef long long ll;
const int maxn=10000010;
int vis[maxn],pri[maxn/10],mu[maxn],f[maxn],cnt;
void solve(int n) {
vis[1]=mu[1]=1;
rep(i,2,n) {
if(!vis[i]) pri[++cnt]=i,mu[i]=-1;
rep(j,1,cnt) {
if(i*pri[j]>n) break;
vis[i*pri[j]]=1;
if(i%pri[j]==0) {mu[i*pri[j]]=0;break;}
mu[i*pri[j]]=-mu[i];
}
}
rep(i,1,cnt) {
int x=pri[i];
for(int j=x;j<=n;j+=x) f[j]+=mu[j/x];
}
ll ans=0;
rep(i,1,n) ans+=(ll)(n/i)*(n/i)*(ll)f[i];
printf("%lld\n",ans);
}
int main() {
solve(read());
return 0;
}

View Code