【bzoj2818】 Gcd

http://www.lydsy.com/JudgeOnline/problem.php?id=2818 (题目链接)

题意　　

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

Solution

　　对于gcd(x,y)=p的数对个数，就相当于x/p和y/p互质。

细节

　　前缀和开LL

代码

// poj2478
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#define LL long long
#define inf 2147483640
#define MOD 10000
#define Pi acos(-1.0)
#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
using namespace std;
 
const int maxn=10000010;
int phi[maxn],p[maxn],vis[maxn],n;
LL s[maxn];

void calphi() {
	phi[1]=1;
	for (int i=2;i<=n;i++) {
		if (!vis[i]) {p[++p[0]]=i;phi[i]=i-1;}
		for (int j=1;j<=p[0];j++) {
			if (i*p[j]>n) break;
			vis[p[j]*i]=1;
			if (i%p[j]==0) {phi[p[j]*i]=phi[i]*p[j];break;}
			else phi[p[j]*i]=phi[p[j]]*phi[i];
		}
	}
	for (int i=2;i<=n;i++) s[i]=s[i-1]+phi[i];
}
int main() {
	scanf("%d",&n);
	calphi();
	LL ans=p[0];
	for (int i=1;i<=p[0];i++) {
		if (n/p[i]==1) break;
		ans+=2*s[n/p[i]];
	}
	printf("%lld",ans);
    return 0;
}