bzoj 2818 gcd 线性欧拉函数

2818: Gcd

Time Limit: 10 Sec  Memory Limit: 256 MB
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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

思路：gcd（x,y)=p;p为素数；

　　　gcd（x/p,y/p)==1;

　　　一个p的贡献为1-（N/p)，求前缀和意思；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mod 1000000009
#define inf 999999999
#define esp 0.00000000001
//#pragma comment(linker, "/STACK:102400000,102400000")
const int N=1e5,M=1e7+1000;
ll p[M],ji;
bool vis[M];
ll phi[M];
void get_eular(int n)
{
    ji = 0;
    memset(vis, true, sizeof(vis));
    for(int i = 2; i <= n; i++)
    {
        if(vis[i])
        {
            p[ji ++] = i;
            phi[i] = i - 1;
        }
        for(int j = 0; j < ji && i * p[j] <= n; j++)
        {
            vis[i * p[j]] = false;
            if(i % p[j] == 0)
            {
                phi[i * p[j]] = phi[i] * p[j];
                break;
            }
            else
            phi[i * p[j]] = phi[i] * phi[p[j]];
        }
    }
}
ll sumphi[M];
int main()
{
    ll x,y,z,i,t;
    get_eular(10000100);
    sumphi[0]=1;
    for(i=1;i<=10000000;i++)
    sumphi[i]=sumphi[i-1]+phi[i];
    while(~scanf("%lld",&x))
    {
        ll ans=0;
        for(i=0;p[i]<=x;i++)
        ans+=2*sumphi[x/p[i]]-1;
        printf("%lld\n",ans);
    }
    return 0;
}