POJ2478 - Farey Sequence(法雷级数&&欧拉函数)

题目大意

直接看原文吧。。。。

The Farey Sequence Fn for any integer n with n >= 2 is the set of irreducible rational numbers a/b with 0 < a < b <= n and gcd(a,b) = 1 arranged in increasing order. The first few are
F2 = {1/2}
F3 = {1/3, 1/2, 2/3}
F4 = {1/4, 1/3, 1/2, 2/3, 3/4}
F5 = {1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5}
You task is to calculate the number of terms in the Farey sequence Fn.

题解

欧拉函数的一个应用~~对于输入n，答案就等于phi[1]+phi[2]+……phi[n]

代码：

#include<iostream>

#include<cmath>

#include<cstdio>

#include<cstring>

#define MAXN 1000005

using namespace std;

int phi[MAXN],prime[MAXN],check[MAXN];

void euler_phi()

{

    int cnt=0;

    memset(check,false,sizeof(check));

    phi[1]=1;

    for(int i=2; i<MAXN; i++)

    {

        if(!check[i])

        {

            prime[cnt++]=i;

            phi[i]=i-1;

        }

        for(int j=0; j<cnt&&i*prime[j]<MAXN; j++)

        {

            check[i*prime[j]]=true;

            if(i%prime[j]==0)

            {

                phi[i*prime[j]]=phi[i]*prime[j];

                break;

            }

            else

                phi[i*prime[j]]=phi[i]*(prime[j]-1);

        }

    }

}

int main()

{

    int n;

    euler_phi();

    while(cin>>n&&n)

    {

        long long  sum=0;

        for(int i=2; i<=n; i++)

            sum+=phi[i];

        cout<<sum<<endl;

    }

    return 0;

}