【poj2487】Farey Sequence 欧拉函数

Farey Sequence

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 17399 Accepted: 6980

Description

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.

Input

There are several test cases. Each test case has only one line, which contains a positive integer n (2 <= n <= 10⁶). There are no blank lines between cases. A line with a single 0 terminates the input.

Output

For each test case, you should output one line, which contains N(n) ---- the number of terms in the Farey sequence Fn. 

Sample Input

23450

Sample Output

1359

欧拉函数模板

欧拉函数的三个性质

1 当q为质数时f(q)=q-1

2 如果i/p是p的倍数时f(i)=f(i/p)*p

3 如果i/p不是p的倍数时f(i)=f(i/p)*(p-1)

#include<iostream>#include<cstdio>#include<cstring>#define N 1000500using namespace std;long long a[N],res[N],f[N],prime[N];int n,tot;void init(){    tot=0;    memset(a,0,sizeof(a));    memset(f,0,sizeof(f));    memset(res,0,sizeof(res));    n=1000010;}void getprime(){    for (int i=2;i<=n;i++)    {        if (!a[i])        {            a[i]=i;//a数组存的是最小质因子，如果为0就是质数            prime[tot++]=i;        }        for (int j=0;j<tot && i*prime[j]<=n;j++)        {            a[i*prime[j]]=prime[j];            if (i%prime[j] == 0) break;        }    }}void getphi(){    f[1]=1;    for (int i=2;i<=n;i++)    {        if (a[i/a[i]] == a[i]) f[i]=f[i/a[i]]*a[i];//性质2        else f[i]=f[i/a[i]]*(a[i]-1);//性质3    }}int main(){    init();    getprime();    getphi();    for (int i=2;i<=n;i++)    res[i]=res[i-1]+f[i];    while(cin>>n && n) cout<<res[n]<<endl;    return 0;}