2478 Farey Sequence //欧拉函数

Farey Sequence

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 7076	Accepted: 2610

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 <= 10 6). 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

2
3
4
5
0      

Sample Output

1
3
5
9      

Source

POJ Contest,Author:[email protected]

#include<cstdio>

#include<cstring>

#include<iostream>

using namespace std;

//基于素数筛选法，计算欧拉函数phi[1] to phi[MAX],复杂度约为n，很快

//传入三个数组:

//phi[]用于存放欧拉函数

//prime[]用来存放小于i的所有素数，这里其实是一个模拟堆栈

//isprime[]用来标志该数是不是素数，初始值为0

/*

这个算法的核心，基于以下这个定理：

若(N%a==0 && (N/a)%a==0) 则有:E(N)=E(N/a)*a;

若(N%a==0 && (N/a)%a!=0) 则有:E(N)=E(N/a)*(a-1);

*/

#define MAX 1000000

__int64 phi[MAX+1]={0};

__int64 prime[MAX+1]={0};

bool isprime[MAX+1]={0};

void get_phi()//这是一个基于素数筛选的线性算法，很快

{

__int64 i,j;

__int64 len=0;

for(i=2;i<=MAX;i++)

{

if(isprime[i]==false) //false代表是质数

{

prime[++len]=i;

phi[i]=i-1;

}

for(j=1;j<=len&&prime[j]*i<=MAX;j++)

{

isprime[prime[j]*i]=true;//true代表是合数

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

{

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

break;

}

else

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

}

}

}

/**//END_TEMPLATE_BY_ABILITYTAO_ACM///

__int64 s[MAX];

int main()

{

__int64 i;

phi[1]=0;

get_phi();

for(i=2;i<=MAX;i++)

s[i]=s[i-1]+phi[i];

while(1)

{

scanf("%I64d",&i);

if(i==0) break;

printf("%I64d/n",s[i]);

}

return 0;

}