欧拉函数

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#C语言

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

Sample Output

1
3
5
9




这是一个欧拉函数的模板题，但是要longlong类型；
#include <iostream>


using namespace std;
long long p[1000005],ans[1000005];
/*int phi(int n)
{
    int i,rea=n;
    for(int i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            rea=rea-rea/i;
            while(n%i==0) n=n/i;
        }
    }
    if(n>1)
    rea=rea-rea/n;
    return rea;
}*/这也是一种欧拉函数的写法；


int main()
{
    for(int i=1;i<1000005;i++)
    p[i]=i;
    for(int i=2;i<1000005;i=i+2)
    p[i]=p[i]/2;
    ans[1]=0;ans[2]=1;
    for(int i=3;i<1000005;i=i+2)
    {
        if(p[i]==i)
        {
            for(int j=i;j<1000005;j=j+i)
            p[j]=p[j]-p[j]/i;
        }
    }
     for(int i=3;i<1000005;i++)
     {
         ans[i]=ans[i-1]+p[i];
     }
    int n;
    while(cin>>n)
    {
        if(n==0)
        break;
        else
        {
            cout<<ans[n]<<endl;
        }


    }


    return 0;
}