【数论 欧拉函数】poj 2478 Farey Sequence

Farey Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 23423		Accepted: 9425

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

2
3
4
5
0

Sample Output

1
3
5
9

欧拉函数线性筛法

代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;
const int maxn=1e6+5;
typedef long long ll;
int tot,prime[maxn],phi[maxn],mark[maxn];
void init()
{
    phi[1]=1;
    for(int i=2;i<=maxn;i++)
    {
        if(!mark[i])
        {
            prime[++tot]=i;
            phi[i]=i-1;
        }
        for(int j=1;j<=tot;j++)
        {
            int x=prime[j];
            if(x*i>maxn) break;
            mark[i*x]=1;
            if(i%x==0)
            {
                phi[i*x]=phi[i]*x;
                break;
            }
            else phi[i*x]=phi[i]*phi[x];
        }
    }
}
int main()
{
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
    init();
    int n;
    while(scanf("%d",&n) && n)
    {
        ll ans=0;
        for(int i=2;i<=n;i++)
            ans+=phi[i];
        printf("%lld\n",ans);
    }
    return 0;
}