Poj 2478-Farey Sequence 欧拉函数,素数,线性筛

Farey Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14291		Accepted: 5647

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

Source

POJ Contest,Author:Mathematica@ZSU

 

题解：

直接欧拉函数，求个前缀和，就好了。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
#define LL long long
LL qz[1000010];
int phi[1000010],prime[80010],tot;
bool vis[1000010];
void Eular()
{
    int i,j;
    phi[1]=1;tot=0;
    for(i=2;i<=1000000;i++)
    {
        if(vis[i]==false)
        {
            prime[++tot]=i;
            phi[i]=i-1;
        }
        for(j=1;j<=tot&&prime[j]*i<=1000000;j++)
        {
            vis[prime[j]*i]=true;
            if(i%prime[j]==0)
            {
                phi[prime[j]*i]=phi[i]*prime[j];
                break;
            }
            phi[prime[j]*i]=phi[prime[j]]*phi[i];
        }
    }
}
void Qz()
{
    for(int i=1;i<=1000000;i++)qz[i]=qz[i-1]+phi[i];
}
int main()
{
    int n;
    Eular();
    //for(int i=1;i<=100;i++)printf("%d ",phi[i]);
    //printf("\n");
    Qz();
    while(1)
    {
        scanf("%d",&n);
        if(n==0)break;
        printf("%lld\n",qz[n]-1);
    }
    return 0;
}