POJ 2478

Farey Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11772		Accepted: 4585

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

 

题目题意很明白，思路也很清晰

用欧拉函数去做就是

问题在于总是超时

直接euler_phi()

void euler_phi(int n)
{
    memset(phi,0,sizeof(phi));
    phi[1] = 1;
    for(int i=2;i<=n;i++)
    {
        int ans=i;
        int h = i;
        int m = sqrt(i+0.5);
        if(h % 2 ==0)
        {
            ans = ans/2;
            while(h%2==0)
                h/=2;
        }
        for(int j=2;j<=m;j+=2)
        {
            if(h%j==0)
            {
                ans=ans/j*(j-1);
                while(h%j==0)h/=j;
             }
         }
        if(h>1)ans=ans/h*(h-1);
        phi[i]=ans;
    }
}

超时

然后看最短用时16ms = =

天惹

可以先把素数求出来

然后再求欧拉函数

不知道为什么好像我的被我写得更慢了

找到一个直接求欧拉函数就过了的

看了好久才看懂

#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
const int maxn=1000001;
int phi[maxn];

void getphi(){//这里算法减少了一个while循环
    memset(phi,0,sizeof(phi));
    phi[1]=1;
    for(int i=2;i<maxn;i++){
        if(phi[i]==0){
            for(int j=i;j<maxn;j+=i){
                if(phi[j]==0)
                    phi[j]=j;
                phi[j]=phi[j]/i*(i-1);
            }
        }
    }
}
int main(){
    getphi();
    int n;
    while(cin>>n&&n){
        long long sum=0;
        for(int i=2;i<=n;i++){
            sum+=(long long)phi[i];
        }
       cout<<sum<<endl;
    }
    return 0;
}