poj 2478 Farey Sequence

最新推荐文章于 2020-07-21 15:08:37 发布

luoxiaozheng

最新推荐文章于 2020-07-21 15:08:37 发布

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文章标签： mathematica integer output numbers each input

本文链接：https://blog.youkuaiyun.com/luoxiaozheng/article/details/6327930

Farey Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7298		Accepted: 2724

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

Source

POJ Contest ,Author:Mathematica@ZSU

法拉雷数列：通过观察可以看出其实是NUM（Fn）=（phi（1）+phi（2）+phi（3）+……+phi（n））；

也就是连续区间上的欧拉函数值求和；

对于这个问题欧拉函数的值总是要重复用到的所以我们希望一次求出1……n的phi（i）；那么怎么样才能高效的求出呢？

通过欧拉函数的性质：

所以在进行线性时间筛法时我们就可以把欧拉函数值求出来；

通过一次求出前缀和便解决了问题；

代码如下：

#include <iostream>
#define maxlen 1000050
using namespace std;
bool is_prime[maxlen]={0};
int prime[maxlen];
int phi[maxlen];
long long sum[maxlen];
void linear_prime()
{
    int i=2;
    int j=0;
    int k;
    phi[1]=1;
    for(i=2;i<=maxlen;i++)
    {
        if(!is_prime[i])
        {
            prime[j++]=i;
            phi[i]=i-1;
            //if(i==2)
                
            //cout<<prime[j-1]<<endl;
        }
        //cout<<j<<endl;
        for(k=0;k<j;k++)
        {
            if(i*prime[k]>maxlen)
                break;
            is_prime[i*prime[k]]=1;
            //cout<<i*prime[k]<<endl;
            if(i%prime[k]==0)
            {
                phi[i*prime[k]]=phi[i]*prime[k];
                break;
            }
            else
            {
                phi[i*prime[k]]=phi[i]*(prime[k]-1);
            }
        }
    }
}
    
int main()
{
    int i,j;
    //system("pause");
    linear_prime();
    for(i=2;i<=maxlen;i++)
        sum[i]=sum[i-1]+phi[i];
    //cout<<prime[0]<<endl;
    //system("pause");
    //system("pause");
    int k;
    while(cin>>k&&k)
        cout<<sum[k]<<endl;
    return 0;
}