POJ 2478 欧拉公式应用筛选素数

最新推荐文章于 2021-01-30 18:37:02 发布

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最新推荐文章于 2021-01-30 18:37:02 发布

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分类专栏： POJ 文章标签：欧拉公式素数筛法 poj

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POJ 2478 - Farey Sequence

Farey Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15789		Accepted: 6280

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

Sample Output

Source

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int phi[1000005];//设置phi[i]为小于i中与之互质（互素）的数的个数，然后根据欧拉公式算出。
void solve()
{
memset(phi,0,sizeof(phi));
for (int i=2;i<=1000000;i++)//筛选求phi
{
if (!phi[i])//当phi[i]未访问
{
for (int j=i;j<=1000000;j+=i)
{
if (phi[j]==0)//当phi[j]未访问
phi[j]=j;
phi[j]=phi[j]-phi[j]/i;//算出phi[j]的个数(欧拉公式)。
}
}
}
}
int main()
{
int n;
solve();
while (cin>>n)
{
if(n==0) break;
long long sum = 0 ;
for (int i=2;i<=n;i++)
sum+=phi[i];
cout<<sum<<endl;
}
return 0;
}