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 <= 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

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。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<cstdlib>
 4 #include<cmath>
 5 #include<iostream>
 6 #include<algorithm>
 7 using namespace std;
 8 #define LL long long
 9 LL qz[1000010];
10 int phi[1000010],prime[80010],tot;
11 bool vis[1000010];
12 void Eular()
13 {
14     int i,j;
15     phi[1]=1;tot=0;
16     for(i=2;i<=1000000;i++)
17     {
18         if(vis[i]==false)
19         {
20             prime[++tot]=i;
21             phi[i]=i-1;
22         }
23         for(j=1;j<=tot&&prime[j]*i<=1000000;j++)
24         {
25             vis[prime[j]*i]=true;
26             if(i%prime[j]==0)
27             {
28                 phi[prime[j]*i]=phi[i]*prime[j];
29                 break;
30             }
31             phi[prime[j]*i]=phi[prime[j]]*phi[i];
32         }
33     }
34 }
35 void Qz()
36 {
37     for(int i=1;i<=1000000;i++)qz[i]=qz[i-1]+phi[i];
38 }
39 int main()
40 {
41     int n;
42     Eular();
43     //for(int i=1;i<=100;i++)printf("%d ",phi[i]);
44     //printf("\n");
45     Qz();
46     while(1)
47     {
48         scanf("%d",&n);
49         if(n==0)break;
50         printf("%lld\n",qz[n]-1);
51     }
52     return 0;
53 }

View Code

 

转载于:https://www.cnblogs.com/Var123/p/5288048.html