poj 2478: Farey Sequence

Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 13984   Accepted: 5526

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
 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cmath>
 5 #include<algorithm>
 6 #include<cstring>
 7 #include<queue>
 8 #include<vector>
 9 using namespace std;
10 typedef long long LL;
11 const LL maxn=1e6+5;
12 LL N,tot,ANS;
13 LL prime[maxn],phi[maxn];
14 bool not_p[maxn];
15 void shai(){
16     not_p[1]=1;
17     for(LL i=1;i<maxn;i++){
18         if(not_p[i]==false){
19             prime[++tot]=i;
20             phi[i]=i-1;
21         }
22         for(LL j=1;j<=tot;j++){
23             LL k=prime[j]*i;
24             if(k>maxn) break;
25             not_p[k]=true;
26             if(i%prime[j]!=0){
27                 phi[k]=phi[i]*phi[prime[j]];
28             }
29             else{
30                 phi[k]=phi[i]*prime[j];
31                 break;
32             }
33         }
34     }
35 }
36 int main(){
37     shai();
38     while(scanf("%lld",&N)&&N){
39         ANS=0;
40         for(LL i=2;i<=N;i++){
41             ANS+=phi[i];
42         }
43         printf("%lld\n",ANS);
44     }
45     return 0;
46 } 

 

 
posted @ 2016-02-29 10:15  CXCXCXC  阅读(125)  评论(0编辑  收藏  举报