POJ-2478 Farey Sequence(欧拉函数)

Farey Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15449		Accepted: 6149

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

 

欧拉函数φ( )是指不超过n且与n互素的正整数的个数

φ函数的值 通式：φ(n)=n(1-1/p1)(1-1/p2)(1-1/p3)(1-1/p4)…..(1-1/pk),其中p1, p2……pk为n的所有质因数

#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <queue>
#include <stack>
#include <vector>
#include <iostream>
#include "algorithm"
using namespace std;
typedef long long LL;
const int MAX=1000005;
int n,m;
int phi[MAX];
LL sum[MAX];
void init(){
    int i,j;
    for (i=1;i<MAX;i++) phi[i]=i;
    for (i=2;i<MAX;i+=2) phi[i]/=2;
    for (i=3;i<MAX;i+=2) 
     if (phi[i]==i){
         for (j=i;j<MAX;j+=i)
          phi[j]=phi[j]/i*(i-1);
     }
    sum[1]=0;
    sum[2]=phi[2];
    for (i=3;i<MAX;i++){
        sum[i]=sum[i-1]+(LL)phi[i];
    }
}
int main(){
    freopen ("sequence.in","r",stdin);
    freopen ("sequence.out","w",stdout);
    int i,j;
    init();
    while (1){
        scanf("%d",&j);
        if (j==0)
         break;
        printf("%lld\n",sum[j]);
    }
    return 0;
}