Farey Sequence

Time Limit: 1000MS	Memory Limit: 65536K
Total Submissions: 13204	Accepted: 5181

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:[email protected]

簡單的尤拉函式模板題。

所謂尤拉函式：對於一個正整數n，小於n且和n 互質的正整數（包括1 ）的個數，記做φ(n) 。

通式：φ(x)=x*(1-1/p1)*(1-1/p2)*(1-1/p3)*(1-1/p4)…..(1-1/pn),其中p1, p2……pn為x的所有質因數，x是不為0的整數。φ(1)=1（唯一和1互質的數就是1本身）。

尤拉函式程式碼實現：   

//直接求解尤拉函式
int euler(int n){ //返回euler(n) 
     int res=n,a=n;
     for(int i=2;i*i<=a;i++){
         if(a%i==0){
             res=res/i*(i-1);//先進行除法是為了防止中間資料的溢位 
             while(a%i==0) a/=i;
         }
     }
     if(a>1) res=res/a*(a-1);
     return res;
}

//篩選法打尤拉函式表 
#define Max 1000001
int euler[Max];
void Init(){ 
     euler[1]=1;
     for(int i=2;i<Max;i++)
       euler[i]=i;
     for(int i=2;i<Max;i++)
        if(euler[i]==i)
           for(int j=i;j<Max;j+=i)
              euler[j]=euler[j]/i*(i-1);//先進行除法是為了防止中間資料的溢位 
}
 

本題就是尤拉函式的直接使用：

#include<stdio.h>
#include<string.h>
#include<math.h>
#define LL  __int64
#define Max 1005000
LL sum[1005000];
void init(){
	sum[1]=1;
	for(LL i=2;i<Max;i++)
	sum[i]=i;
	for(LL i=2;i<Max;i++)
	if(sum[i]==i)
	for(LL j=i;j<Max;j+=i)
	sum[j]=sum[j]/i*(i-1);
}
	
int main()
{
	LL n;
    LL i,t;
	init();
	while(scanf("%I64d",&n)!=EOF)
	{
		t=0;
		if(n==0)break;
		for(i=2;i<=n;i++)
		t+=sum[i];
		printf("%I64d\n",t);
	}
	return 0;
}