poj 2478 Farey Sequence

poj 2478 Farey Sequence

Farey Sequence Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 11996 Accepted: 4657

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

歐拉函數：

#include  
#include    
#define N 1000001 
int prime[N]; 
__int64 ans[N];  
  
int main()  
{  
    int i,j,n;
    prime[0]=0;
	prime[1]=0;        
    for(i=2;i<=N;i++)   // 生成素數表
        prime[i]=1;  
    for (i=2;i*i<=N;i++)  
    {  
        if(prime[i])  
        {  
            for(j=i*i;j<=N;j+=i)  
                prime[j]=0;  
        }  
    }  
  
    for(i=1;i<=N;i++)   // 生成歐拉函數值
        ans[i]=i;  
    for(i=2;i<=N;i++)  
    {  
        if(prime[i])  
        {  
            for(j=i;j<=N;j+=i)  
                ans[j]=ans[j]/i*(i-1);  //
        }  
    }  


    for(i=3;i<=N;i++)  //前n個數中兩兩互質的組合數
        ans[i]+=ans[i-1];  
    while(~scanf("%d",&n)&&n!=0)  
        printf("%I64d\n",ans[n]);  
    return 0;  
}