POJ 2478 Farey Sequence

 名字是法雷数列其实是欧拉phi函数

             Farey Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11246		Accepted: 4363

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

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

typedef unsigned long long int LL;

LL sum[1001000],phi[1001000];

void phi_table()
{
   phi[1]=1;
   for(int i=2;i<=1000010;i++)if(!phi[i])
   {
       for(int j=i;j<=1000010;j+=i)
       {
           if(!phi[j]) phi[j]=j;
           phi[j]=phi[j]/i*(i-1);
       }
   }
   sum[2]=1;
   for(int i=3;i<=1000010;i++)
   {
       sum[i]=sum[i-1]+phi[i];
   }
}

int main()
{
   phi_table();
   int n;
   while(scanf("%d",&n)!=EOF&&n)
   {
       printf("%llu
",sum[n]);
   }
   return 0;
}