Poj 2478 Farey Sequence

Farey Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12815		Accepted: 5003

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

题解：其实就是求出2-N的欧拉函数之和，弄个前缀和即可= =（PS：最近才学的欧拉函数，拿来用用^_^）

var
   i,j,l,m,n:longint;
   a,b:array[0..1000005] of int64;
begin
     for i:=2 to 1000000 do
         begin
              if a[i]=0 then
                 begin
                      inc(b[0]);
                      b[b[0]]:=i;
                      a[i]:=i-1;
                 end;
              for j:=1 to b[0] do
                  begin
                       if (b[j])>(1000000 div i) then break;
                       if (i mod b[j])=0 then a[i*b[j]]:=a[i]*b[j] else a[i*b[j]]:=a[i]*(b[j]-1);
                  end;
         end;
     a[1]:=0;
     for i:=2 to 1000000 do a[i]:=a[i-1]+a[i];
     while not(eof) do
           begin
                readln(n);
                if n=0 then halt;
                writeln(a[n]);
           end;
end.