【NJU749D】triple（莫比乌斯反演）

题意：

cas<=100 n<=10^5

思路：与两个数的没什么区别

F(d)=(n div d)*(n div d-1)*(n div d-2) div 6

再加上喜闻乐见的下底函数分块

又是一个不能测P的萎靡题库……

const max=1000000;
var mu,flag,prime:array[0..max]of longint;
    sum:array[0..max]of int64;
    n,m,v,pos,i,j,t,cas:longint;
    ans,x:int64;
begin
 assign(input,'triple.in'); reset(input);
 assign(output,'triple.out'); rewrite(output);
 mu[1]:=1;
 for i:=2 to max do
 begin
  if flag[i]=0 then
  begin
   inc(m); prime[m]:=i;
   mu[i]:=-1;
  end;
  j:=1;
  while (j<=m)and(prime[j]*i<=max) do
  begin
   t:=prime[j]*i; flag[t]:=1;
   if i mod prime[j]=0 then
   begin
    mu[t]:=0;
    break;
   end;
   mu[t]:=-mu[i];
   inc(j);
  end;
 end;
 for i:=1 to max do sum[i]:=sum[i-1]+mu[i];
 readln(cas);
 for v:=1 to cas do
 begin
  readln(n,m);
  ans:=0; i:=1;
  n:=n div m;
  while i<=n do
  begin
   x:=n div i; pos:=n div x;
   ans:=ans+x*(x-1)*(x-2) div 6*(sum[pos]-sum[i-1]);
   i:=pos+1;
  end;
  writeln(ans);
 end;
 close(input);
 close(output);
end.