【BZOJ2820】YY的GCD（莫比乌斯反演）

题意：给定N, M,求1<=x<=N, 1<=y<=M且gcd(x, y)为质数的(x, y)有多少对，多组数据

T = 10000
N, M <= 10000000

思路：

const max=10000000;
var sum:array[0..max]of int64;
    prime,flag,f,mu:array[0..max]of longint;
    n,m,i,j,t,v,cas:longint;

function clac(n,m:longint):int64;
var t,x,y,t1,t2,i,pos:longint;
begin
 if n>m then
 begin
  t:=n; n:=m; m:=t;
 end;
 clac:=0; i:=1;
 while i<=n do
 begin
  x:=n div i; y:=m div i;
  t1:=n div x; t2:=m div y;
  if t1<t2 then pos:=t1
   else pos:=t2;
  clac:=clac+(sum[pos]-sum[i-1])*x*y;
  i:=pos+1;
 end;
end;

begin
 assign(input,'bzoj2820.in'); reset(input);
 assign(output,'bzoj2820.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 m do
  for j:=1 to max div prime[i] do
  begin
   t:=prime[i]*j;
   f[t]:=f[t]+mu[j];
  end;
 for i:=1 to max do sum[i]:=sum[i-1]+f[i];
 read(cas);
 for v:=1 to cas do
 begin
  read(n,m);
  writeln(clac(n,m));
 end;
 close(input);
 close(output);
end.