【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：

 1 const max=10000000;
 2 var sum:array[0..max]of int64;
 3     prime,flag,f,mu:array[0..max]of longint;
 4     n,m,i,j,t,v,cas:longint;
 5 
 6 function clac(n,m:longint):int64;
 7 var t,x,y,t1,t2,i,pos:longint;
 8 begin
 9  if n>m then
10  begin
11   t:=n; n:=m; m:=t;
12  end;
13  clac:=0; i:=1;
14  while i<=n do
15  begin
16   x:=n div i; y:=m div i;
17   t1:=n div x; t2:=m div y;
18   if t1<t2 then pos:=t1
19    else pos:=t2;
20   clac:=clac+(sum[pos]-sum[i-1])*x*y;
21   i:=pos+1;
22  end;
23 end;
24 
25 begin
26  assign(input,'bzoj2820.in'); reset(input);
27  assign(output,'bzoj2820.out'); rewrite(output);
28  mu[1]:=1;
29  for i:=2 to max do
30  begin
31   if flag[i]=0 then
32   begin
33    inc(m); prime[m]:=i;
34    mu[i]:=-1;
35   end;
36   j:=1;
37   while (j<=m)and(prime[j]*i<=max) do
38   begin
39    t:=prime[j]*i; flag[t]:=1;
40    if i mod prime[j]=0 then
41    begin
42     mu[t]:=0; break;
43    end;
44    mu[t]:=-mu[i];
45    inc(j);
46   end;
47  end;
48  for i:=1 to m do
49   for j:=1 to max div prime[i] do
50   begin
51    t:=prime[i]*j;
52    f[t]:=f[t]+mu[j];
53   end;
54  for i:=1 to max do sum[i]:=sum[i-1]+f[i];
55  read(cas);
56  for v:=1 to cas do
57  begin
58   read(n,m);
59   writeln(clac(n,m));
60  end;
61  close(input);
62  close(output);
63 end.

 

posted on 2017-04-05 09:36  myx12345  阅读(125)  评论(0)  编辑  收藏  举报

导航