【BZOJ2301】Problem b（莫比乌斯反演）

题意：对于给出的n个询问，每次求有多少个数对(x,y)，满足a≤x≤b，c≤y≤d，

且gcd(x,y) = k，gcd(x,y)函数为x和y的最大公约数。

1≤n≤50000，1≤a≤b≤50000，1≤c≤d≤50000，1≤k≤50000

思路：第一题反演……

利用容斥原理将原询问拆成4个，问题就转化为：

1<=i<=trunc(a div k)，1<=j<=trunc(b div k)，gcd(i,j)=1的(i,j)数对个数

令f(i)表示满足gcd(x,y)=i时(x,y)的对数，F(i)表示满足i|gcd(x,y)的(x,y)的对数

显然F(i)=trunc(n div i)*trunc(m div i)

f(1)=sigma u(d)*trunc(n div d)*trunc(m div d) (1<=d<=n)

观察可得trunc(n div d)*trunc(m div d)只有2根号n个取值，且每个取值对应的u(i)是连续的一段

然后就可以记录u的前缀和来优化

From  http://m.blog.csdn.net/article/details?id=50590197

 1 //uses sysutils;
 2 const max=50000;
 3 var mu,sum,prime:array[0..max]of longint;
 4     flag:array[0..max]of longint;
 5     a,b,c,d,k,i,j,t,m,cas,v:longint;
 6     tmp:double;
 7
 8 procedure swap(var x,y:longint);
 9 var t:int64;
10 begin
11  t:=x; x:=y; y:=t;
12 end;
13
14 function clac(n,m:longint):int64;
15 var i,t1,t2,pos:longint;
16     x,y:int64;
17 begin
18  if n>m then swap(n,m);
19  clac:=0; i:=1;
20  while i<=n do
21  begin
22   x:=n div i; y:=m div i;
23   t1:=n div x;
24   t2:=m div y;
25   if t1<t2 then pos:=t1
26    else pos:=t2;
27   clac:=clac+x*y*(sum[pos]-sum[i-1]);
28   i:=pos+1;
29  end;
30 end;
31
32 begin
33  assign(input,‘bzoj2301.in‘); reset(input);
34  assign(output,‘bzoj2301.out‘); rewrite(output);
35 // tmp:=now;
36  read(cas);
37  mu[1]:=1;
38  for i:=2 to max do
39  begin
40   if flag[i]=0 then
41   begin
42    inc(m); prime[m]:=i;
43    mu[i]:=-1;
44   end;
45   j:=1;
46   while (j<=m)and(prime[j]*i<=max) do
47   begin
48    t:=prime[j]*i;
49    flag[t]:=1;
50    if i mod prime[j]=0 then
51    begin
52     mu[t]:=0;
53     break;
54    end;
55    mu[t]:=-mu[i];
56    inc(j);
57   end;
58  end;
59  for i:=1 to max do sum[i]:=sum[i-1]+mu[i];
60  for v:=1 to cas do
61  begin
62   read(a,b,c,d,k);
63   dec(a); dec(c);
64   a:=a div k; b:=b div k; c:=c div k; d:=d div k;
65   writeln(clac(b,d)-clac(b,c)-clac(a,d)+clac(a,c));
66  end;
67  //writeln((now-tmp)*86400:0:2);
68  close(input);
69  close(output);
70 end.