【BZOJ2440】完全平方数(莫比乌斯函数,容斥原理)

题意:求第k个无平方因子数

k<=10^9

思路:

感觉这东西和欧拉筛差不多……活到老学到老,退役前学点知识也是好的

为什么二分答案的上界是2*n?连LYY都证不出来

话说约大爷一年之前就已经AC此题……%%%

 1 const max=50000;
 2 var mu,flag,prime:array[1..max]of int64;
 3     cas,v,i,m,j,n:longint;
 4     ans,l,r,mid,last,tmp:int64;
 5 
 6 function clac(n:int64):int64;
 7 var i:longint;
 8 begin
 9  clac:=0;
10  for i:=1 to trunc(sqrt(n)) do clac:=clac+mu[i]*n div (i*i);
11 end;
12 
13 begin
14  assign(input,'bzoj2440.in'); reset(input);
15  assign(output,'bzoj2440.out'); rewrite(output);
16  readln(cas);
17  mu[1]:=1;
18  for i:=2 to max do
19  begin
20   if flag[i]=0 then
21   begin
22    inc(m); prime[m]:=i;
23    mu[i]:=-1;
24   end;
25   j:=1;
26   while prime[j]*i<=max do
27   begin
28    flag[prime[j]*i]:=1;
29    if i mod prime[j]=0 then
30    begin
31     mu[prime[j]*i]:=0;
32     break;
33    end;
34    mu[prime[j]*i]:=-mu[i];
35    inc(j);
36   end;
37  end;
38  for v:=1 to cas do
39  begin
40   readln(n);
41   l:=1; r:=2*n; last:=1;
42   while l<=r do
43   begin
44    mid:=(l+r)>>1;
45    tmp:=clac(mid);
46    if tmp=n then begin last:=mid; r:=mid-1; end
47     else if tmp<n then l:=mid+1
48      else r:=mid-1;
49   end;
50   writeln(last);
51  end;
52  close(input);
53  close(output);
54 end.

 

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posted on 2017-04-01 16:16  myx12345  阅读(180)  评论(0编辑  收藏  举报

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