3301: [USACO2011 Feb] Cow Line

The N (1 <= N <= 20) cows conveniently numbered 1...N are playing 
yet another one of their crazy games with Farmer John. The cows 
will arrange themselves in a line and ask Farmer John what their 
line number is. In return, Farmer John can give them a line number 
and the cows must rearrange themselves into that line. 
A line number is assigned by numbering all the permutations of the 
line in lexicographic order. 

Consider this example: 
Farmer John has 5 cows and gives them the line number of 3. 
The permutations of the line in ascending lexicographic order: 
1st: 1 2 3 4 5 
2nd: 1 2 3 5 4 
3rd: 1 2 4 3 5 
Therefore, the cows will line themselves in the cow line 1 2 4 3 5. 

The cows, in return, line themselves in the configuration "1 2 5 3 4" and 
ask Farmer John what their line number is. 

Continuing with the list: 
4th : 1 2 4 5 3 
5th : 1 2 5 3 4 
Farmer John can see the answer here is 5 

Farmer John and the cows would like your help to play their game. 
They have K (1 <= K <= 10,000) queries that they need help with. 
Query i has two parts: C_i will be the command, which is either ‘P‘ 
or ‘Q‘. 

If C_i is ‘P‘, then the second part of the query will be one integer 
A_i (1 <= A_i <= N!), which is a line number. This is Farmer John 
challenging the cows to line up in the correct cow line. 

If C_i is ‘Q‘, then the second part of the query will be N distinct 
integers B_ij (1 <= B_ij <= N). This will denote a cow line. These are the 
cows challenging Farmer John to find their line number. 

有N头牛，分别用1……N表示，排成一行。 
将N头牛，所有可能的排列方式，按字典顺序从小到大排列起来。 
例如：有5头牛 
1st: 1 2 3 4 5 
2nd: 1 2 3 5 4 
3rd: 1 2 4 3 5 
4th : 1 2 4 5 3 
5th : 1 2 5 3 4 
…… 
现在，已知N头牛的排列方式，求这种排列方式的行号。 
或者已知行号，求牛的排列方式。 
所谓行号，是指在N头牛所有可能排列方式，按字典顺序从大到小排列后，某一特定排列方式所在行的编号。 
如果，行号是3，则排列方式为1 2 4 3 5 
如果，排列方式是 1 2 5 3 4 则行号为5 

有K次问答，第i次问答的类型，由C_i来指明，C_i要么是‘P’要么是‘Q’。 
当C_i为P时，将提供行号，让你答牛的排列方式。当C_i为Q时，将告诉你牛的排列方式，让你答行号。 

* Line 1: Two space-separated integers: N and K 
* Lines 2..2*K+1: Line 2*i and 2*i+1 will contain a single query. 
Line 2*i will contain just one character: ‘Q‘ if the cows are lining 
up and asking Farmer John for their line number or ‘P‘ if Farmer 
John gives the cows a line number. 

If the line 2*i is ‘Q‘, then line 2*i+1 will contain N space-separated 
integers B_ij which represent the cow line. If the line 2*i is ‘P‘, 
then line 2*i+1 will contain a single integer A_i which is the line 
number to solve for. 

第1行：N和K 
第2至2*K+1行：Line2*i ，一个字符‘P’或‘Q’，指明类型。 
如果Line2*i是P，则Line2*i+1，是一个整数，表示行号； 
如果Line2*i+1 是Q ，则Line2+i，是N个空格隔开的整数，表示牛的排列方式。

* Lines 1..K: Line i will contain the answer to query i. 

If line 2*i of the input was ‘Q‘, then this line will contain a 
single integer, which is the line number of the cow line in line 
2*i+1. 

If line 2*i of the input was ‘P‘, then this line will contain N 
space separated integers giving the cow line of the number in line 
2*i+1. 
第1至K行：如果输入Line2*i 是P，则输出牛的排列方式；如果输入Line2*i是Q，则输出行号

Silver

题解：这道题嘛。。。一开始想到的是生成法全排列，不过看N<=20，对于O(N!)的算法必挂无疑（生成法神马的感觉立刻让我回到小学的时光啊有木有，事实上小学时用QB跑全排列时N=12就已经需要相当长的时间了）

本题我在某某地方看到了一个新的很神奇的算法——康托展开（传送门在此，具体算法在此处不再赘述），于是开始瞎搞，一开始Q类问题求出初始序列后还弄了个树状数组进行维护，再看到N<=20时立刻感觉自己膝盖上中了来自USACO的鄙视之箭，于是P类询问我也开始暴力模拟，反正才N<=20，只要不真的瞎写都问题不大的

 1 /**************************************************************
 2     Problem: 3301
 3     User: HansBug
 4     Language: Pascal
 5     Result: Accepted
 6     Time:192 ms
 7     Memory:228 kb
 8 ****************************************************************/
 9  
10 var
11    list:array[0..20] of int64;
12    i,j,k,l,m,n:longint;
13    a1,a2,a3,a4,a5:int64;
14    a,b,c,d:array[0..100] of int64;
15    ch:char;
16 procedure add(x:longint);
17           begin
18                if x=0 then exit;
19                while x<=n do
20                      begin
21                           inc(c[x]);
22                           inc(x,x and -x);
23                      end;
24           end;
25 function sum(x:longint):int64;
26          begin
27               if x=0 then exit(0);
28               sum:=0;
29               while x>0 do
30                     begin
31                          inc(sum,c[x]);
32                          dec(x,x and -x)
33                     end;
34          end;
35 begin
36      list[0]:=1;
37      for i:=1 to 20 do list[i]:=list[i-1]*i;
38      readln(n,m);
39      for i:=1 to m do
40          begin
41               readln(ch);
42               case upcase(ch) of
43                    ‘P‘:begin
44                             readln(a1);
45                             a1:=a1-1;
46                             for j:=1 to n do
47                                 begin
48                                      a[j]:=a1 div list[n-j];
49                                      a1:=a1 mod list[n-j];
50                                 end;
51                             fillchar(c,sizeof(c),0);
52                             for j:=1 to n do
53                                 begin
54                                      l:=0;
55                                      for k:=1 to n do
56                                          begin
57                                               if c[k]=1 then continue;
58                                               if a[j]=l then
59                                                  begin
60                                                       d[j]:=k;
61                                                       c[k]:=1;
62                                                  end;
63                                               inc(l);
64                                          end;
65                                 end;
66                             for j:=1 to n do if j<n then write(d[j],‘ ‘) else writeln(d[j]);
67                    end;
68                    ‘Q‘:begin
69                             for j:=1 to n do read(b[j]);
70                             readln;a1:=0;
71                             fillchar(c,sizeof(c),0);
72                             for j:=1 to n do
73                                 begin
74                                      add(b[j]);
75                                      inc(a1,(b[j]-sum(b[j]))*list[n-j]);
76                                 end;
77                             writeln(a1+1);
78                    end;
79               end;
80          end;
81 end.