3301: [USACO2011 Feb] Cow Line
3301: [USACO2011 Feb] Cow Line
3301: [USACO2011 Feb] Cow Line
Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 82 Solved: 49
Description
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。
Input
* 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。
Output
* 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
Sample Input
5 2 P 3 Q 1 2 5 3 4
Sample Output
1 2 4 3 5 5
HINT
Source
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.