3301: [USACO2011 Feb] Cow Line
Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 82 Solved: 49
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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.