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（，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 var11    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           begin18                if x=0 then exit;19                while x<=n do20                      begin21                           inc(c[x]);22                           inc(x,x and -x);23                      end;24           end;25 function sum(x:longint):int64;26          begin27               if x=0 then exit(0);28               sum:=0;29               while x>0 do30                     begin31                          inc(sum,c[x]);32                          dec(x,x and -x)33                     end;34          end;35 begin36      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 do40          begin41               readln(ch);42               case upcase(ch) of43                    'P':begin44                             readln(a1);45                             a1:=a1-1;46                             for j:=1 to n do47                                 begin48                                      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 do53                                 begin54                                      l:=0;55                                      for k:=1 to n do56                                          begin57                                               if c[k]=1 then continue;58                                               if a[j]=l then59                                                  begin60                                                       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