康托展开

　　裸的康托展开&逆康托展开

　　康托展开就是一种特殊的hash，且是可逆的……

　　康托展开计算的是有多少种排列的字典序比这个小，所以编号应该+1；逆运算同理（-1）。

　　序列->序号：（康托展开）

　　　　对于每个数a[i]，数比它小的数有多少个在它之前没出现，记为b[i]，$ans=1+sum b[i]* (n-i)!$

　　序号->序列：（逆康托展开）

　　　　求第x个排列所对应的序列，先将x-1，然后对于a[i]，$leftlfloor frac{x}{(n-i)!} ight floor $即为在它之后出现的比它小的数的个数，所以从小到大数一下有几个没出现的数，就知道a[i]是第几个数了。

然而这题在比较答案的时候不忽略行末空格……大家小心一点……

 1 /**************************************************************
 2     Problem: 3301
 3     User: Tunix
 4     Language: C++
 5     Result: Accepted
 6     Time:84 ms
 7     Memory:1276 kb
 8 ****************************************************************/
 9  
10 //BZOJ 3301
11 #include<cstdio>
12 #include<cstring>
13 #include<cstdlib>
14 #include<iostream>
15 #include<algorithm>
16 #define rep(i,n) for(int i=0;i<n;++i)
17 #define F(i,j,n) for(int i=j;i<=n;++i)
18 #define D(i,j,n) for(int i=j;i>=n;--i)
19 #define pb push_back
20 using namespace std;
21 typedef long long LL;
22 inline LL getint(){
23     LL r=1,v=0; char ch=getchar();
24     for(;!isdigit(ch);ch=getchar()) if (ch=='-') r=-1;
25     for(; isdigit(ch);ch=getchar()) v=v*10-'0'+ch;
26     return r*v;
27 }
28 const int N=25;
29 /*******************template********************/
30 int n,m;
31 LL fac[N];
32 int a[N];
33 bool vis[N];
34 void pailie(LL x){
35     memset(vis,0,sizeof vis);
36     F(i,1,n){
37         int t=x/fac[n-i],j,k;
38         for(k=1,j=0;j<=t;k++) if (!vis[k]) j++;
39         vis[k-1]=1; a[i]=k-1;
40         x%=fac[n-i];
41     }
42     F(i,1,n-1) printf("%d ",a[i]);
43     printf("%d
",a[n]);
44 }
45 void hanghao(){
46     LL ans=1;
47     memset(vis,0,sizeof vis);
48     F(i,1,n){
49         int j=0,k;
50         vis[a[i]]=1;
51         F(k,1,a[i]) if (!vis[k]) j++;
52         ans+=j*fac[n-i];
53     }
54     printf("%lld
",ans);
55 }
56 int main(){
57 #ifndef ONLINE_JUDGE
58     freopen("3301.in","r",stdin);
59     freopen("3301.out","w",stdout);
60 #endif
61     n=getint(); m=getint();
62     fac[0]=1;
63     F(i,1,20) fac[i]=fac[i-1]*i;
64 //  F(i,0,20) printf("%lld ",fac[i]); puts("");
65     char cmd[5];
66     while(m--){
67         scanf("%s",cmd);
68         if (cmd[0]=='P'){
69             LL x=getint()-1;
70             pailie(x);
71         }else{
72             F(i,1,n) a[i]=getint();
73             hanghao();
74         }
75     }
76     return 0;
77 }
78

View Code

3301: [USACO2011 Feb] Cow Line

Time Limit: 10 Sec Memory Limit: 128 MB
Submit: 84 Solved: 50
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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

Silver

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【BZOJ】【3301】【USACO2011 Feb】Cow Line