#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << (#x) << " = " << (x) << endl
#define printarr2(a, b, c) for1(_, 1, b) { for1(__, 1, c) cout << a[_][__]; cout << endl; }
#define printarr1(a, b) for1(_, 1, b) cout << a[_] << '\t'; cout << endl
inline const ll getint() { ll r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }int vis[105], n, cs, a[105];
ll m, p[50], ans;int main() {read(n); read(cs);p[0]=1;for1(i, 1, 25) p[i]=p[i-1]*(ll)i;char s[3];rep(tt, cs) {int j, k, t;scanf("%s", s);CC(vis, 0);if(s[0]=='P') {read(m); --m;for3(i, n, 1) {k=m/p[i-1]; m%=p[i-1];for(j=1, t=0; t<=k; ++j) if(!vis[j]) ++t;a[i]=j-1;vis[j-1]=1;}for3(i, n, 2) printf("%d ", a[i]); printf("%d\n", a[1]);}else {ans=0;for3(i, n, 1) read(a[i]);for3(i, n, 1) {for(j=1, t=0; j<a[i]; ++j) if(!vis[j]) ++t;ans+=(ll)t*p[i-1];vis[a[i]]=1;}printf("%lld\n", ans+1);}}return 0;
}

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