BZOJ3301: [USACO2011 Feb] Cow Line

3301: [USACO2011 Feb] Cow Line

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 67  Solved: 39
[Submit][Status]

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

题解：

我还是太sb。。。

裸的康托展开和逆康托展开。

没开long long 一直WA，搞了两小时。。。

代码：

 #include<cstdio>
 #include<cstdlib>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 #include<iostream>
 #include<vector>
 #include<map>
 #include<set>
 #include<queue>
 #include<string>
 #define inf 1000000000
 #define maxn 500+100
 #define maxm 500+100
 #define eps 1e-10
 #define ll long long
 #define pa pair<int,int>
 #define for0(i,n) for(int i=0;i<=(n);i++)
 #define for1(i,n) for(int i=1;i<=(n);i++)
 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
 #define mod 1000000007
 using namespace std;
 inline ll read()
 {
     ll x=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=*x+ch-'';ch=getchar();}
     return x*f;
 }
 ll n,m,a[],b[],fac[];
 int main()
 {
     freopen("input.txt","r",stdin);
     freopen("output.txt","w",stdout);
     n=read();m=read();
     fac[]=;
     for(ll i=;i<n;i++)fac[i]=fac[i-]*i;
     char ch;
     while(m--)
     {
         ch=' ';
         while(ch!='P'&&ch!='Q')ch=getchar();
         for1(i,n)a[i]=;
         if(ch=='P')
         {
             ll x=read()-;
             for1(i,n)
             {
                 ll t=x/fac[n-i]+,j=,k;
                 for(k=;j<t;k++)if(!a[k])j++;
                 a[k-]=;b[i]=k-;
                 x%=fac[n-i];
             }
             for1(i,n-)printf("%d ",b[i]);printf("%d\n",b[n]);
         }
         else
         {
             for1(i,n)b[i]=read();
             ll x=;
             for1(i,n)
             {
                 ll j=,k;
                 for(k=;k<b[i];k++)if(!a[k])j++;
                 a[k]=;
                 x+=j*fac[n-i];
             }
             printf("%lld\n",x);
         }
     }
     return ;
 }