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

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 ;
}

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