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=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
return x*f;
}
ll n,m,a[25],b[25],fac[25];
int main()
{
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
n=read();m=read();
fac[0]=1;
for(ll i=1;i<n;i++)fac[i]=fac[i-1]*i;
char ch;
while(m--)
{
ch=' ';
while(ch!='P'&&ch!='Q')ch=getchar();
for1(i,n)a[i]=0;
if(ch=='P')
{
ll x=read()-1;
for1(i,n)
{
ll t=x/fac[n-i]+1,j=0,k;
for(k=1;j<t;k++)if(!a[k])j++;
a[k-1]=1;b[i]=k-1;
x%=fac[n-i];
}
for1(i,n-1)printf("%d ",b[i]);printf("%d\n",b
);
}
else
{
for1(i,n)b[i]=read();
ll x=1;
for1(i,n)
{
ll j=0,k;
for(k=1;k<b[i];k++)if(!a[k])j++;
a[k]=1;
x+=j*fac[n-i];
}
printf("%lld\n",x);
}
}
return 0;
}

View Code