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

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:
 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<algorithm>
 6 #include<iostream>
 7 #include<vector>
 8 #include<map>
 9 #include<set>
10 #include<queue>
11 #include<string>
12 #define inf 1000000000
13 #define maxn 500+100
14 #define maxm 500+100
15 #define eps 1e-10
16 #define ll long long
17 #define pa pair<int,int>
18 #define for0(i,n) for(int i=0;i<=(n);i++)
19 #define for1(i,n) for(int i=1;i<=(n);i++)
20 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
21 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
22 #define mod 1000000007
23 using namespace std;
24 inline ll read()
25 {
26     ll x=0,f=1;char ch=getchar();
27     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
28     while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
29     return x*f;
30 }
31 ll n,m,a[25],b[25],fac[25];
32 int main()
33 {
34     freopen("input.txt","r",stdin);
35     freopen("output.txt","w",stdout);
36     n=read();m=read();
37     fac[0]=1;
38     for(ll i=1;i<n;i++)fac[i]=fac[i-1]*i;
39     char ch;
40     while(m--)
41     {
42         ch=' ';
43         while(ch!='P'&&ch!='Q')ch=getchar();
44         for1(i,n)a[i]=0;
45         if(ch=='P')
46         {
47             ll x=read()-1;
48             for1(i,n)
49             {
50                 ll t=x/fac[n-i]+1,j=0,k;
51                 for(k=1;j<t;k++)if(!a[k])j++;
52                 a[k-1]=1;b[i]=k-1;
53                 x%=fac[n-i];
54             }
55             for1(i,n-1)printf("%d ",b[i]);printf("%d\n",b[n]); 
56         }
57         else
58         {
59             for1(i,n)b[i]=read();
60             ll x=1;
61             for1(i,n)
62             {
63                 ll j=0,k;
64                 for(k=1;k<b[i];k++)if(!a[k])j++;
65                 a[k]=1;
66                 x+=j*fac[n-i];
67             }
68             printf("%lld\n",x);
69         }
70     }
71     return 0;
72 }
View Code

 

posted @ 2014-10-03 13:17  ZYF-ZYF  Views(210)  Comments(0Edit  收藏