【BZOJ】3301: [USACO2011 Feb] Cow Line(康托展开)

http://www.lydsy.com/JudgeOnline/problem.php?id=3301

其实这一题很早就a过了,但是那时候看题解写完也是似懂非懂的。。。。

听zyf神犇说是康托展开,然后拖到今天才来看看。。。

sad。。

从不知道那里来的文档里边抄的:

康托展开就是一种特殊的哈希函数,它的使用范围是对于n个数的排列进行状态的压缩和存储。
X=a[n]*(n-1)!+a[n-1]*(n-2)!+…+a[i]*(i-1)!+…+a[2]*1!+a[1]*0!
其中,a为整数,并且0<=a[i]<i(1<=i<=n)
再举个例子:1324是{1,2,3,4}排列数中第几个大的数:第一位是1小于1的数没有,是0个 0*3! 第二位是3小于3的数有1和2,但1已经在第一位了,所以只有一个数2 1*2! 。第三位是2小于2的数是1,但1在第一位,所以有0个数 0*1! ,所以比1324小的排列有0*3!+1*2!+0*1!=2个,1324是第三个大数。
例:{1,2,3,4,5}的全排列
(1)找出第96个数
首先用96-1得到95
用95去除4! 得到3余23
用23去除3! 得到3余5
用5去除2!得到2余1
用1去除1!得到1余0
有3个数比它小的数是4,所以第一位是4
有3个数比它小的数是4,但4已经在之前出现过了,所以是5(因为4在之前出现过了所以实际比5小的数是3个)
有2个数比它小的数是3
有1个数比它小的数是2
最后一个数只能是1
所以这个数是45321
(2)找出第16个数
首先用16-1得到15
用15去除4!得到0余15
用15去除3!得到2余3
用3去除2!得到1余1
用1去除1!得到1余0
有0个数比它小的数是1
有2个数比它小的数是3, 但由于1已经在之前出现过了,所以是4(因为1在之前出现过了所以实际比4小的数是2)
有1个数比它小的数是2, 但由于1已经在之前出现过了,所以是3(因为1在之前出现过了所以实际比3小的数是1)
有1个数比它小得数是2, 但由于1,3,4已经在之前出现过了,所以是5(因为1,3,4在之前出现过了所以实际比5小的数是1)
最后一个数只能是2
所以这个数是14352。

所以我们暴力模拟找第k个没有使用过的值然后搞搞就行了。

注意read要开longlong啊啊啊啊

#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;
}

 

 


 

 

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

posted @ 2014-10-15 17:32  iwtwiioi  阅读(184)  评论(0编辑  收藏