【BZOJ】【3301】【USACO2011 Feb】Cow Line

康托展开

---

　　裸的康托展开&逆康托展开

　　康托展开就是一种特殊的hash，且是可逆的……

　　康托展开计算的是有多少种排列的字典序比这个小，所以编号应该+1；逆运算同理（-1）。

　　序列->序号：（康托展开）

　　　　对于每个数a[i]，数比它小的数有多少个在它之前没出现，记为b[i]，$ans=1+\sum b[i]* (n-i)!$

　　序号->序列：（逆康托展开）

　　　　求第x个排列所对应的序列，先将x-1，然后对于a[i]，$\left\lfloor \frac{x}{(n-i)!} \right\rfloor $即为在它之后出现的比它小的数的个数，所以从小到大数一下有几个没出现的数，就知道a[i]是第几个数了。

 

 

然而这题在比较答案的时候不忽略行末空格……大家小心一点……

/**************************************************************
    Problem: 3301
    User: Tunix
    Language: C++
    Result: Accepted
    Time:84 ms
    Memory:1276 kb
****************************************************************/
 
//BZOJ 3301
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define rep(i,n) for(int i=0;i<n;++i)
#define F(i,j,n) for(int i=j;i<=n;++i)
#define D(i,j,n) for(int i=j;i>=n;--i)
#define pb push_back
using namespace std;
typedef long long LL;
inline LL getint(){
    LL r=1,v=0; char ch=getchar();
    for(;!isdigit(ch);ch=getchar()) if (ch=='-') r=-1;
    for(; isdigit(ch);ch=getchar()) v=v*10-'0'+ch;
    return r*v;
}
const int N=25;
/*******************template********************/
int n,m;
LL fac[N];
int a[N];
bool vis[N];
void pailie(LL x){
    memset(vis,0,sizeof vis);
    F(i,1,n){
        int t=x/fac[n-i],j,k;
        for(k=1,j=0;j<=t;k++) if (!vis[k]) j++;
        vis[k-1]=1; a[i]=k-1;
        x%=fac[n-i];
    }
    F(i,1,n-1) printf("%d ",a[i]);
    printf("%d\n",a[n]);
}
void hanghao(){
    LL ans=1;
    memset(vis,0,sizeof vis);
    F(i,1,n){
        int j=0,k;
        vis[a[i]]=1;
        F(k,1,a[i]) if (!vis[k]) j++;
        ans+=j*fac[n-i];
    }
    printf("%lld\n",ans);
}
int main(){
#ifndef ONLINE_JUDGE
    freopen("3301.in","r",stdin);
    freopen("3301.out","w",stdout);
#endif
    n=getint(); m=getint();
    fac[0]=1;
    F(i,1,20) fac[i]=fac[i-1]*i;
//  F(i,0,20) printf("%lld ",fac[i]); puts("");
    char cmd[5];
    while(m--){
        scanf("%s",cmd);
        if (cmd[0]=='P'){
            LL x=getint()-1;
            pailie(x);
        }else{
            F(i,1,n) a[i]=getint();
            hanghao();
        }
    }
    return 0;
}

3301: [USACO2011 Feb] Cow Line

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 84  Solved: 50
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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

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