BZOJ 3301: [USACO2011 Feb] Cow Line

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

分析

康托展开的模板题

代码

#include <bits/stdc++.h>

#define N 50
#define ll long long

ll jc[N];
int n,k,num[N];
bool vis[N];

void KTN()
{
    ll x;
    scanf("%lld",&x);
    x--;
    std::memset(vis,0,sizeof(vis));
    for (int i = n - 1; i >= 0; i--)
    {
        int num = x / jc[i];
        int now = 0;
        x %= jc[i];
        for (int j = 1; j <= n; j++)
            if (!vis[j])
            {
                if (now == num)
                {
                    vis[j] = 1;
                    printf("%d",j);
                    if (i)
                        printf(" ");
                    break;
                }
                else now++;
            }
    }
    printf("\n");
}

void KT()
{
    for (int i = n; i > 0; i--)
        scanf("%d",&num[i]);
    ll ans = 0;
    for (int i = n - 1; i >= 0; i--)
    {
        ll sum = 0;
        for (int j = 1; j <= i; j++)
            if (num[j] < num[i + 1])
                sum++;
        ans += sum * jc[i];
    }
    printf("%lld\n",ans + 1);
}

void getJC()
{
    for (int i = 1; i <= n; i++)
        jc[i] = jc[i - 1] * i;
}

int main()
{
    scanf("%d%d",&n,&k);
    jc[0] = 1;
    getJC();
    for (int i = 1; i <= k; i++)
    {
        char ch;
        std::cin>>ch;
        if (ch == 'P')
            KTN();
        else KT();
    }
}