Cow Line（洛谷-P3014）

题目描述

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.

输入输出格式

输入格式：

* 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.

输出格式：

* 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.

输入输出样例

输入样例#1：

5 2 
P 
3 
Q 
1 2 5 3 4 

输出样例#1：

1 2 4 3 5 
5 

题意：n 头牛编号从 1~n 排成一行，行号按照字典序分配，有 k 组查询，P x 代表询问行号为 x 的序列，Q a1 a2 ... an 代表询问在 a1 a2 ... an 排列下的行号

思路：可以发现，P 查询是询问 1~n 的逆康托展开，Q 询问是询问 1~n 的康托展开，套用模版即可，需要注意的是由于 n 最大到 20，要用 long long

源代码

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#define EPS 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LL long long
#define Pair pair<int,int>
const int MOD = 1E9+7;
const int N = 1000+5;
const int dx[] = {-1,1,0,0,-1,-1,1,1};
const int dy[] = {0,0,-1,1,-1,1,-1,1};
using namespace std;

LL a[N];
LL fac[N];
bool vis[N];
void getFactor(LL n) { //计算阶乘
    fac[0]=1;
    for(int i=1; i<=n; i++)
        fac[i]=fac[i-1]*i;
}
LL contor(LL n) { //康托展开
    LL X=0;
    for(int i=1;i<n;i++){
        LL cnt=0;
        for(int j=i+1;j<=n;j++)
            if(a[j]<a[i])
                cnt++;
        X+=cnt*fac[n-i];
    }
    return X+1;
}
vector<LL> revContor(LL n, LL X) {//逆康托展开
    memset(vis,false,sizeof(vis));
    vector<LL> res(n,-1);
    X--;

    LL residue=X;//除数
    for (int i=0; i<=n-1; i++) {
        LL cnt=residue/(fac[n-i-1]);
        residue=residue%(fac[n-i-1]);

        for(int j=1;j<=n;j++) {
            if (!vis[j]) {
                if(!cnt){
                    vis[j]=true;
                    res[i]=j;
                    break;
                }
                cnt--;
            }
        }
    }
    return res;
}
int main() {
    LL n;
    scanf("%lld",&n);
    getFactor(n);

    LL k;
    scanf("%lld",&k);
    while(k--){
        getchar();
        char ch;
        scanf("%c",&ch);
        if(ch=='P'){
            LL num;
            scanf("%lld",&num);
            vector<LL> res=revContor(n,num);
            for(int i=0;i<res.size();i++)
                printf("%lld ",res[i]);
            printf("\n");
        }
        else if(ch=='Q'){
            for(int i=1;i<=n;i++)
                scanf("%lld",&a[i]);
            LL res=contor(n);
            printf("%lld\n",res);
        }
    }
    return 0;
}