洛谷 P3014 [USACO11FEB]牛线Cow Line

Posted 2020-10-05 一蓑烟雨任生平

P3014 [USACO11FEB]牛线Cow Line

题目背景

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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 

#include<map>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int n,m,vis[21];
long long num[21];
int main(){
    //freopen("permutation.in","r",stdin);
    //freopen("permutation.out","w",stdout);
    scanf("%d%d",&n,&m);
    for(int i=1;i<=21;i++)    num[i]=1;
    for(int i=2;i<=n;i++)    num[n-i+1]=i*num[n-i+2];
    for(int k=1;k<=m;k++){
        char c;long long x;
        cin>>c;
        memset(vis,0,sizeof(vis));
        if(c==‘P‘){
            cin>>x;
            long long k=x;
            int flag=1;
            for(int i=1;i<=n;i++){
                int ans;
                if(k>=num[i+1]){
                    long long a=k/num[i+1];
                    long long nn=k%num[i+1];
                    if(nn)    flag=1;
                    else flag=0;
                    int bns=0;
                    for(int j=1;j<=n;j++){
                        if(!vis[j])    bns++;
                        if(bns-flag==a){
                            ans=j;
                            vis[j]=1;
                            break;
                        }
                    }
                    k=nn;
                }
                else{
                    if(flag==1){
                        for(int j=1;j<=n;j++)
                            if(!vis[j]){
                                ans=j;
                                vis[j]=1;
                                break;
                            }    
                    }
                    else{
                        for(int j=n;j>=1;j--)
                            if(!vis[j]){
                                ans=j;
                                vis[j]=1;
                                break;
                            }
                    }
                }
                cout<<ans<<" ";
            }
            cout<<endl;
        }
        else{
            long long ans=0;
            for(int i=1;i<=n;i++){
                cin>>x;int bns=0;
                for(int j=1;j<=n;j++){
                    if(!vis[j])    bns++;
                    if(j==x)    break;
                }
                bns--;
                vis[x]=1;
                ans+=(long long)bns*num[i+1];
            }
            cout<<ans+1<<endl;
        }
    }
}