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  • [USACO11FEB] Cow Line

    https://www.luogu.org/problem/show?pid=3014

    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:
    询问这是第几个全排列,询问第k个全排列是什么

    #include<cstdio>
    #include<cstring>
    using namespace std;
    long long jc[21];
    bool v[21];
    int p[21];
    int main()
    {
        int n,q,cnt,now,sum;
        long long x,tot;
        char c[5];
        scanf("%d%d",&n,&q);
        jc[1]=1;
        for(int i=2;i<=n;i++) jc[i]=jc[i-1]*i;
        while(q--)
        {
            scanf("%s",c);
            if(c[0]=='P')
            {
                scanf("%lld",&x);
                memset(v,0,sizeof(v));
                for(int i=1;i<n;i++)
                {
                    cnt=now=0;
                    while(x>jc[n-i]) x-=jc[n-i],cnt++;
                    for(int i=1;i<=20;i++)
                        if(!v[i]) 
                        {
                            now++;
                            if(now==cnt+1) 
                            {
                                printf("%d ",i);
                                v[i]=true;
                                continue;
                            }
                        }
                }
                for(int i=1;i<=20;i++) 
                    if(!v[i])
                    {
                        printf("%d
    ",i);
                        break;
                    }
            }
            else
            {
                tot=0,sum;
                for(int i=1;i<=n;i++) scanf("%d",&p[i]);
                memset(v,0,sizeof(v));
                for(int i=1;i<n;i++)
                {
                    sum=0;
                    for(now=1;now<=n;now++)
                        if(!v[now] && now<p[i]) sum++;
                    v[p[i]]=true;
                    tot+=sum*jc[n-i];
                }
                printf("%lld
    ",tot+1);
            }
        }
    }
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  • 原文地址:https://www.cnblogs.com/TheRoadToTheGold/p/7429472.html
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