P3014 [USACO11FEB]牛线Cow Line

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:

有N头牛，和K个询问

为P时，读入一个数（x），输出第x个全排列；

为Q时，读入N个数，求这是第几个全排列。 

通过康托展开 求第x个全排列 求某个全排列是第几个 

对于n个数的全排列
康托展开 

ai代表第i个元素在未出现的元素中是第几大 即在第i~n位的元素中排第几

对与 2 1 3
对于 2 只有1比它大 所以排第1大
对于 1 没有比它大的 所以排第0大 
对于 3 没有和它比较的 所以排第0大
a3=3 a2=0a1=0 

ans=3;

康托展开逆运算 
已知某一全排列在所有排列中排第x
可以求得 这个全排列 

例如 ans=20; 求它的全排列 
核心是这样的一张图


第一次 20/(3!) =3……2 说明在第一个元素后面有3个比现在的元素小 这个元素就是4
第二次 2/(2!) =1……0 说明在这个元素后面有一个比它小 这只能是2
第三次 0/(1!) =0……0 说明这个元素后面没有比它小的 只能是 1
第四次 0/(0!) =0……0 只剩3了

/*
    把1 2 3 4 5 视为第0种排列
      1 2 3 5 4 视为第1种排列 
    所以寻找第x种排列时 x 要减1
    
    同理 ans要加1 
*/
#include <vector>
#include <cctype>
#include <cstdio>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;

const int MAXN=30;

int n,m;

int s[MAXN];

LL a[MAXN];

char c[5];

inline void read(int&x) {
    int f=1;register char c=getchar();
    for(x=0;!isdigit(c);c=='-'&&(f=-1),c=getchar());
    for(;isdigit(c);x=x*10+c-48,c=getchar());
    x=x*f;
}

inline LL The_num() {
    LL ans=0;
    for(int i=1;i<=n;++i) {
        int t=0;
        for(int j=i+1;j<=n;++j) 
          if(s[i]>s[j]) ++t;
        ans+=t*a[n-i];
    }
    return ans;
}

inline void The_permutation(LL k) {
    --k;
    vector<int> v,ans;
    for(int i=1;i<=n;++i) 
      v.push_back(i);
    for(int i=n;i>=1;--i) {
        LL r=k%a[i-1];
        LL b=k/a[i-1];
        k=r;
        sort(v.begin(),v.end());
        ans.push_back(v[b]);
        v.erase(v.begin()+b);
    }
    vector<int>::iterator it;
    for(it=ans.begin();it!=ans.end();++it) 
      printf("%d ",*it);
    printf("
");
    return;
}

int hh() {
//    freopen("permutation.in","r",stdin);
//    freopen("permutation.out","w",stdout);
    LL x;
    read(n);read(m);
    a[0]=1;
    for(int i=1;i<=n;++i) a[i]=a[i-1]*i;
    for(int i=1;i<=m;++i) {
        scanf("%s",c);
        if(c[0]=='P')  {
            cin>>x;
            The_permutation(x);
        }
        else if(c[0]=='Q') {
            for(int j=1;j<=n;++j) read(s[j]);
            LL ans=The_num();
            printf("%lld
",ans+1);
        }
    }
    return 0;
}

int sb=hh();
int main(int argc,char**argv) {;}