3301: [USACO2011 Feb] Cow Line
2015-04-13 23:18
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3301: [USACO2011 Feb] Cow Line
Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 82 Solved: 49
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Description
The N (1 <= N <= 20) cows conveniently numbered 1...N are playingyet 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 2P
3
Q
1 2 5 3 4
Sample Output
1 2 4 3 55
HINT
Source
Silver题解:这道题嘛。。。一开始想到的是生成法全排列,不过看N<=20,对于O(N!)的算法必挂无疑(生成法神马的感觉立刻让我回到小学的时光啊有木有,事实上小学时用QB跑全排列时N=12就已经需要相当长的时间了)
本题我在某某地方看到了一个新的很神奇的算法——康托展开(传送门在此,具体算法在此处不再赘述),于是开始瞎搞,一开始Q类问题求出初始序列后还弄了个树状数组进行维护,再看到N<=20时立刻感觉自己膝盖上中了来自USACO的鄙视之箭,于是P类询问我也开始暴力模拟,反正才N<=20,只要不真的瞎写都问题不大的
/************************************************************** Problem: 3301 User: HansBug Language: Pascal Result: Accepted Time:192 ms Memory:228 kb ****************************************************************/ var list:array[0..20] of int64; i,j,k,l,m,n:longint; a1,a2,a3,a4,a5:int64; a,b,c,d:array[0..100] of int64; ch:char; procedure add(x:longint); begin if x=0 then exit; while x<=n do begin inc(c[x]); inc(x,x and -x); end; end; function sum(x:longint):int64; begin if x=0 then exit(0); sum:=0; while x>0 do begin inc(sum,c[x]); dec(x,x and -x) end; end; begin list[0]:=1; for i:=1 to 20 do list[i]:=list[i-1]*i; readln(n,m); for i:=1 to m do begin readln(ch); case upcase(ch) of 'P':begin readln(a1); a1:=a1-1; for j:=1 to n do begin a[j]:=a1 div list[n-j]; a1:=a1 mod list[n-j]; end; fillchar(c,sizeof(c),0); for j:=1 to n do begin l:=0; for k:=1 to n do begin if c[k]=1 then continue; if a[j]=l then begin d[j]:=k; c[k]:=1; end; inc(l); end; end; for j:=1 to n do if j<n then write(d[j],' ') else writeln(d[j]); end; 'Q':begin for j:=1 to n do read(b[j]); readln;a1:=0; fillchar(c,sizeof(c),0); for j:=1 to n do begin add(b[j]); inc(a1,(b[j]-sum(b[j]))*list[n-j]); end; writeln(a1+1); end; end; end; end.
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