【BZOJ】【3301】【USACO2011 Feb】Cow Line
2015-05-18 08:45
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康托展开
裸的康托展开&逆康托展开康托展开就是一种特殊的hash,且是可逆的……
康托展开计算的是有多少种排列的字典序比这个小,所以编号应该+1;逆运算同理(-1)。
序列->序号:(康托展开)
对于每个数a[i],数比它小的数有多少个在它之前没出现,记为b[i],$ans=1+\sum b[i]* (n-i)!$
序号->序列:(逆康托展开)
求第x个排列所对应的序列,先将x-1,然后对于a[i],$\left\lfloor \frac{x}{(n-i)!} \right\rfloor $即为在它之后出现的比它小的数的个数,所以从小到大数一下有几个没出现的数,就知道a[i]是第几个数了。
然而这题在比较答案的时候不忽略行末空格……大家小心一点……
/************************************************************** Problem: 3301 User: Tunix Language: C++ Result: Accepted Time:84 ms Memory:1276 kb ****************************************************************/ //BZOJ 3301 #include<cstdio> #include<cstring> #include<cstdlib> #include<iostream> #include<algorithm> #define rep(i,n) for(int i=0;i<n;++i) #define F(i,j,n) for(int i=j;i<=n;++i) #define D(i,j,n) for(int i=j;i>=n;--i) #define pb push_back using namespace std; typedef long long LL; inline LL getint(){ LL r=1,v=0; char ch=getchar(); for(;!isdigit(ch);ch=getchar()) if (ch=='-') r=-1; for(; isdigit(ch);ch=getchar()) v=v*10-'0'+ch; return r*v; } const int N=25; /*******************template********************/ int n,m; LL fac ; int a ; bool vis ; void pailie(LL x){ memset(vis,0,sizeof vis); F(i,1,n){ int t=x/fac[n-i],j,k; for(k=1,j=0;j<=t;k++) if (!vis[k]) j++; vis[k-1]=1; a[i]=k-1; x%=fac[n-i]; } F(i,1,n-1) printf("%d ",a[i]); printf("%d\n",a ); } void hanghao(){ LL ans=1; memset(vis,0,sizeof vis); F(i,1,n){ int j=0,k; vis[a[i]]=1; F(k,1,a[i]) if (!vis[k]) j++; ans+=j*fac[n-i]; } printf("%lld\n",ans); } int main(){ #ifndef ONLINE_JUDGE freopen("3301.in","r",stdin); freopen("3301.out","w",stdout); #endif n=getint(); m=getint(); fac[0]=1; F(i,1,20) fac[i]=fac[i-1]*i; // F(i,0,20) printf("%lld ",fac[i]); puts(""); char cmd[5]; while(m--){ scanf("%s",cmd); if (cmd[0]=='P'){ LL x=getint()-1; pailie(x); }else{ F(i,1,n) a[i]=getint(); hanghao(); } } return 0; }
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3301: [USACO2011 Feb] Cow Line
Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 84 Solved: 50
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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
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