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BZOJ3301: [USACO2011 Feb] Cow Line
3301: [USACO2011 Feb] Cow Line
Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 67 Solved: 39
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Description
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.
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
P
3
Q
1 2 5 3 4
Sample Output
1 2 4 3 5
5
HINT
Source
Silver
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #include<iostream> 7 #include<vector> 8 #include<map> 9 #include<set>10 #include<queue>11 #include<string>12 #define inf 100000000013 #define maxn 500+10014 #define maxm 500+10015 #define eps 1e-1016 #define ll long long17 #define pa pair<int,int>18 #define for0(i,n) for(int i=0;i<=(n);i++)19 #define for1(i,n) for(int i=1;i<=(n);i++)20 #define for2(i,x,y) for(int i=(x);i<=(y);i++)21 #define for3(i,x,y) for(int i=(x);i>=(y);i--)22 #define mod 100000000723 using namespace std;24 inline ll read()25 {26 ll x=0,f=1;char ch=getchar();27 while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}28 while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}29 return x*f;30 }31 ll n,m,a[25],b[25],fac[25];32 int main()33 {34 freopen("input.txt","r",stdin);35 freopen("output.txt","w",stdout);36 n=read();m=read();37 fac[0]=1;38 for(ll i=1;i<n;i++)fac[i]=fac[i-1]*i;39 char ch;40 while(m--)41 {42 ch=‘ ‘;43 while(ch!=‘P‘&&ch!=‘Q‘)ch=getchar();44 for1(i,n)a[i]=0;45 if(ch==‘P‘)46 {47 ll x=read()-1;48 for1(i,n)49 {50 ll t=x/fac[n-i]+1,j=0,k;51 for(k=1;j<t;k++)if(!a[k])j++;52 a[k-1]=1;b[i]=k-1;53 x%=fac[n-i];54 }55 for1(i,n-1)printf("%d ",b[i]);printf("%d\n",b[n]); 56 }57 else58 {59 for1(i,n)b[i]=read();60 ll x=1;61 for1(i,n)62 {63 ll j=0,k;64 for(k=1;k<b[i];k++)if(!a[k])j++;65 a[k]=1;66 x+=j*fac[n-i];67 }68 printf("%lld\n",x);69 }70 }71 return 0;72 }
BZOJ3301: [USACO2011 Feb] Cow Line