Krypton Factor

Krypton Factor

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 181    Accepted Submission(s): 60

Problem Description

You
have been employed by the organisers of a Super Krypton Factor Contest
in which contestants have very high mental and physical abilities. In
one section of the contest the contestants are tested on their ability
to recall a sequenace of characters which has been read to them by the
Quiz Master. Many of the contestants are very good at recognising
patterns. Therefore, in order to add some difficulty to this test, the
organisers have decided that sequences containing certain types of
repeated subsequences should not be used. However, they do not wish to
remove all subsequences that are repeated, since in that case no single
character could be repeated. This in itself would make the problem too
easy for the contestants. Instead it is decided to eliminate all
sequences containing an occurrence of two adjoining identical
subsequences. Sequences containing such an occurrence will be called
``easy''. Other sequences will be called ``hard''.

For example,
the sequence ABACBCBAD is easy, since it contains an adjoining
repetition of the subsequence CB. Other examples of easy sequences are:

BB
ABCDACABCAB
ABCDABCD

Some examples of hard sequences are:

D
DC
ABDAB
CBABCBA

 

Input

In
order to provide the Quiz Master with a potentially unlimited source of
questions you are asked to write a program that will read input lines
that contain integers n and L (in that order), where n > 0 and L is
in the range , and for each input line prints out the nth hard sequence
(composed of letters drawn from the first L letters in the alphabet),
in increasing alphabetical order (alphabetical ordering here corresponds
to the normal ordering encountered in a dictionary), followed (on the
next line) by the length of that sequence. The first sequence in this
ordering is A. You may assume that for given n and L there do exist at
least n hard sequences.

For example, with L = 3, the first 7 hard sequences are:

A
AB
ABA
ABAC
ABACA
ABACAB
ABACABA
As
each sequence is potentially very long, split it into groups of four
(4) characters separated by a space. If there are more than 16 such
groups, please start a new line for the 17th group.

Therefore, if the integers 7 and 3 appear on an input line, the output lines produced should be

ABAC ABA
7
Input is terminated by a line containing two zeroes. Your program may assume a maximum sequence length of 80.

 

Sample Input

30 3

0 0

 

Sample Output

ABAC ABCA CBAB CABA CABC ACBA CABA

28

 

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <string>
 #include <vector>
 #include <set>
 #include <map>
 #include <stack>
 #include <queue>
 using namespace std;
 const int INF=0x7fffffff;
 const double EXP=1E-;
 const int MS=;
 int a[MS];
 int cnt,l,n;
 
 int dfs(int cur)
 {
     if(cnt++==n)
     {
         for(int i=;i<cur;i++)
         {
             if(i&&(i%)==)   //第k*64+1个的话,需要换行
                 printf("\n");
             else if(i&&i%==)   //第k*4+1个的话，需要空格
                 printf(" ");
             printf("%c",a[i]+'A');
         }
         printf("\n%d\n",cur);
         return ;    //为了得到了答案立即退出方便
     }
     else
     {
         for(int i=;i<l;i++)
         {
             a[cur]=i;
             int ok=;
             for(int j=;j*<=cur+;j++)
             {
                 int is=;
                 for(int k=;k<j;k++)
                     if(a[cur-k]!=a[cur-j-k])
                 {
                     is=;
                     break;
                 }
                 if(is)
                 {
                     ok=;
                     break;
                 }
             }
             if(ok)
             {
                 if(!dfs(cur+))
                     return ;
             }
         }
     }
     return ;
 }
 
 int main()
 {
     while(cin>>n>>l&&(n+l))
     {
         cnt=;
         dfs();
     }
 }