【Uva 129】Krypton Factor（困难的串）-原创手记-慕课网

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

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 from standard input and will write to standard output.

Input

Each input line contains integers n and L (in that order), where n > 0 and L is in the range 1 ≤ L ≤ 26.

Input is terminated by a line containing two zeroes.

Output

For each input line prints out the n-th 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.

As such a 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.

Your program may assume a maximum sequence length of 80.

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

ABA

ABAC

ABACA

ABACAB

ABACABA

Sample Input

7 3

30 3

0 0

Sample Output

ABAC ABA

ABAC ABCA CBAB CABA CABC ACBA CABA

大致的题意：就是你要输入一个n和L，分别代表前L个字母输出第n个小的困难的串，而困难的串就是其中没有连续重复的字符串

利用的方法就是回溯法

#include<bits/stdc++.h>

using namespace std;

int S[100],cnt;

int n,L;

int dfs(int cur){

//输出当前的hard sequence

//判断的条件是cnt为n 也就是第n个小的hard sequence

if(cnt++==n){

//cur是当前坐标长度

for(int i=0;i<cur;i++){

//注意格式

printf("%c",'A'+S[i]);

if(i%64==63 && i!=cur-1) printf("\n");

else if(i%4==3 && i!=cur-1) printf(" ");

}

//输出长度

printf("\n%d\n",cur);

return 0;

}

//接下来的内容就是判断当前字符串是不是hard sequence

for(int i=0;i<L;i++){

S[cur]=i;

int ok=1;

for(int j=1;j*2<=cur+1;j++){ //后缀长度为j a(bcd)(bcd)

//内循环检查 flag为equal

//外循环检查 flag为 ok

int equal=1;

for(int k=0;k<j;k++) if(S[cur-k]!=S[cur-k-j]) {equal=0;break;}

if(equal) {ok=0;break;}

}

if(ok) if(!dfs(cur+1)) return 0;

}

return 1;

}

int main(){

while(cin>>n>>L && (n!=0 && L!=0)){

cnt=0;

dfs(0);

}

return 0;

}