BNU17268 UVA129 Krypton Factor

Krypton Factor

Time Limit: 3000ms

Memory Limit: 131072KB

This problem will be judged on UVA. Original ID: 129
64-bit integer IO format: %lld      Java class name: Main

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 Krypton Factor 

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 sequence 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 and Output

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 ABA7

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

Sample Input

30 30 0

Sample Output

ABAC ABCA CBAB CABA CABC ACBA CABA28

https://www.bnuoj.com/v3/problem_show.php?pid=17268

如果一个字符串包含两个相邻的重复字串,则称它是容易的串,其他的串为困难的串

输出由前L个字符组成的,字典序第N小的困难的串.

输出的答案保证不超过80个字符

dfs(但要合理的暴力)

#include<bits/stdc++.h>using namespace std;string hh,gg,str;int check,cnt,cur;int n,l;void dfs(int k){    if(cnt==n)    {        check=1;        cur=k;        return ;    }    int kk,i,flag,j;    for(i=0; i<l; i++)    {        int ggg=str.length()-1;        if(k>ggg)//k>str.length()-1;这样写可以吗                     //已经被坑死详情见        <a target=_blank href="http://blog.csdn.net/xky140610205/article/details/52947494">http://blog.csdn.net/xky140610205/article/details/52947494</a>            str+=(i+65);        else str[k]=i+65;        flag=0;        for(j=k-1,kk=1; j>=0; j=j-2,kk++)        {            gg=str.substr(j,kk);            hh=str.substr(j+kk,kk);            if(gg==hh)            {                flag=1;                break;            }        }        if(flag)            continue;        cnt++;        dfs(k+1);        if(check) return ;    }}int main(){    int i;    while(scanf("%d%d",&n,&l)!=EOF)    {        if(n+l==0) break;        check=0;        cnt=0;        str="";//清零        dfs(0);        for(i=0; i<cur; i++)        {            if(i==cur-1)            {                cout<<str[i]<<endl;                continue;            }            if((1+i)%64==0)            {                cout<<str[i]<<endl;                continue;            }            if((1+i)%4==0)                cout<<str[i]<<" ";            else cout<<str[i];        }        printf("%d\n",cur);    }    return 0;}