UVA129 —— Krypton Factor (氪因素)

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 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

题意：如果有一个字符串中包含两个相邻的重复子串，则称作容易的串，其他的则称为困难的串，要求不包含容易的串

#include <iostream>
#include <cstdio>
#include <vector>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
int cnt ;
char ma[1100][1100];
int n, len;
int  p[100];
int dfs(int cur)
{
    if(cnt == n)
    {
        for(int i = 1; i < cur; i++)
        {
            printf("%c",p[i] + 'A' -1);
            if(i == cur - 1)
                break;
            if(i%4==0)
            {
                if(i%64==0)
                    putchar('
');
                else
                    putchar(' ');
            }
        }
        printf("
%d
", cur-1);
        return 0;
    }
    for(int i = 1; i <= len; i++)
    {
        p[cur] = i;
        int ok = 1;
        for(int j = 1; j * 2 <= cur+1; j++)  //枚举进行比较
        {
            int flag =1;
            for(int k = 0; k < j; k++)
            {
                if(p[cur- k] != p[cur-j-k])
                {
                    flag = 0;
                    break;
                }
            }
            if(flag )
            {
                ok =0;
                break;
            }
        }
        if(ok)
        {
            cnt ++;
            if(!dfs(cur + 1))
                return 0;
        }
    }
    return 1;
}

int main()
{
    while(scanf("%d%d",&n,&len) != EOF)
    {
        cnt = 0;
        if(!n && !len)
            break;
        dfs(1);
    }
    return 0;
}