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:



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('\n');
else
putchar(' ');
}
}
printf("\n%d\n", 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;
}