例题 7-5 困难的串 UVa129

#include<iostream>
#include<cstdio>
#include<math.h>
#include<string>
#include<algorithm>
using namespace std;
int cnt ,n , L;
char s[1000];
int dfs(int cur)
{
	if (cnt++ == n)
	{
		for (int i = 0; i < cur; i++)
		{
			if (i){
				if (i % 64){
					if (i % 4 == 0)putchar(' ');
				}
				else putchar(10);
			}
			putchar('A' + s[i]);
		}
		printf("\n%d\n", cur);
		return 0;
	}
	for (int i = 0; i < L; i++){
		s[cur] = i;
		int ok = 1;
		for (int j = 1; j * 2 <= cur + 1; j++){
			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&&!dfs(cur + 1))return 0;
	}
	return 1;
}
int main()
{
	while (cin >> n >> L,n)
	{
		cnt = 0;
		dfs(0);
	}
    return 0;
}