UVa 129 Krypton Factor

#include <iostream>
#include <cstdio>
using namespace std;
int n, L, ans[80], cnt(0), ove = false;;
bool isHard(int cur) {
	int bor = (cur + 1) / 2;
	for (int i = 1; i != bor + 1; ++i) {
		bool ok = false;
		for (int j = 0; j != i ; ++j) {
			if (ans[cur - j] != ans[cur - i - j]) {
				ok = true;
				break;
			}
		}
		if (!ok) return false;
		if (i == bor && ok) return true;
	}
}
void dfs(int cur) {
	if (ove) return;
	if (cnt == n) {
		ove = true;
		for (int i = 0; i != cur + 1; ++i) {
			if (i && !(i % 4) && i != 64) printf(" ");
			if (i == 64) printf("\n");
			printf("%c", ans[i] + ‘A‘);
		}
		printf("\n%d\n", cur + 1);
	} else {
		for (int i = 0; i != L; ++i) {
			ans[cur + 1] = i;
			if (isHard(cur + 1)) {
				++cnt;
				dfs(cur + 1);
			}
		}
	}
}
int main() {
	while (scanf("%d%d", &n, &L) == 2 && n) {
		ove = false;
		cnt = 0;
		for (int i = 0; i != L; ++i) {
			ans[0] = i;
			++cnt;
			dfs(0);
		}
	}
	return 0;
}