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#include <bits/stdc++.h>
using namespace std;
const int N = 110;
int a[N];
int n;
int ways;
/**
* @param step 走了几次
* @param level 现在站在第几阶台阶上
* N阶楼梯上楼问题：一次可以走两阶或一阶，请把所有行走方式打印出来。
* 测试数据: 5 输出结果 一共有8种走法
* 测试数据: 15 输出结果 一共有987种走法
* 方案 ：回溯法+递归
*/
void dfs(int level, int step) {
if (level == n) {
ways++;
//输出结果
for (int i = 0; i < step; i++)printf("%d\t", a[i]);
printf("\n");
}
//2种分枝
for (int i = 1; i <= 2; i++) {
//可行
if (level + i <= n) {
a[step] = i;//记录解向量
dfs(level + i, step + 1);
int main() {
cin >> n;
dfs(0, 0);
printf("一共 %d 种方法。\n", ways);
return 0;
