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#include <bits/stdc++.h>
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using namespace std;
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const int N = 110;
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int a[N];
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int n;
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int ways;
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/**
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* @param step 走了几次
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* @param level 现在站在第几阶台阶上
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* 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。
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* 测试数据: 5 输出结果 一共有8种走法
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* 测试数据: 15 输出结果 一共有987种走法
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* 方案 :回溯法+递归
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*/
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void dfs(int level, int step) {
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if (level == n) {
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ways++;
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//输出结果
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for (int i = 0; i < step; i++)printf("%d\t", a[i]);
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printf("\n");
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}
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//2种分枝
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for (int i = 1; i <= 2; i++) {
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//可行
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if (level + i <= n) {
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a[step] = i;//记录解向量
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dfs(level + i, step + 1);
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}
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}
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}
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int main() {
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cin >> n;
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dfs(0, 0);
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printf("一共 %d 种方法。\n", ways);
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return 0;
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}
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