我们知道包含 N 个元素的堆可以看成是一棵包含 N 个节点的完全二叉树。
每个节点有一个权值。对于小根堆来说，父节点的权值一定小于其子节点的权值。
假设 N 个节点的权值分别是 1~N，你能求出一共有多少种不同的小根堆吗？
例如对于 N=4 有如下 3 种：
1
/
2 3
/
4
1
/
3 2
/
4
1
/
2 4
/
3
由于数量可能超过整型范围，你只需要输出结果除以 1000000009 的余数。
【输入格式】 一个整数 N。
对于 40% 的数据，1 <= N <= 1000
对于 70% 的数据，1 <= N <= 10000
对于 100% 的数据，1 <= N <= 100000
【输出格式】 一个整数表示答案。
【输入样例】 4
【输出样例】 3
资源约定： 峰值内存消耗（含虚拟机） < 256M CPU 消耗 < 1000ms
请严格按要求输出，不要画蛇添足地打印类似：“请您输入…” 的多余内容。
所有代码放在同一个源文件中，调试通过后，拷贝提交该源码。 不要使用 package 语句。不要使用 jdk1.7 及以上版本的特性。
主类的名字必须是：Main，否则按无效代码处理。
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1 |
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keyword">for</span><span class="token punctuation">(</span><span class="token keyword">int</span> i<span class="token operator">=</span><span class="token number">1</span><span class="token punctuation">;</span>i<span class="token operator"><=</span>n<span class="token punctuation">;</span>i<span class="token operator">++</span><span class="token punctuation">)</span><span class="token punctuation">{<!-- --></span> dp<span class="token punctuation">[</span>i<span class="token punctuation">]</span><span class="token operator">=</span><span class="token number">1</span><span class="token punctuation">;</span> <span class="token punctuation">}</span> <span class="token keyword">for</span><span class="token punctuation">(</span><span class="token keyword">int</span> i<span class="token operator">=</span>n<span class="token punctuation">;</span>i<span class="token operator">>=</span><span class="token number">1</span><span class="token punctuation">;</span>i<span class="token operator">--</span><span class="token punctuation">)</span><span class="token punctuation">{<!-- --></span> <span class="token keyword">if</span><span class="token punctuation">(</span>i<span class="token operator">*</span><span class="token number">2</span><span class="token operator">+</span><span class="token number">1</span><span class="token operator"><=</span>n<span class="token punctuation">)</span><span class="token punctuation">{<!-- --></span> dp<span class="token punctuation">[</span>i<span class="token punctuation">]</span><span class="token operator">=</span><span class="token punctuation">(</span><span class="token function">C</span><span class="token punctuation">(</span>s<span class="token punctuation">[</span>i<span class="token punctuation">]</span><span class="token operator">-</span><span class="token number">1</span><span class="token punctuation">,</span>s<span class="token punctuation">[</span>i<span class="token operator">*</span><span class="token number">2</span><span class="token operator">+</span><span class="token number">1</span><span class="token punctuation">]</span><span class="token punctuation">)</span><span class="token operator">*</span>dp<span class="token punctuation">[</span>i<span class="token operator">*</span><span class="token number">2</span><span class="token operator">+</span><span class="token number">1</span><span class="token punctuation">]</span><span class="token punctuation">)</span><span class="token operator">%</span>mod<span class="token operator">*</span>dp<span class="token punctuation">[</span>i<span class="token operator">*</span><span class="token number">2</span><span class="token punctuation">]</span><span class="token operator">%</span>mod<span class="token punctuation">;</span> <span class="token punctuation">}</span> <span class="token punctuation">}</span> cout <span class="token operator"><<</span> dp<span class="token punctuation">[</span><span class="token number">1</span><span class="token punctuation">]</span> <span class="token operator"><<</span> endl<span class="token punctuation">;</span> <span class="token keyword">return</span> <span class="token number">0</span><span class="token punctuation">;</span> <span class="token punctuation">}</span> |
学习的这位大佬的思路：
思路：假设d[i]是以完全二叉树i号位置为根结点的二叉子堆个数，则考虑我们现在需要把n个点放入这个完全二叉树里，显然根节点已经被确定，只能放最小的，然后假设左子树的节点个数为lsize，则我们需要从n-1个节点中选出lsize个节点放入左子树，选法一共组合数C(n-1,lsize)种，剩余的放在右子树中，所以d[i]=C(n-1,lsize)*d[i的左儿子]*d[i的右儿子]；
注意：求组合数需要用快速幂，乘法逆元的知识。以i为根节点个数可以先用动态规划算出来，s[i]=s[i的左儿子]+s[i的右儿子];
求阶乘逆元O（nlongn），动态规划O（n）；
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