<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[dihangmeng9104的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/dihangmeng9104</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; dihangmeng9104]]></copyright><item><title><![CDATA[Codeforces Round #529 (Div. 3) C. Powers Of Two]]></title><link>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289648</link><guid>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289648</guid><author>dihangmeng9104</author><pubDate>Sun, 30 Dec 2018 00:18:00 +0800</pubDate><description><![CDATA[http://codeforces.com/contest/1095/problem/C
题意:给n找出k个2的幂，加起来正好等于n。例如 9，4：9 = 1 + 2 + 2 + 4
思路:首先任何数都能表示成2的次幂的和，其次很容易发现，n和k都是二的次幂的情况是最基础的，因为可以分成k个n/k，而n/k一定是二的次幂。
所以，可以得出结论，只要n是2的次幂，且k&lt;=n，一...]]></description><category></category></item><item><title><![CDATA[lintcode - 不同的子序列]]></title><link>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289646</link><guid>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289646</guid><author>dihangmeng9104</author><pubDate>Sat, 30 Sep 2017 01:04:00 +0800</pubDate><description><![CDATA[给出字符串S和字符串T，计算S的不同的子序列中T出现的个数。
子序列字符串是原始字符串通过删除一些(或零个)产生的一个新的字符串，并且对剩下的字符的相对位置没有影响。(比如，“ACE”是“ABCDE”的子序列字符串,而“AEC”不是)。
样例

给出S ="rabbbit", T ="rabbit"
返回 3



 1 class Solution {
 ...]]></description><category></category></item><item><title><![CDATA[lintcode - 删除数字]]></title><link>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289642</link><guid>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289642</guid><author>dihangmeng9104</author><pubDate>Fri, 29 Sep 2017 10:44:00 +0800</pubDate><description><![CDATA[1 class Solution {
 2 public:
 3     /*
 4      * @param A: A positive integer which has N digits, A is a string
 5      * @param l: Remove k digits
 6      * @return: A string
 7      *...]]></description><category></category></item><item><title><![CDATA[lintcode - 统计比给定整数小的数的个数(两种方法)]]></title><link>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289637</link><guid>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289637</guid><author>dihangmeng9104</author><pubDate>Tue, 26 Sep 2017 22:41:00 +0800</pubDate><description><![CDATA[1 class Solution {
 2 public:
 3     /*
 4      * @param A: An integer array
 5      * @param queries: The query list
 6      * @return: The number of element in the array that are smal...]]></description><category></category></item><item><title><![CDATA[lintcode - 生成括号]]></title><link>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289634</link><guid>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289634</guid><author>dihangmeng9104</author><pubDate>Mon, 25 Sep 2017 19:02:00 +0800</pubDate><description><![CDATA[1 class Solution {
 2 public:
 3     /*
 4      * @param n: n pairs
 5      * @return: All combinations of well-formed parentheses
 6      */
 7     vector&lt;string&gt; generateParenthe...]]></description><category></category></item><item><title><![CDATA[lintcode - 房屋染色]]></title><link>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289632</link><guid>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289632</guid><author>dihangmeng9104</author><pubDate>Sun, 24 Sep 2017 10:06:00 +0800</pubDate><description><![CDATA[1 class Solution {
 2 public:
 3     /*
 4      * @param costs: n x 3 cost matrix
 5      * @return: An integer, the minimum cost to paint all houses
 6      */
 7      const int inf = 0...]]></description><category></category></item><item><title><![CDATA[lintcode - 恢复ip地址]]></title><link>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289641</link><guid>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289641</guid><author>dihangmeng9104</author><pubDate>Mon, 18 Sep 2017 20:11:00 +0800</pubDate><description><![CDATA[1 class Solution {
 2 public:
 3     /*
 4      * @param s: the IP string
 5      * @return: All possible valid IP addresses
 6      */
 7     vector&lt;string&gt; restoreIpAddresses(str...]]></description><category></category></item><item><title><![CDATA[lintcode - 被围绕的区域]]></title><link>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289627</link><guid>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289627</guid><author>dihangmeng9104</author><pubDate>Mon, 18 Sep 2017 20:09:00 +0800</pubDate><description><![CDATA[1 class Solution {
 2 public:
 3     /*
 4      * @param board: board a 2D board containing 'X' and 'O'
 5      * @return: nothing
 6      */
 7     bool flag[220][220];
 8     vector&l...]]></description><category></category></item><item><title><![CDATA[Lintcode 摊平嵌套的列表]]></title><link>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289636</link><guid>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289636</guid><author>dihangmeng9104</author><pubDate>Wed, 13 Sep 2017 00:45:00 +0800</pubDate><description><![CDATA[1 /**
 2  * // This is the interface that allows for creating nested lists.
 3  * // You should not implement it, or speculate about its implementation
 4  * class NestedInteger {
 5  *   ...]]></description><category></category></item><item><title><![CDATA[codeforces 620D Professor GukiZ and Two Arrays]]></title><link>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289500</link><guid>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289500</guid><author>dihangmeng9104</author><pubDate>Tue, 29 Aug 2017 23:59:00 +0800</pubDate><description><![CDATA[1 #include &lt;bits/stdc++.h&gt;
  2 
  3 using namespace std;
  4 
  5 const int maxn = 2000 + 50;
  6 
  7 const long long inf = 1e18;
  8 
  9 int n, m;
 10 
 11 long long suma, ...]]></description><category></category></item><item><title><![CDATA[lintcode 解码方法]]></title><link>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289624</link><guid>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289624</guid><author>dihangmeng9104</author><pubDate>Wed, 16 Aug 2017 18:49:00 +0800</pubDate><description><![CDATA[简单的动态规划

 1 class Solution {
 2 public:
 3     
 4     /*
 5      * @param s: a string,  encoded message
 6      * @return: an integer, the number of ways decoding
 7      */
 8     int ...]]></description><category></category></item><item><title><![CDATA[Tensorflow安装教程-Win10环境下]]></title><link>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289633</link><guid>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289633</guid><author>dihangmeng9104</author><pubDate>Sat, 08 Jul 2017 09:29:00 +0800</pubDate><description><![CDATA[背景：最新版的Tensoflow已经支持Python3.6
首先，下载并安装Anaconda3 内置Python3.6的版本https://www.continuum.io/downloads安装时不要修改它的推荐选项
接着下载并安装CUDA 8.0https://developer.nvidia.com/cuda-downloads
然后下载并安装cuDNN 5.1(官方推...]]></description><category></category></item><item><title><![CDATA[lintcode-dfs实现二叉树的层序遍历]]></title><link>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289620</link><guid>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289620</guid><author>dihangmeng9104</author><pubDate>Sat, 10 Jun 2017 01:06:00 +0800</pubDate><description><![CDATA[1 class Solution {
 2     /**
 3      * @param root: The root of binary tree.
 4      * @return: Level order a list of lists of integer
 5      */
 6     struct node{
 7         TreeNode...]]></description><category></category></item><item><title><![CDATA[交叉字符串]]></title><link>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289631</link><guid>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289631</guid><author>dihangmeng9104</author><pubDate>Wed, 07 Jun 2017 12:02:00 +0800</pubDate><description><![CDATA[给出三个字符串:s1、s2、s3，判断s3是否由s1和s2交叉构成。
样例
比如 s1 ="aabcc"s2 ="dbbca"

- 当 s3 ="aadbbcbcac"，返回 true.
- 当 s3 ="aadbbbaccc"， 返回 false.
dp[i][j][k] 代表 当到了s3的第i位时，s1的到了第j位 s2到了第k位。 因为i是一直向前循环...]]></description><category></category></item><item><title><![CDATA[二叉树非递归方式遍历]]></title><link>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289617</link><guid>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289617</guid><author>dihangmeng9104</author><pubDate>Wed, 07 Jun 2017 00:33:00 +0800</pubDate><description><![CDATA[1 #include &lt;bits/stdc++.h&gt;
 2 
 3 using namespace std;
 4 
 5 /**
 6  * Definition of TreeNode:
 7  * class TreeNode {
 8  * public:
 9  *     int val;
10  *     TreeNode *left,...]]></description><category></category></item><item><title><![CDATA[Uva1149]]></title><link>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289626</link><guid>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289626</guid><author>dihangmeng9104</author><pubDate>Mon, 22 May 2017 23:47:00 +0800</pubDate><description><![CDATA[每个bin最多只能放两个，所以最佳的贪心策略是从大的开始放，如果有空间放第二个，尽量放最大的。

 1 #include &lt;bits/stdc++.h&gt;
 2 
 3 using namespace std;
 4 
 5 const int maxn = 100000 + 10;
 6 
 7 int t;
 8 
 9 int n,l;
10 
1...]]></description><category></category></item><item><title><![CDATA[Uva1608]]></title><link>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289613</link><guid>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289613</guid><author>dihangmeng9104</author><pubDate>Thu, 18 May 2017 23:44:00 +0800</pubDate><description><![CDATA[如果一个序列的所有子序列中均存在至少一个元素，这个元素在该子序列中只出现一次，则这个序列non-boring。
当一个序列[x,y]中没有元素只出现一次，那么该序列不符合要求，如果有的话，设为第i个元素，则 只要[x,i-1]和 [i+1,n] 符合要求，则该序列符合要求。
在O(N)的时间内预处理一个元素左右离它最近的相同元素的位置，即可在O(1)时间内查询是否这个元素在给定序列中...]]></description><category></category></item><item><title><![CDATA[Uva12174]]></title><link>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289623</link><guid>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289623</guid><author>dihangmeng9104</author><pubDate>Tue, 16 May 2017 23:49:00 +0800</pubDate><description><![CDATA[1 #include &lt;bits/stdc++.h&gt;
 2 
 3 using namespace std;
 4 
 5 const int maxn = 5000000;
 6 
 7 int t; 
 8 
 9 int s,n;
10 
11 int a[maxn+10];
12 
13 int res[maxn+10];
14 
1...]]></description><category></category></item><item><title><![CDATA[UVa11093]]></title><link>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289619</link><guid>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289619</guid><author>dihangmeng9104</author><pubDate>Mon, 24 Apr 2017 19:08:00 +0800</pubDate><description><![CDATA[//当汽车从第i个加油站到第j个加油站无法继续走下去的时候，这时候[i,j]区间的所有加油站都无法作为起点，因为当我们到第k个加油站的时候，起码是带着&gt;=0的油去的，现在不带油直接从第k个开始肯定更不行了。 1 #include&lt;bits/stdc++.h&gt;
 2 
 3 using namespace std;
 4 
 5 const int maxn...]]></description><category></category></item><item><title><![CDATA[合并果子思路]]></title><link>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289610</link><guid>https://blog.youkuaiyun.com/dihangmeng9104/article/details/101289610</guid><author>dihangmeng9104</author><pubDate>Fri, 21 Apr 2017 23:55:00 +0800</pubDate><description><![CDATA[有N堆数量不同的果子，我们每次最多可以合并K堆果子，消耗体力值为合并这些堆的果子数之和，设计耗费体力最少的方案把N堆果子合并为一堆。
如果我们能每次合并K堆的话，显然按照哈弗曼的思想，每次合并最小的K堆直到合并完是最优解。但很显然有不能每次合并K堆的情况，比如N=8，K=3的情况，最后剩下两堆。
从N堆到1堆，减少的堆数是N-1堆，我们每次最多只能让整体减少K-1堆，所以，当 (N-...]]></description><category></category></item></channel></rss>