<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[yangdelu855的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/yangdelu855</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; yangdelu855]]></copyright><item><title><![CDATA[随机变量均值、方差和期望]]></title><link>https://blog.youkuaiyun.com/yangdelu855/article/details/134441578</link><guid>https://blog.youkuaiyun.com/yangdelu855/article/details/134441578</guid><author>yangdelu855</author><pubDate>Thu, 16 Nov 2023 15:56:39 +0800</pubDate><description><![CDATA[假设进行一次实验，发生概率为p，未发生为1-p，期望= 1*p + 0*(1-p) = p  n次实验则为np。这个概率质量函数称为几何分布，在一系列独立的伯努利试验中，需要进行多少次试验才能获得第一次成功的情况。推广来看，如果我们希望n次实验中，有k次发生，对于离散变量 我们得到二项分布概率质量函数。如果p = 0.2 则 第一次发生事件的平均实验次数为5次。经过k次实验 第一次发生的概率为。，利用求导和求和的可交换性，方差 = n*p*(1-p)假设一个事件发生概率为。]]></description><category></category></item><item><title><![CDATA[补题 | 给定前序中序遍历，求二叉树右视图]]></title><link>https://blog.youkuaiyun.com/yangdelu855/article/details/129846776</link><guid>https://blog.youkuaiyun.com/yangdelu855/article/details/129846776</guid><author>yangdelu855</author><pubDate>Wed, 29 Mar 2023 22:12:37 +0800</pubDate><description><![CDATA[2.层次遍历取最右边元素。]]></description><category></category></item><item><title><![CDATA[补题 | 给定L、R翻转这个范围内的链表]]></title><link>https://blog.youkuaiyun.com/yangdelu855/article/details/129845466</link><guid>https://blog.youkuaiyun.com/yangdelu855/article/details/129845466</guid><author>yangdelu855</author><pubDate>Wed, 29 Mar 2023 21:02:47 +0800</pubDate><description><![CDATA[翻转部分链表]]></description><category></category></item><item><title><![CDATA[Numpy实现两个矩阵欧式距离最小下标返回]]></title><link>https://blog.youkuaiyun.com/yangdelu855/article/details/129198376</link><guid>https://blog.youkuaiyun.com/yangdelu855/article/details/129198376</guid><author>yangdelu855</author><pubDate>Fri, 24 Feb 2023 12:34:19 +0800</pubDate><description><![CDATA[numpy使用]]></description><category></category></item><item><title><![CDATA[大脑康复训练 | 从今天起每天写一次快排]]></title><link>https://blog.youkuaiyun.com/yangdelu855/article/details/120239209</link><guid>https://blog.youkuaiyun.com/yangdelu855/article/details/120239209</guid><author>yangdelu855</author><pubDate>Sat, 11 Sep 2021 16:12:29 +0800</pubDate><description><![CDATA[def quick_sort(start,end,nums):
    if (start &gt;=end):
        return
    mid = nums[start]
    left,right = start,end
    while left &lt; right:
        while nums[right] &gt;= mid and left &lt; right:
            right -= 1
        nums[left] = nums[r.]]></description><category></category></item><item><title><![CDATA[大脑康复训练 | 马拉车算法 python实现]]></title><link>https://blog.youkuaiyun.com/yangdelu855/article/details/120189945</link><guid>https://blog.youkuaiyun.com/yangdelu855/article/details/120189945</guid><author>yangdelu855</author><pubDate>Wed, 08 Sep 2021 21:59:00 +0800</pubDate><description><![CDATA[s = list(s)
        s = "\0#"+"#".join(s) +"#\1"
        def manacher(sss):
            mx = -1
            id0 = -1
            length = len(sss)
            p = [0]*length
            maxstr = ""
            for i in range(1,length-1):
         ...]]></description><category></category></item><item><title><![CDATA[补题| 大脑康复训练]]></title><link>https://blog.youkuaiyun.com/yangdelu855/article/details/119767904</link><guid>https://blog.youkuaiyun.com/yangdelu855/article/details/119767904</guid><author>yangdelu855</author><pubDate>Thu, 19 Aug 2021 16:33:38 +0800</pubDate><description><![CDATA[最深公共父节点

递归写法


class Solution:
    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -&gt; 'TreeNode':
        
        def dfs(root,p,q):
            if(root == None):
                return None
            if(root.val == p.]]></description><category></category></item><item><title><![CDATA[铁废物补题]]></title><link>https://blog.youkuaiyun.com/yangdelu855/article/details/119153518</link><guid>https://blog.youkuaiyun.com/yangdelu855/article/details/119153518</guid><author>yangdelu855</author><pubDate>Tue, 27 Jul 2021 21:20:42 +0800</pubDate><description><![CDATA[每次参加公司组织的类似icpc的个人赛都心态炸裂

有意思的一道题 A,B 有两幅扑克牌标了1-n，随机打乱分堆，最少两堆，A先手，每次只能拿牌堆顶的数字相同的两张牌，使对方没有牌拿的那方获胜，AB已知每堆的排列，都采取最优的策略

直接拓扑排序就行了emmm


...]]></description><category></category></item><item><title><![CDATA[The name ‘save/Const:0‘ refers to a Tensor which does not exist]]></title><link>https://blog.youkuaiyun.com/yangdelu855/article/details/118788970</link><guid>https://blog.youkuaiyun.com/yangdelu855/article/details/118788970</guid><author>yangdelu855</author><pubDate>Fri, 16 Jul 2021 11:38:49 +0800</pubDate><description><![CDATA[被这个问题折磨了两天

load模型报了这个错误

只需要把这两句的顺序这么写就可以了


tf.compat.v1.reset_default_graph()
sess = tf.Session()


]]></description><category></category></item><item><title><![CDATA[9个点放在4*4的矩阵中构成三条线]]></title><link>https://blog.youkuaiyun.com/yangdelu855/article/details/115545178</link><guid>https://blog.youkuaiyun.com/yangdelu855/article/details/115545178</guid><author>yangdelu855</author><pubDate>Fri, 09 Apr 2021 12:00:20 +0800</pubDate><description><![CDATA[题意如题目

暴力



toint = {"#": 1, ".": 0}
mmaps = [[[1, 1, 1, 1], [1, 0, 1, 0], [1, 1, 0, 0], [1, 0, 0, 0]],
         [[1, 1, 1, 1], [0, 1, 0, 1], [0, 0, 1, 1], [0, 0, 0, 1]],
         [[1, 1, 1, 1], [0, 1, 1, 0], [0, 1, 0, 0], [1, 1, 0, 0]],
         [[1, 1, ]]></description><category></category></item><item><title><![CDATA[小Q的减法]]></title><link>https://blog.youkuaiyun.com/yangdelu855/article/details/115380507</link><guid>https://blog.youkuaiyun.com/yangdelu855/article/details/115380507</guid><author>yangdelu855</author><pubDate>Thu, 01 Apr 2021 18:08:42 +0800</pubDate><description><![CDATA[# -*- coding: utf-8 -*-
n,k=map(int,input().split())
ll=list(set(map(int,input().split())))
ll.sort()
cnt = 0
ind = 0
nowid = 0
lim = n
sum_value = 0
for i in range(n):
    if(ll[i]!=0):
        ind=i
        nowid=i+1
        break
print(ll[ind])
cnt+=1
.]]></description><category></category></item><item><title><![CDATA[截取数组逆序插入到数组前，原数组开始位置]]></title><link>https://blog.youkuaiyun.com/yangdelu855/article/details/114410201</link><guid>https://blog.youkuaiyun.com/yangdelu855/article/details/114410201</guid><author>yangdelu855</author><pubDate>Fri, 05 Mar 2021 19:37:30 +0800</pubDate><description><![CDATA[昨天看到道题
有一个严格递增或递减数组
截取后半段逆序加到前面

比如
100 90 80 5 6 7 8
结果5

以为二分可做 然而感觉不好做

三分太好理解了 求凹凸函数最值


# -*- coding: utf-8 -*-
def tri_search(nums_list,is_max):    #求凹凸函数最值
    l,r,mid,mmid = 0,len(nums_list)-1,0,0

    while(l&lt;r-2):  #剩三个点

        mid = (l+r)/.]]></description><category></category></item><item><title><![CDATA[ES查询某字段大于或小于某个值的数据]]></title><link>https://blog.youkuaiyun.com/yangdelu855/article/details/110391383</link><guid>https://blog.youkuaiyun.com/yangdelu855/article/details/110391383</guid><author>yangdelu855</author><pubDate>Mon, 30 Nov 2020 16:31:34 +0800</pubDate><description><![CDATA[GET /user/_search
{
  "query":{
  "range":{
      "level":{
        "gte": 5
      }
    }
  }
}

query -&gt; range 语句 字段是level

gte 是大于等于 gt是大于 lte是小于等于 lt是小于
]]></description><category></category></item><item><title><![CDATA[如何使用hadoop进行Bert tokenize]]></title><link>https://blog.youkuaiyun.com/yangdelu855/article/details/107080486</link><guid>https://blog.youkuaiyun.com/yangdelu855/article/details/107080486</guid><author>yangdelu855</author><pubDate>Thu, 02 Jul 2020 11:24:32 +0800</pubDate><description><![CDATA[任务是统计bert tokenize的后的word count

需要代码mapper，reducer，Shell脚本

首先是实现Bert tokenizer 通过sys.stdin 读取文件，将结果直接输出


# coding=utf-8
# Copyright 2018 The Google AI Language Team Authors and The HuggingFace Inc. team.
#
# Licensed under the Apache License, Version ..]]></description><category></category></item><item><title><![CDATA[五位以内的对称素数]]></title><link>https://blog.youkuaiyun.com/yangdelu855/article/details/106717816</link><guid>https://blog.youkuaiyun.com/yangdelu855/article/details/106717816</guid><author>yangdelu855</author><pubDate>Fri, 12 Jun 2020 16:09:33 +0800</pubDate><description><![CDATA[好久不做这种签到题，突然发现写了小bug的能力简直突飞猛进Orz，




#include&lt;iostream&gt;
#include &lt;string.h&gt;
#include&lt;stdio.h&gt;
using namespace std;
typedef long long ll;
int prime[100005];
int visit[100005];
int cnt = 0;
void init_prime() {

    memset(visit,0,sizeof(vi]]></description><category></category></item><item><title><![CDATA[小Q的歌单 | 卢卡斯定理]]></title><link>https://blog.youkuaiyun.com/yangdelu855/article/details/106406713</link><guid>https://blog.youkuaiyun.com/yangdelu855/article/details/106406713</guid><author>yangdelu855</author><pubDate>Thu, 28 May 2020 15:50:13 +0800</pubDate><description><![CDATA[好久不见数论的题，碰到组合数+取模 马上想到卢卡斯定理 （其实组合数的递推式也可以做

Ac代码


#include &lt;iostream&gt;
#include&lt;algorithm&gt;

using namespace std;
typedef long long ll;
ll f[101];
ll mod = 1000000007;

void init() {
    f[0] = 1;
    for (ll i = 1; i &lt;= 100; i++)
        f[i]]></description><category></category></item><item><title><![CDATA[假的数学游戏]]></title><link>https://blog.youkuaiyun.com/yangdelu855/article/details/106259117</link><guid>https://blog.youkuaiyun.com/yangdelu855/article/details/106259117</guid><author>yangdelu855</author><pubDate>Thu, 21 May 2020 15:37:54 +0800</pubDate><description><![CDATA[题目如图 看到阶乘马上想到斯大林公式  

防止溢出对两遍取对数

代码如下


import math
def solve(x):
    l, r = 1, int(1e12)
    aim = x * math.log(x)
    mid = 0
    while (l &lt; r):
        mid = (l + r) // 2
        value = const_ + mid * (math.log(mid) - 1) + 0.5 * math.log(m...]]></description><category></category></item><item><title><![CDATA[如何在服务器间用ssh跳来跳去]]></title><link>https://blog.youkuaiyun.com/yangdelu855/article/details/103427675</link><guid>https://blog.youkuaiyun.com/yangdelu855/article/details/103427675</guid><author>yangdelu855</author><pubDate>Fri, 06 Dec 2019 18:14:00 +0800</pubDate><description><![CDATA[首先查看当前机器的authorized_keys


vi .ssh/authorized_keys


ssh-rsa 
xxxxxxxxx1 YangDeLu@x
ssh-rsa 
xxxxxxxxx2 YangDeLu@y
ssh-rsa 
xxxxxxxxx3 YangDeLu@z

可以看到有几个之前添加的别的机器的公钥

这里可以添加多个其他机器的公钥

然后


vi .ssh/c...]]></description><category></category></item><item><title><![CDATA[Emoji表情正则匹配]]></title><link>https://blog.youkuaiyun.com/yangdelu855/article/details/100533999</link><guid>https://blog.youkuaiyun.com/yangdelu855/article/details/100533999</guid><author>yangdelu855</author><pubDate>Wed, 04 Sep 2019 10:33:50 +0800</pubDate><description><![CDATA[emoji表情其实就是四位的unicode

所以我们可以通过unicode来识别emoji表情


&lt;U+1F300&gt; - &lt;U+1F5FF&gt;      # symbols &amp; pictographs
&lt;U+1F600&gt; - &lt;U+1F64F&gt;      # emoticons
&lt;U+1F680&gt; - &lt;U+1F6FF&g...]]></description><category></category></item><item><title><![CDATA[淘口令正则匹配]]></title><link>https://blog.youkuaiyun.com/yangdelu855/article/details/100120873</link><guid>https://blog.youkuaiyun.com/yangdelu855/article/details/100120873</guid><author>yangdelu855</author><pubDate>Wed, 28 Aug 2019 15:49:34 +0800</pubDate><description><![CDATA[匹配带淘口令的Query


# coding:utf-8
import re
f=open("query","r")
w=open("tkl_in_query","w")
readline=f.readlines()
pat_list=["₳","$","¢","₴","€","₤","￥","＄","《"]
patt=[]
for key in pat_list:
    pat=re.co...]]></description><category></category></item></channel></rss>