<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[退役ACMer]]></title><description><![CDATA[新博客更新LeetCode题解: https://www.cnblogs.com/CodingDreamer/]]></description><link>https://blog.youkuaiyun.com/ADjky</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; ADjky]]></copyright><item><title><![CDATA[代码狂欢复赛]]></title><link>https://blog.youkuaiyun.com/ADjky/article/details/143826429</link><guid>https://blog.youkuaiyun.com/ADjky/article/details/143826429</guid><author>ADjky</author><pubDate>Sun, 17 Nov 2024 02:01:26 +0800</pubDate><description><![CDATA[离散化树状数组 + 二分，现场没有加离散化超时。]]></description><category></category></item><item><title><![CDATA[LRU]]></title><link>https://blog.youkuaiyun.com/ADjky/article/details/132993659</link><guid>https://blog.youkuaiyun.com/ADjky/article/details/132993659</guid><author>ADjky</author><pubDate>Wed, 31 Aug 2022 10:01:00 +0800</pubDate><description><![CDATA[#include &lt;iostream&gt;
#include &lt;list&gt;
#include &lt;unordered_map&gt;

struct Node
{
  int v_;
  Node(int v) : v_(v) {}
};
class LRU
{
public:
  LRU(int capacity) : capacity_(capacity), size...]]></description><category></category></item><item><title><![CDATA[51nod 1449 砝码称重]]></title><link>https://blog.youkuaiyun.com/ADjky/article/details/132993658</link><guid>https://blog.youkuaiyun.com/ADjky/article/details/132993658</guid><author>ADjky</author><pubDate>Wed, 22 Sep 2021 15:13:00 +0800</pubDate><description><![CDATA[Solution

按位判断
考虑w进制借位

Sample Code
#include &lt;bits/stdc++.h&gt;
using namespace std;

int w, m;

void Do(){
    if(w == 1) {
        cout &lt;&lt; "YES" &lt;&lt; endl;
        return;
    }
    in...]]></description><category></category></item><item><title><![CDATA[LeetCode 21-29题解]]></title><link>https://blog.youkuaiyun.com/ADjky/article/details/133043589</link><guid>https://blog.youkuaiyun.com/ADjky/article/details/133043589</guid><author>ADjky</author><pubDate>Fri, 17 Sep 2021 16:09:00 +0800</pubDate><description><![CDATA[21 简单
Solution

模拟

Sample Code
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int ...]]></description><category></category></item><item><title><![CDATA[LeetCode 11-20题解]]></title><link>https://blog.youkuaiyun.com/ADjky/article/details/132993657</link><guid>https://blog.youkuaiyun.com/ADjky/article/details/132993657</guid><author>ADjky</author><pubDate>Thu, 16 Sep 2021 16:18:00 +0800</pubDate><description><![CDATA[11 中等
Solution


用i, j(i &lt; j)表示选取的两条线位置, 则水量为\((j - i)* min(a_j, a_i)\)


我们考虑固定选取的右侧j, 则左侧可选为1...j - 1


设\(i \in [1, j - 1]\), 若存在 \(k \in [1, j - 1] \ \&amp; \ k &lt; i \ \&amp;  \ a_k &gt;= a_...]]></description><category></category></item><item><title><![CDATA[LeetCode 6-10 题解]]></title><link>https://blog.youkuaiyun.com/ADjky/article/details/132993656</link><guid>https://blog.youkuaiyun.com/ADjky/article/details/132993656</guid><author>ADjky</author><pubDate>Wed, 15 Sep 2021 10:03:00 +0800</pubDate><description><![CDATA[6 中等
Soluion

模拟规律题
找同一行相邻两个字符再原字符串中的间隔

Sample Code (map)
class Solution {
public:
    string convert(string s, int numRows) {
        if(numRows == 1) return s;
        string res = "";
        int...]]></description><category></category></item><item><title><![CDATA[LeetCode 1-5题解]]></title><link>https://blog.youkuaiyun.com/ADjky/article/details/120289511</link><guid>https://blog.youkuaiyun.com/ADjky/article/details/120289511</guid><author>ADjky</author><pubDate>Tue, 14 Sep 2021 16:31:06 +0800</pubDate><description><![CDATA[1 简单
Soluion

map
排序 + 双指针

Sample Code (map)
class Solution {
public:
    map&lt;int, int&gt; mp;
    map&lt;int, int&gt; pos;
    vector&lt;int&gt; twoSum(vector&lt;int&gt;&amp; nums, int target) {
        int len = nums.size();
        for(int i = 0; i ]]></description><category></category></item><item><title><![CDATA[LeetCode刷题重启博客]]></title><link>https://blog.youkuaiyun.com/ADjky/article/details/120289087</link><guid>https://blog.youkuaiyun.com/ADjky/article/details/120289087</guid><author>ADjky</author><pubDate>Tue, 14 Sep 2021 15:49:34 +0800</pubDate><description><![CDATA[Motivation

近两年没怎么刷题
研二实践教学，目前任务还算轻松
打算整点时间把LeetCode刷一遍，保持手感
题解形式: 代码 + 口胡(活多的话各位就将就着看)
尽量一天5道题, 就按序来
工作日更新，也调整下目前懒散的学习状态


...]]></description><category></category></item><item><title><![CDATA[数据结构－树]]></title><link>https://blog.youkuaiyun.com/ADjky/article/details/80149621</link><guid>https://blog.youkuaiyun.com/ADjky/article/details/80149621</guid><author>ADjky</author><pubDate>Mon, 30 Apr 2018 15:24:44 +0800</pubDate><description><![CDATA[


数据结构－树(上)
定义
特点
基本术语
二叉树(Binary Tree)
二叉树的性质
操作集










数据结构－树(上)



定义

Tree = (D, H), D是具有相同特性的数据元素的集合, H是Ｄ上二元关系的集合, T满足如下定义

ADT Tree{

​   数据对象: D

​   数据关系: H

​   若D=∅D=∅D = \emptyset, H也为...]]></description><category></category></item><item><title><![CDATA[数据结构-串 & KMP]]></title><link>https://blog.youkuaiyun.com/ADjky/article/details/79787926</link><guid>https://blog.youkuaiyun.com/ADjky/article/details/79787926</guid><author>ADjky</author><pubDate>Mon, 02 Apr 2018 13:22:19 +0800</pubDate><description><![CDATA[


串
串的定义及子串
空白字符
特点
C语言字符数组和字符型指针
串定义
串的模式匹配算法
AC自动机









串


  字符串的简称, 字符串本身就是一种数据结构, 由零个或者多个字符组成的有限序列(顺序存储结构)




串的定义及子串


定义 ：在C语言中, 串可以用字符数组来定义：

char str[] = &quot;adrui&quot;;

这里str 是串的名字, 串长为5 ，而s...]]></description><category></category></item><item><title><![CDATA[数据结构-栈]]></title><link>https://blog.youkuaiyun.com/ADjky/article/details/79750303</link><guid>https://blog.youkuaiyun.com/ADjky/article/details/79750303</guid><author>ADjky</author><pubDate>Thu, 29 Mar 2018 22:21:24 +0800</pubDate><description><![CDATA[


栈(Stack)
栈顶和栈底
栈的特性
栈的存储结构
顺序栈
链栈
栈的应用
CodeUp 1743









栈(Stack)


  是一种限制存取点的线性表, 只能在一端进行插入或删除操作




栈顶和栈底


栈顶:  指栈中允许进行插入和删除操作的一端
栈底:  栈的另一端




栈的特性

典型的先入后出



栈的存储结构


顺序存储结构: 顺序栈
链式存储结构:...]]></description><category></category></item><item><title><![CDATA[数据结构-队列]]></title><link>https://blog.youkuaiyun.com/ADjky/article/details/79750278</link><guid>https://blog.youkuaiyun.com/ADjky/article/details/79750278</guid><author>ADjky</author><pubDate>Thu, 29 Mar 2018 22:19:14 +0800</pubDate><description><![CDATA[


队列(Queue)
队首和队尾
队列的特性
队列的存储结构
顺序队
链队








队列(Queue)


  也是一种限制存取点的线性表




队首和队尾


队首: 队列允许进行删除操作的一端
队尾: 队列允许进行插入操作的一端




队列的特性

​    先入先出



队列的存储结构


顺序存储结构: 顺序队
链式存储结构: 链队




顺序队


结构定义:

ty...]]></description><category></category></item><item><title><![CDATA[线性表]]></title><link>https://blog.youkuaiyun.com/ADjky/article/details/79679541</link><guid>https://blog.youkuaiyun.com/ADjky/article/details/79679541</guid><author>ADjky</author><pubDate>Sat, 24 Mar 2018 16:56:42 +0800</pubDate><description><![CDATA[


线性表
分类
顺序表
链表








线性表


  是一种逻辑结构, 不特指某一具体在内存中以某种物理结构存储的线性结构
  
  线性表是指具有相同特性数据元素的有限序列




分类


[顺序表]
[链表]




顺序表


物理结构(在内存中存储方式):

连续存储, 且静态分配
数据域

存放数据的data数组
记录数据长度的length
结构定义:

/*
*ElemT...]]></description><category></category></item><item><title><![CDATA[About Me]]></title><link>https://blog.youkuaiyun.com/ADjky/article/details/79182185</link><guid>https://blog.youkuaiyun.com/ADjky/article/details/79182185</guid><author>ADjky</author><pubDate>Sat, 27 Jan 2018 18:15:19 +0800</pubDate><description><![CDATA[My Info



教育

本科计算机在读(2015.9至今), ACM竞赛退役选手



技能


熟悉C++、Java等编程语言
了解javaWeb并有一定开发经验
熟悉数据结构与算法
ACM相关算法, 擅(只)长(会)数论




竞赛经历(ACM-ICPC)


2015年10月接触Online Judge, 入坑
2016年3月参加ACM校赛(超常发挥), 入校集训队
2016年5月参]]></description><category></category></item><item><title><![CDATA[Codecraft-18 and Codeforces Round #458 (Div. 1 + Div. 2, combined) C】 Travelling Salesman and Specia]]></title><link>https://blog.youkuaiyun.com/ADjky/article/details/79121254</link><guid>https://blog.youkuaiyun.com/ADjky/article/details/79121254</guid><author>ADjky</author><pubDate>Sun, 21 Jan 2018 17:00:17 +0800</pubDate><description><![CDATA[数位dp

预处理1-1000中经过刚好经过k-1次变换到1的数. 
然后搞一个数位dp, 当limit为false的时候可以在后面用组合数放置1. 
要特判k==0(只有1满足)k==1(1不满足要减去)

#include 
using namespace std;
vectorint> vec;
templateclass T>
T read()
{
    T x = 0, f = 1;]]></description><category></category></item><item><title><![CDATA[Codeforces 893E 数论]]></title><link>https://blog.youkuaiyun.com/ADjky/article/details/78679090</link><guid>https://blog.youkuaiyun.com/ADjky/article/details/78679090</guid><author>ADjky</author><pubDate>Thu, 30 Nov 2017 18:03:47 +0800</pubDate><description><![CDATA[#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<int> VI;const int N = 1000050;
const ll mod = 1e9 + 7;template<class T>
T read()
{
    T x = 0, f = 1; char ch = getch]]></description><category></category></item><item><title><![CDATA[Wannafly挑战赛4 E 二次剩余]]></title><link>https://blog.youkuaiyun.com/ADjky/article/details/78629135</link><guid>https://blog.youkuaiyun.com/ADjky/article/details/78629135</guid><author>ADjky</author><pubDate>Fri, 24 Nov 2017 23:10:19 +0800</pubDate><description><![CDATA[题解
p =2 特判a,b
p!=2, 左边乘4-> 
(2x+a)2=a2−4b(modp)(2x+a)^2=a^2-4b(mod p) 
所以有解的话a^2-4*b是模p的二次剩余,  
此时设t^2 = a^2-4*b(mod p) 
则2*x+a=t(mod p) 
因为gcd(2, p) = 1, 扩展欧几里得定理可得一定有解
#include <bits/stdc++.h>
using]]></description><category></category></item><item><title><![CDATA[Wannafly模拟赛5 F Course]]></title><link>https://blog.youkuaiyun.com/ADjky/article/details/78467694</link><guid>https://blog.youkuaiyun.com/ADjky/article/details/78467694</guid><author>ADjky</author><pubDate>Tue, 07 Nov 2017 14:33:22 +0800</pubDate><description><![CDATA[退役后刷的第一道题…
题解:
n!内某个因子a的指数: 
每次轮流除以a, 当前商+剩余贡献即可 
例: 
9! : 9 / 3 = 3; 3 / 3 = 1 
ans = 4
对于[ka, ka + a - 1], 
阶乘中因子a的指数相同
定义cal(i,n)是i!中因子n的指数cal(i, n)是i!中因子n的指数 
ans=∑i=1nmcal(i,n)=cal(nm,n)+∑i=1nm−1c]]></description><category></category></item><item><title><![CDATA[2017 CCPC 秦皇岛 G题]]></title><link>https://blog.youkuaiyun.com/ADjky/article/details/78395776</link><guid>https://blog.youkuaiyun.com/ADjky/article/details/78395776</guid><author>ADjky</author><pubDate>Mon, 30 Oct 2017 18:22:01 +0800</pubDate><description><![CDATA[题意给定非负整数n, m(n < 10^1000, 1<= m  < 10^100), 把n分成m个非负整数的和 
∑i=1mai=n(ai>=0)\sum_{i = 1} ^m a_i = n(a_i >= 0) 
且使 
a1 or a2 or a3 … or am最小题解1.对于任意i in [1, m] 
 a1 or a2 or a3 … or am >= ai 
当n % m == 0时]]></description><category></category></item><item><title><![CDATA[数论-模板 (待更新)]]></title><link>https://blog.youkuaiyun.com/ADjky/article/details/78208089</link><guid>https://blog.youkuaiyun.com/ADjky/article/details/78208089</guid><author>ADjky</author><pubDate>Wed, 11 Oct 2017 20:38:56 +0800</pubDate><description><![CDATA[质因数分解


  在求解数论问题时, 经常会把一个整数分解为多个质数的乘积(形式唯一), 
  还有一些题目需要从它的质因数分解形式来考虑.


质因数分解:



x=∏i=1npaii(pi为不可分解的质因数)x = \prod_{i=1}^np_i^{a_i}(p_i为不可分解的质因数)

eg: 
12=2∗2∗3=22∗312 = 2 * 2 * 3 = 2^2*3 
注: 不含的质因]]></description><category></category></item></channel></rss>