<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[DM010maker的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/DM010maker</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; DM010maker]]></copyright><item><title><![CDATA[笔记本推荐]]></title><link>https://blog.youkuaiyun.com/DM010maker/article/details/127586433</link><guid>https://blog.youkuaiyun.com/DM010maker/article/details/127586433</guid><author>DM010maker</author><pubDate>Sat, 29 Oct 2022 14:02:57 +0800</pubDate><description><![CDATA[联想小新Air14Plus 2022 R76800HS联想ThinkBook 14+ 2022 i5-12500H联想小新pro14 R76800HS华硕A豆Pro i5-12500H]]></description><category></category></item><item><title><![CDATA[AcWing食用指南]]></title><link>https://blog.youkuaiyun.com/DM010maker/article/details/127350111</link><guid>https://blog.youkuaiyun.com/DM010maker/article/details/127350111</guid><author>DM010maker</author><pubDate>Sun, 16 Oct 2022 18:32:32 +0800</pubDate><description><![CDATA[​	AcWing是一个计算机编程相关知识学习网站，其中有算法课程与工程课程。适合计算机相关专业以及想要从事计算机相关专业工作的同学食用。这是一个很适合学习算法的平台，学习过程非常平滑，适合所有阶段的同学。]]></description><category></category></item><item><title><![CDATA[# CF #808 Div.2(A - C)]]></title><link>https://blog.youkuaiyun.com/DM010maker/article/details/125830378</link><guid>https://blog.youkuaiyun.com/DM010maker/article/details/125830378</guid><author>DM010maker</author><pubDate>Sun, 17 Jul 2022 12:08:29 +0800</pubDate><description><![CDATA[​既然是这样分布，我们不如直接令gcd(i,ai)==i，因为ai没有说不能取重复的，所以直接对于每个i去[l,r]里找i的第一个倍数即可。​由于i的取值是[1,n]，所以gcd若是结果不同的话，一定是分布在[1,n]（gcd(a,b)...]]></description><category></category></item><item><title><![CDATA[# CF #807 Div.2(A - D)]]></title><link>https://blog.youkuaiyun.com/DM010maker/article/details/125829836</link><guid>https://blog.youkuaiyun.com/DM010maker/article/details/125829836</guid><author>DM010maker</author><pubDate>Sun, 17 Jul 2022 11:22:42 +0800</pubDate><description><![CDATA[​给定2n个数和k，请问是否存在n对(i,j)abs(ai-aj)>=k​先排序然后对1和n+1，2和n+2连续比对。​赛时拖了节奏因为看成了==k下次注意读题。]]></description><category></category></item><item><title><![CDATA[AtCoder ABC 131]]></title><link>https://blog.youkuaiyun.com/DM010maker/article/details/125690757</link><guid>https://blog.youkuaiyun.com/DM010maker/article/details/125690757</guid><author>DM010maker</author><pubDate>Sat, 09 Jul 2022 10:51:11 +0800</pubDate><description><![CDATA[Description:​	给你一段区间[a, b] (]]></description><category></category></item><item><title><![CDATA[AtCoder ABC 128(C ~ E)]]></title><link>https://blog.youkuaiyun.com/DM010maker/article/details/125118443</link><guid>https://blog.youkuaiyun.com/DM010maker/article/details/125118443</guid><author>DM010maker</author><pubDate>Sat, 04 Jun 2022 08:13:49 +0800</pubDate><description><![CDATA[Description:​	有n盏灯和m组设定，要求输出同时符合m组设定的灯开关的方案数​	设定：s_i中亮灯的数量%2 == p_iSolution:​	因为跟开关灯有关也就是01有关 再加上N只有10 于是我们就用01串枚举一下灯的状态​	用位运算来check是否合法Code:
D equeue(暴力 贪心)
Description:​	给出长度为n的序列 可以在头部或者尾部进行k次四种操作​	1 从头部拿出一个  2 把从头部拿的任意一个放回去​	3 从尾部拿出一个  4 把从尾部拿的任意一个放回去S]]></description><category></category></item><item><title><![CDATA[AtCoder ABC 127(C ~ E)]]></title><link>https://blog.youkuaiyun.com/DM010maker/article/details/125117767</link><guid>https://blog.youkuaiyun.com/DM010maker/article/details/125117767</guid><author>DM010maker</author><pubDate>Sat, 04 Jun 2022 00:21:24 +0800</pubDate><description><![CDATA[Description:​	给出若干个区间m 请问有多少个点被覆盖了m次Solution:​	差分求解即可Code:
D Integer Cards(贪心 排序)
Description:​	给出n个值 给出m组数对（可用次数，值） 可用任意选数对中的值来替换原来值 求替换后最大和Solution:​	贪心地想 最大的值我一定要给他换上 于是结构体排序值大优先 小根堆维护一下n个值 替换即可Code:
E Cell Distance(枚举 + 排列组合)
Description:​	给你三个数 n,m,k，]]></description><category></category></item><item><title><![CDATA[AtCoder ABC 126(C ~ E)]]></title><link>https://blog.youkuaiyun.com/DM010maker/article/details/125115425</link><guid>https://blog.youkuaiyun.com/DM010maker/article/details/125115425</guid><author>DM010maker</author><pubDate>Fri, 03 Jun 2022 20:05:13 +0800</pubDate><description><![CDATA[Description:Solution:​	枚举[1, n] 每个需要cnt次才能大于等于k 对答案贡献为(1 / 2) ^ kCode:
D Even Relationhttps://atcoder.jp/contests/abc126/tasks/abc126_c)（DFS + 染色 + 贪心)
Description:​Solution:​	随便从一个地方开始遍历 如果连着是偶数就染同一个颜色，若不是偶数就换​	a - b 偶 && a - c 偶 那么b - c 是偶​	a - b 奇 && a ]]></description><category></category></item><item><title><![CDATA[AtCoder ABC 251 （D ~ E）]]></title><link>https://blog.youkuaiyun.com/DM010maker/article/details/124780967</link><guid>https://blog.youkuaiyun.com/DM010maker/article/details/124780967</guid><author>DM010maker</author><pubDate>Sun, 15 May 2022 12:55:48 +0800</pubDate><description><![CDATA[AtCoder ABC 251
D - At Most 3 (Contestant ver.)(构造 + 思维)
Description:
​	构造一个长度为300的序列 任取1或者2或者3个数 使其能构成1~1e6的所有数
Solution:
​	将1e6想办法分为三分 如果是想着/3的话 那就走上了歪路
​	应该想的是 将六位数分割成六位 一个数可以影响两位
​	三个数相加就可以影响6位了 于是构造方案出现
Code:
signed main()
{
    int w;
    cin &gt;&gt]]></description><category></category></item><item><title><![CDATA[AtCoder ABC 249]]></title><link>https://blog.youkuaiyun.com/DM010maker/article/details/124373864</link><guid>https://blog.youkuaiyun.com/DM010maker/article/details/124373864</guid><author>DM010maker</author><pubDate>Sun, 24 Apr 2022 01:15:48 +0800</pubDate><description><![CDATA[A Jogging
没看清楚题 a不是移速 所以卡了十分钟 语法题
int main()
{
    int a, b, c, d, e, f, x;
    cin &gt;&gt; a &gt;&gt; b &gt;&gt; c &gt;&gt; d &gt;&gt; e &gt;&gt; f &gt;&gt; x; // 速度 b 和 e
    int aa1 = b * a, bb1 = e * d;
    
    int res1 = aa1 * (x / (a + c));
    int]]></description><category></category></item><item><title><![CDATA[大连大学2022 4月程序设计竞赛]]></title><link>https://blog.youkuaiyun.com/DM010maker/article/details/124239415</link><guid>https://blog.youkuaiyun.com/DM010maker/article/details/124239415</guid><author>DM010maker</author><pubDate>Sun, 17 Apr 2022 23:46:31 +0800</pubDate><description><![CDATA[E   数圆圈(规律 模拟)
Description:
​	定义f(x) = x里各位圆圈数量之和
​	现在给定x 对x进行k次f(x)操作 返回结果
​	0, 6, 9 = 1;  8 = 2;
Solution:
​	那我们考虑最坏的情况 那就是17位上面都是8 那也是17*2 会化成一个很小的数
​	我们就可以知道 对于一个x 我们可以通过数十次次迭代之内 将他变成01的循环态
​	于是我们模拟即可 （用字符串不会挂 但是用整数是会挂的 因为x可能是负数
Code:
LL get(LL x)
{
  ]]></description><category></category></item><item><title><![CDATA[AtCoder ABC 247]]></title><link>https://blog.youkuaiyun.com/DM010maker/article/details/124111492</link><guid>https://blog.youkuaiyun.com/DM010maker/article/details/124111492</guid><author>DM010maker</author><pubDate>Mon, 11 Apr 2022 22:29:40 +0800</pubDate><description><![CDATA[A Move Right(水)
字符串右移一位 然后在左边补0
void solve()
{
    string s;
    cin &gt;&gt; s;
    s = '0' + s;
    s.erase(s.end() - 1);
    cout &lt;&lt; s;
}

B Unique Nicknames(读题)
Description:
​	每个人有两个备选名字 然后要给每个人从这两个里选一个命名
​	要求任一人的名字 不能跟其他人的备选名字相同
Solution:
​	map存]]></description><category></category></item><item><title><![CDATA[CF Div2 781]]></title><link>https://blog.youkuaiyun.com/DM010maker/article/details/124075702</link><guid>https://blog.youkuaiyun.com/DM010maker/article/details/124075702</guid><author>DM010maker</author><pubDate>Sun, 10 Apr 2022 11:57:50 +0800</pubDate><description><![CDATA[A GCD vs LCM(数论 + 思维)
Description:
​	求4个和为n的数 要求 gcd(a, b) == lcm(c, d)
Solution:
​	令lcm(c, d) = 1 则 c = 1 d = 1
​	因为x取任意正整数 gcd(1, x)  = 1始终成立
​	令gcd(a, b) = 1 则 a = 1 b = n - 3
Code:
void solve()
{
	int n; cin &gt;&gt; n;
	cout &lt;&lt; 1 &lt;&lt; ' ' &l]]></description><category></category></item><item><title><![CDATA[2022 04 05学习记录]]></title><link>https://blog.youkuaiyun.com/DM010maker/article/details/123986320</link><guid>https://blog.youkuaiyun.com/DM010maker/article/details/123986320</guid><author>DM010maker</author><pubDate>Wed, 06 Apr 2022 11:42:47 +0800</pubDate><description><![CDATA[早上补牛客比赛题
下午一觉睡到4点 复习会数学吧
gk的树
题目链接：https://ac.nowcoder.com/acm/contest/30825/F
反省 在队友做出了数学题的时候自己放弃了 其实这题应该在我的能力范围内 没有写出这题很不应该Description:
​	给定一棵树，每次操作可以删一条边 最少需要多少次操作使得每个节点的度数都&lt;=k
Solution:
​	从叶子节点开始处理 若叶子节点度数大于k 就删边 模拟一下倒推回根节点 可以用dfs 也可以dp
Code:
const ]]></description><category></category></item><item><title><![CDATA[竞赛中的STL库函数使用]]></title><link>https://blog.youkuaiyun.com/DM010maker/article/details/123897675</link><guid>https://blog.youkuaiyun.com/DM010maker/article/details/123897675</guid><author>DM010maker</author><pubDate>Fri, 01 Apr 2022 14:41:59 +0800</pubDate><description><![CDATA[简单的介绍一下STL的用法	看起来内容很多 只要遇到对应的题写个几题 很容易就熟能生巧了
为何cpp是竞赛使用最为广泛的语言呢 1 速度快 2 STL强大 	因此掌握好STL 一定能让你做题的效率更上一层楼
Vector
介绍
​	你可以把vector当做一个动态的数组 长度可扩容	且在main里开vector是开在堆空间的
​	初始化
//需要头文件 #include &lt;vector&gt;

vector&lt;int&gt; q;
vector&lt;node&gt; q;

//定义长度
ve]]></description><category></category></item><item><title><![CDATA[AtCoder #ABC 245]]></title><link>https://blog.youkuaiyun.com/DM010maker/article/details/123832658</link><guid>https://blog.youkuaiyun.com/DM010maker/article/details/123832658</guid><author>DM010maker</author><pubDate>Tue, 29 Mar 2022 21:25:07 +0800</pubDate><description><![CDATA[A Good morning(模拟)
过水
void solve()
{
    int h1, m1, h2, m2;
	cin &gt;&gt; h1 &gt;&gt; m1 &gt;&gt; h2 &gt;&gt; m2;
    if(h1 == h2)
    {
        if(m1 == m2)    puts("Takahashi");
        else if(m1 &lt; m2)    puts("Takahashi");
        else puts("Aoki")]]></description><category></category></item><item><title><![CDATA[CF #125 Edu(A ~ D)]]></title><link>https://blog.youkuaiyun.com/DM010maker/article/details/123733776</link><guid>https://blog.youkuaiyun.com/DM010maker/article/details/123733776</guid><author>DM010maker</author><pubDate>Fri, 25 Mar 2022 13:45:20 +0800</pubDate><description><![CDATA[CF Edu125 Div.2
A - Integer Moves(思维)
Description:
​	给定一个坐标x y 从0 0开始走 每次走的欧几里得距离必须是整数 问几步能到
Solution
​	0 0到0 0 	0步
​	如果是直角三角形 就可以一步到
​	如果不是直角三角形 那就先走一步横再走一步竖 两步到
Code:
void solve()
{
	int x, y;
	cin &gt;&gt; x &gt;&gt; y;
	if(x == 0 &amp;&amp; y == 0)	{co]]></description><category></category></item><item><title><![CDATA[AtCoder ABC 244(A ~ E F待补)]]></title><link>https://blog.youkuaiyun.com/DM010maker/article/details/123649921</link><guid>https://blog.youkuaiyun.com/DM010maker/article/details/123649921</guid><author>DM010maker</author><pubDate>Tue, 22 Mar 2022 00:10:18 +0800</pubDate><description><![CDATA[AtCoder ABC 244
A - Last Letter(语法)
Description:
​	输出给定字符串的最后一个字符
Resolution
​	过水
Code:
void solve()
{
	int n;
    string s;
    cin &gt;&gt; n;
    cin &gt;&gt; s;
    cout &lt;&lt; s[n - 1];
}

B - Go Straight and Turn Right(模拟)
Description:
​	对于给定操作序列 S]]></description><category></category></item><item><title><![CDATA[AtCoder ABC 243题解]]></title><link>https://blog.youkuaiyun.com/DM010maker/article/details/123456929</link><guid>https://blog.youkuaiyun.com/DM010maker/article/details/123456929</guid><author>DM010maker</author><pubDate>Sun, 13 Mar 2022 11:50:56 +0800</pubDate><description><![CDATA[AtCoder ABC 243
A Shampoo(水)
Description:
​	三个人轮流用水 请问用到谁的时候水就用完了
Method:
​	水题 但是因为%和/用错了 导致蛙了好几次
​	分类讨论 如果一轮用不完 就取余再判断 如果一轮用得完 直接判断
Code:
void solve()
{
	int v, a, b, c;
    cin &gt;&gt; v &gt;&gt; a &gt;&gt; b &gt;&gt; c;
    int sum = a + b + c;
    int]]></description><category></category></item><item><title><![CDATA[CF #776 Div.3题解(A ~ E)]]></title><link>https://blog.youkuaiyun.com/DM010maker/article/details/123388947</link><guid>https://blog.youkuaiyun.com/DM010maker/article/details/123388947</guid><author>DM010maker</author><pubDate>Wed, 09 Mar 2022 22:09:48 +0800</pubDate><description><![CDATA[CF #776 Div.3
A Deletions of Two Adjacent Letters(思维)
Description:
给定一个字符串s和字符c 每次可以从字符串s中消除相邻的两个位置的字符 问是否能通过消除来得到字符c
Method:
枚举字符串中每一个c 若前后都为偶数的话 能
Code:
void solve()
{
	string s;
    char c;
    cin &gt;&gt; s &gt;&gt; c;
    for(int i = 0; i &lt; s.size]]></description><category></category></item></channel></rss>