<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[CreatureNoWords的博客]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/CreatureNoWords</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; CreatureNoWords]]></copyright><item><title><![CDATA[2021年双非计算机意外保研推免经验贴，最终去向西南交通大学计算机与人工智能学院]]></title><link>https://blog.youkuaiyun.com/CreatureNoWords/article/details/120835949</link><guid>https://blog.youkuaiyun.com/CreatureNoWords/article/details/120835949</guid><author>CreatureNoWords</author><pubDate>Mon, 18 Oct 2021 22:07:11 +0800</pubDate><description><![CDATA[一、前言
没经历过的总会觉得保研的都是轻轻松松的人上人，但作为双非边缘人我看到的是一群卑微的舔狗。像把自己明码标价卖给学校和导师，对方可以肆意的挑选，自己还要舔着脸推销。保研人无法抱怨辛苦，因为别人会说你凡尔赛。能保研的人都经历过很多的成功和失败，但这次我仍要说是真的难顶啊。 故作经验贴，补充了一些网上不常见的保研潜规则，希望能帮到给与我相似的后来者，少走弯路。
二、经历
本科成都理工大学，计算机科学与技术专业，成绩排名8/124，六级不算高，竞赛经历比较丰富但含金量一般，有轻度参与水科研项目，挂名四作ei]]></description><category></category></item><item><title><![CDATA[基于距离浓度的人工免疫算法（DCAIA）求函数最值（python实现）]]></title><link>https://blog.youkuaiyun.com/CreatureNoWords/article/details/116177000</link><guid>https://blog.youkuaiyun.com/CreatureNoWords/article/details/116177000</guid><author>CreatureNoWords</author><pubDate>Mon, 26 Apr 2021 23:27:57 +0800</pubDate><description><![CDATA[文章目录基于距离浓度的人工免疫算法（DCAIA）求函数最值（python实现）一、免疫算法理论二、基于距离浓度算法的免疫机理三、免疫算法对应关系四、免疫算法算子五、python实例例（1）例（2）
基于距离浓度的人工免疫算法（DCAIA）求函数最值（python实现）
免疫算法是人工免疫系统形式中应用最广法的一个部分，大多数的免疫算法被用于实现全局优化问题，包括组合优化、函数优化、神经网络优化等。
一、免疫算法理论
免疫算法是模仿生物免疫机制，结合基因的进化机理，人工构造出的一种新型智能优化算法。采用群体搜]]></description><category></category></item><item><title><![CDATA[2020多校第八场Clockwise or Counterclockwise]]></title><link>https://blog.youkuaiyun.com/CreatureNoWords/article/details/108040849</link><guid>https://blog.youkuaiyun.com/CreatureNoWords/article/details/108040849</guid><author>CreatureNoWords</author><pubDate>Sun, 16 Aug 2020 18:47:15 +0800</pubDate><description><![CDATA[思路
利用向量的叉乘判断方向
代码
#include &lt;iostream&gt;
#include &lt;cstdio&gt;
#include &lt;algorithm&gt;
#include &lt;queue&gt;
#include &lt;cmath&gt;
#include &lt;cstring&gt;

#define ll long long
using namespace std;
const int maxn=1e5+5;
const int N=5e6+5;

const]]></description><category></category></item><item><title><![CDATA[2020多校第八场Fluctuation Limit]]></title><link>https://blog.youkuaiyun.com/CreatureNoWords/article/details/108039670</link><guid>https://blog.youkuaiyun.com/CreatureNoWords/article/details/108039670</guid><author>CreatureNoWords</author><pubDate>Sun, 16 Aug 2020 18:00:33 +0800</pubDate><description><![CDATA[思路
从左到右求一遍可行解，再从右到左求一遍可行解。两组解的交集则是全局可行解。输出最小的就是答案。
代码
#include &lt;iostream&gt;
#include &lt;cstdio&gt;
#include &lt;algorithm&gt;
#include &lt;queue&gt;
#include &lt;cmath&gt;
#include &lt;cstring&gt;

#define ll long long
using namespace std;
const int m]]></description><category></category></item><item><title><![CDATA[2020多校第六场]]></title><link>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107922204</link><guid>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107922204</guid><author>CreatureNoWords</author><pubDate>Mon, 10 Aug 2020 21:08:48 +0800</pubDate><description><![CDATA[思路
暴力、模拟
代码
#include &lt;iostream&gt;
#include &lt;cstdio&gt;
#include &lt;algorithm&gt;
#include &lt;queue&gt;
#include &lt;cmath&gt;
#include &lt;cstring&gt;

#define ll long long
using namespace std;
const int maxn=6e6+6;
const ll mod=998244353;
const l]]></description><category></category></item><item><title><![CDATA[2020多校第六场Divisibility]]></title><link>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107921748</link><guid>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107921748</guid><author>CreatureNoWords</author><pubDate>Mon, 10 Aug 2020 20:47:32 +0800</pubDate><description><![CDATA[思路
结论：一个n进制数能被x整除且这个数各位之和也能被x整除，当且仅当x是(n-1)的因子。
代码
#include &lt;iostream&gt;
#include &lt;cstdio&gt;
#include &lt;algorithm&gt;
#include &lt;queue&gt;
#include &lt;cmath&gt;
#include &lt;cstring&gt;

#define ll long long
using namespace std;
const int maxn]]></description><category></category></item><item><title><![CDATA[2020多校第五场Set1]]></title><link>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107900599</link><guid>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107900599</guid><author>CreatureNoWords</author><pubDate>Sun, 09 Aug 2020 21:31:34 +0800</pubDate><description><![CDATA[思路
组合数学+概率。其中的 (n−i)! 是对 i 后面的 (n−i) 个数进行排列，与前面选出的数进行配对。
对于前面的剩余的 i−1−(n−i)=2∗i−n−1 个数，两两进行配对选择，即每次选取 2 两个数，但由于每次配对后，每组的位置就确定了，因此要去除排列。
代码
#include &lt;iostream&gt;
#include &lt;cstdio&gt;
#include &lt;algorithm&gt;
#include &lt;queue&gt;
#include &lt;cmat]]></description><category></category></item><item><title><![CDATA[2020多校第五场Boring Game]]></title><link>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107900363</link><guid>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107900363</guid><author>CreatureNoWords</author><pubDate>Sun, 09 Aug 2020 21:13:15 +0800</pubDate><description><![CDATA[思路
对于每次展开，挑选当前最中间的位置作为 mid ，将上半段翻转一下贴到下半段的左边，直接用 vector 暴力维护整个过程。
代码
#include &lt;iostream&gt;
#include &lt;cstdio&gt;
#include &lt;algorithm&gt;
#include &lt;queue&gt;
#include &lt;cmath&gt;
#include &lt;cstring&gt;

#define ll long long
using namespace s]]></description><category></category></item><item><title><![CDATA[2020多校第五场Paperfolding]]></title><link>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107900195</link><guid>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107900195</guid><author>CreatureNoWords</author><pubDate>Sun, 09 Aug 2020 21:07:41 +0800</pubDate><description><![CDATA[思路
将折痕分为竖直和水平，分别考虑。对折n次，有2^n-1贡献，结合二次项定理将公式进行化简。
代码
#include &lt;iostream&gt;
#include &lt;cstdio&gt;
#include &lt;algorithm&gt;
#include &lt;queue&gt;
#include &lt;cmath&gt;
#include &lt;cstring&gt;

#define ll long long
using namespace std;
const int max]]></description><category></category></item><item><title><![CDATA[2020多校第五场Tetrahedron]]></title><link>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107900128</link><guid>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107900128</guid><author>CreatureNoWords</author><pubDate>Sun, 09 Aug 2020 20:59:31 +0800</pubDate><description><![CDATA[思路
数学计算，要预处理逆元。
代码
#include&lt;bits/stdc++.h&gt;
using namespace std;
typedef long long ll;
const ll mod = 998244353;
const ll Max = 6e6+5 ;
ll inv[Max+7];
void init() {
    inv[1] = 1;
    for (ll i = 2; i &lt;= Max; i++) {
        inv[i] = 1LL*(mod - mod]]></description><category></category></item><item><title><![CDATA[2020多校第四场Go Running]]></title><link>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107778914</link><guid>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107778914</guid><author>CreatureNoWords</author><pubDate>Tue, 04 Aug 2020 09:13:02 +0800</pubDate><description><![CDATA[思路
把横纵坐标分别当做二分图的两边，把点所在的横纵坐标相连。因为我们把横纵坐标当做了二分图的点，二分图中的一个边就代表一个点。即求最小点覆盖，即最大匹配。
代码
#include &lt;bits/stdc++.h&gt;
#define pb push_back
using namespace std;
const int N=1e5+5;
const int inf=0x3f3f3f3f;
const double pi=acos(-1);
vector&lt;int&gt;pic[N&lt;&lt;]]></description><category></category></item><item><title><![CDATA[2020多校第四场]]></title><link>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107775203</link><guid>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107775203</guid><author>CreatureNoWords</author><pubDate>Mon, 03 Aug 2020 22:40:04 +0800</pubDate><description><![CDATA[思路
把L转化为两个L，R转化为两个R，M转为一个R一个L，在扩充图上跑dijkstra。
代码
#include &lt;iostream&gt;
#include &lt;cstdio&gt;
#include &lt;algorithm&gt;
#include &lt;queue&gt;
#include &lt;cmath&gt;
#include &lt;cstring&gt;
#define N 400005
#define M 800005
using namespace std;
int ]]></description><category></category></item><item><title><![CDATA[2020多校第四场Kindergarten Physics]]></title><link>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107768178</link><guid>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107768178</guid><author>CreatureNoWords</author><pubDate>Mon, 03 Aug 2020 16:49:21 +0800</pubDate><description><![CDATA[思路
由于数据过小，到不了误差。。所以直接输出答案。
代码
#include &lt;bits/stdc++.h&gt;
#define hi cout&lt;&lt;"hi\n";
#define inf 0x3f3f3f3f
#define ll long long
#define ull unsigned long long
using namespace std;
const ll mod = 1000000009;
const int maxn = 100005;
ll power(long lo]]></description><category></category></item><item><title><![CDATA[2020多校第四场Equal Sentences]]></title><link>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107768071</link><guid>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107768071</guid><author>CreatureNoWords</author><pubDate>Mon, 03 Aug 2020 16:45:21 +0800</pubDate><description><![CDATA[思路
如果相邻两单词相同，则交换后不改变，则f[i+1]=f[i];
如果不相等，则交换，那么f[i] = f[i-1] + f[i-2];一直到f[n]时算出结果。
代码
#include &lt;bits/stdc++.h&gt;
#define hi cout&lt;&lt;"hi\n";
#define inf 0x3f3f3f3f
#define ll long long
#define ull unsigned long long
using namespace std;
const ll mo]]></description><category></category></item><item><title><![CDATA[2020多校第四场Blow up the Enemy]]></title><link>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107766392</link><guid>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107766392</guid><author>CreatureNoWords</author><pubDate>Mon, 03 Aug 2020 16:04:19 +0800</pubDate><description><![CDATA[思路
暴力。。
代码
#include &lt;bits/stdc++.h&gt;
#define hi cout&lt;&lt;"hi\n";
#define inf 0x3f3f3f3f
#define ll long long
#define ull unsigned long long
using namespace std;
const ll mod = 1000000009;
const int maxn = 100005;
ll power(long long a, long long b, ]]></description><category></category></item><item><title><![CDATA[2020多校第三场Little W and Contest]]></title><link>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107738879</link><guid>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107738879</guid><author>CreatureNoWords</author><pubDate>Sat, 01 Aug 2020 23:56:14 +0800</pubDate><description><![CDATA[思路
用并查集，维护三种连通快中。先求出第一天的ans，每次从不同块中去取不满足的情况，存在四种可能，每次用前一次的减去不满足的情况个数，得到最终ans。
代码
#include &lt;bits/stdc++.h&gt;
#define hi cout&lt;&lt;"hi\n";
#define inf 0x3f3f3f3f
#define ll long long
#define ull unsigned long long
using namespace std;
const ll mod = 1e]]></description><category></category></item><item><title><![CDATA[2020多校第三场Tokitsukaze and Multiple]]></title><link>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107738629</link><guid>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107738629</guid><author>CreatureNoWords</author><pubDate>Sat, 01 Aug 2020 23:28:33 +0800</pubDate><description><![CDATA[思路
先求前缀和，然后把前缀和%p。如果有两个位置的数相等，说明这段区间的和是p的倍数。统计相同的值。
代码
#include &lt;bits/stdc++.h&gt;
#define hi cout&lt;&lt;"hi\n";
#define inf 0x3f3f3f3f
#define ll long long
#define ull unsigned long long
using namespace std;
const ll mod = 1000000009;
const int maxn =]]></description><category></category></item><item><title><![CDATA[2020多校第三场 Parentheses Matching]]></title><link>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107685184</link><guid>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107685184</guid><author>CreatureNoWords</author><pubDate>Thu, 30 Jul 2020 10:56:53 +0800</pubDate><description><![CDATA[思路
贪心+模拟。先进行括号匹配，排除所以已经匹配好的括号。记录未匹配的左括号，右括号，星号的位置。用队列存星号位置可以把复杂度降到O(n)。
考虑贪心策略，先从左开始匹配右括号，要求星号在括号左边且尽量左。然后从右开始匹配左括号，要求星号在括号右边且尽量右。
代码
#include &lt;bits/stdc++.h&gt;
#define hi cout&lt;&lt;"hi\n";
#define inf 0x3f3f3f3f
#define ll long long
#define ull unsi]]></description><category></category></item><item><title><![CDATA[2020多校第二场1001]]></title><link>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107620333</link><guid>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107620333</guid><author>CreatureNoWords</author><pubDate>Mon, 27 Jul 2020 20:18:37 +0800</pubDate><description><![CDATA[思路
逆向思维，从大到小加点来构造图，没个点的贡献为（a[i].w-a[i+1].w）*连通块数。维护一个并查集来计算连通块数。
代码
#include &lt;bits/stdc++.h&gt;
#define hi cout&lt;&lt;"hi\n";
#define inf 0x3f3f3f3f
#define ll long long
#define ull unsigned long long
using namespace std;
const ll mod = 1000000009;
con]]></description><category></category></item><item><title><![CDATA[2020多校第二场1010]]></title><link>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107600099</link><guid>https://blog.youkuaiyun.com/CreatureNoWords/article/details/107600099</guid><author>CreatureNoWords</author><pubDate>Sun, 26 Jul 2020 21:34:40 +0800</pubDate><description><![CDATA[思路
估算复杂度，为k^(n/k)，这个函数在k=3，n=50时最大。此时的运算量并不大，可以dfs暴力。要注意某类型物品为空的情况，需要优化处理。
代码
#include &lt;bits/stdc++.h&gt;
#define hi cout&lt;&lt;"hi\n";
#define inf 0x3f3f3f3f
#define ll long long
#define ull unsigned long long
using namespace std;
const ll mod = 10000]]></description><category></category></item></channel></rss>