<?xml version="1.0" encoding="utf-8" ?><rss version="2.0"><channel><title><![CDATA[小小俊]]></title><description><![CDATA[]]></description><link>https://blog.youkuaiyun.com/weixin_38772011</link><language>zh-cn</language><generator>https://blog.youkuaiyun.com/</generator><copyright><![CDATA[Copyright &copy; weixin_38772011]]></copyright><item><title><![CDATA[~/.vimrc]]></title><link>https://blog.youkuaiyun.com/weixin_38772011/article/details/95755239</link><guid>https://blog.youkuaiyun.com/weixin_38772011/article/details/95755239</guid><author>weixin_38772011</author><pubDate>Sat, 13 Jul 2019 15:18:34 +0800</pubDate><description><![CDATA[https://blog.youkuaiyun.com/amoscykl/article/details/80616688


set nocompatible
filetype off         " required

" set the runtime path to include Vundle and initialize
set rtp+=~/.vim/bundle/Vun...]]></description><category></category></item><item><title><![CDATA[哈希树HashTree  罗堃_吴朝宏]]></title><link>https://blog.youkuaiyun.com/weixin_38772011/article/details/90731309</link><guid>https://blog.youkuaiyun.com/weixin_38772011/article/details/90731309</guid><author>weixin_38772011</author><pubDate>Sat, 01 Jun 2019 17:24:02 +0800</pubDate><description><![CDATA[https://wenku.baidu.com/view/16b2c7abd1f34693daef3e58.html



质数分辨定理：n个不同的质数可以“分辨”的连续整数的个数和他们的乘积相等。“分辨”就是指这些连续的整数不可能有完全相同的余数序列。

余数分辨定理：任意选取n个互不相同的自然数，可以“分辨”的连续整数的个数和LCM相等。


...]]></description><category></category></item><item><title><![CDATA[「LibreOJ β Round」ZQC 的拼图 （dp）]]></title><link>https://blog.youkuaiyun.com/weixin_38772011/article/details/90724786</link><guid>https://blog.youkuaiyun.com/weixin_38772011/article/details/90724786</guid><author>weixin_38772011</author><pubDate>Sat, 01 Jun 2019 00:05:23 +0800</pubDate><description><![CDATA[题意：在m*m的板子上用m块直角三角形（朝左和朝下）相交或相切使得（0，0）被包含在内的最小放大倍数。

http://www.lpoj.cn/problemdetail?problemID=5552

i * a[1].x + j * a[1].y - t &lt;=i' * a[1].x + j' * a[1].y

dp[i][j][k]: 前i块三角形能否到达[j,k]，暴力n*m*m*...]]></description><category></category></item><item><title><![CDATA[「LibreOJ β Round」ZQC 的手办 （线段树+堆）]]></title><link>https://blog.youkuaiyun.com/weixin_38772011/article/details/90724749</link><guid>https://blog.youkuaiyun.com/weixin_38772011/article/details/90724749</guid><author>weixin_38772011</author><pubDate>Fri, 31 May 2019 23:57:30 +0800</pubDate><description><![CDATA[题意：给出n个数，m种操作，1 l r k ,修改[l,r] 区间内，比k小的为k, 2 l r k x,查询[l,r]区间内比k小的最小的x个数，少于k个输出-1.

http://www.lpoj.cn/problemdetail?problemID=5555


#include&lt;bits/stdc++.h&gt;
using namespace std;
const int max...]]></description><category></category></item><item><title><![CDATA[Codeforces Round #562 E. And Reachability(序列自动机 dp)]]></title><link>https://blog.youkuaiyun.com/weixin_38772011/article/details/90641356</link><guid>https://blog.youkuaiyun.com/weixin_38772011/article/details/90641356</guid><author>weixin_38772011</author><pubDate>Tue, 28 May 2019 16:08:23 +0800</pubDate><description><![CDATA[题意：给你n个数和m个询问，对于每个询问l,r是否可以从l跳到r,跳的规则是&amp;不为0，可以输出Shi,相反Fou.


#include&lt;bits/stdc++.h&gt;
using namespace std;
int n,q;
int a[300005];
int nextt[300005][20];
int b[20];
int main()
{
	cin&gt;&gt;n&...]]></description><category></category></item><item><title><![CDATA[Codeforces Round #562 C. Increasing by Modulo (二分)]]></title><link>https://blog.youkuaiyun.com/weixin_38772011/article/details/90641218</link><guid>https://blog.youkuaiyun.com/weixin_38772011/article/details/90641218</guid><author>weixin_38772011</author><pubDate>Tue, 28 May 2019 16:03:40 +0800</pubDate><description><![CDATA[题意：给你n个数和一个模数m，一次操作可选择1-n个数+1%m,求最少的操作次数使得n个数单调不递减。




#include&lt;bits/stdc++.h&gt;
using namespace std;
int n,m;
int a[300005];
int b[300005];
int main()
{
	cin&gt;&gt;n&gt;&gt;m;
	for(int i=1;i&l...]]></description><category></category></item><item><title><![CDATA[codeforces 562 D. Good Triple (暴力)]]></title><link>https://blog.youkuaiyun.com/weixin_38772011/article/details/90605482</link><guid>https://blog.youkuaiyun.com/weixin_38772011/article/details/90605482</guid><author>weixin_38772011</author><pubDate>Tue, 28 May 2019 16:00:30 +0800</pubDate><description><![CDATA[题意：给你一个字符串，询问存在sx=sx+k=sx+2k这样的区间的个数。


#include&lt;bits/stdc++.h&gt;
using namespace std;
int n;
char s[300005];
long long ans;
int main(){
	scanf("%s",s+1);
	n=strlen(s+1);
	for(int i=1,l=0;i&lt;=n...]]></description><category></category></item><item><title><![CDATA[软件测试]]></title><link>https://blog.youkuaiyun.com/weixin_38772011/article/details/90442091</link><guid>https://blog.youkuaiyun.com/weixin_38772011/article/details/90442091</guid><author>weixin_38772011</author><pubDate>Wed, 22 May 2019 13:15:24 +0800</pubDate><description><![CDATA[https://blog.youkuaiyun.com/myvanguard/article/details/84194640
]]></description><category></category></item><item><title><![CDATA[HDU 6540 树形DP]]></title><link>https://blog.youkuaiyun.com/weixin_38772011/article/details/90415564</link><guid>https://blog.youkuaiyun.com/weixin_38772011/article/details/90415564</guid><author>weixin_38772011</author><pubDate>Tue, 21 May 2019 21:15:45 +0800</pubDate><description><![CDATA[http://acm.hdu.edu.cn/showproblem.php?pid=6540

题意：给你一颗树和一些关键点，关键点的子集中，任意两点的距离不大于k的子集个数


#include&lt;bits/stdc++.h&gt;
using namespace std;
int n,m,k;
int a[5005];
vector&lt;int&gt; mp[5005];
const i...]]></description><category></category></item><item><title><![CDATA[D. Cute Sequences]]></title><link>https://blog.youkuaiyun.com/weixin_38772011/article/details/90352501</link><guid>https://blog.youkuaiyun.com/weixin_38772011/article/details/90352501</guid><author>weixin_38772011</author><pubDate>Mon, 20 May 2019 00:19:20 +0800</pubDate><description><![CDATA[https://codeforces.com/contest/1166/problem/D



题意： 给出首尾数，xi = sum xi-1 + k, k属于[1，m]

解; 数学推导下，判断是否可行很简单，还原时从前往后，每加一个乘上(1ll &lt;&lt; (i - j - 1))的贡献值，模拟填充。




#include &lt;bits/stdc++.h&gt;
using...]]></description><category></category></item><item><title><![CDATA[HDU 6534 莫队+树状数组 奇偶优化]]></title><link>https://blog.youkuaiyun.com/weixin_38772011/article/details/90351602</link><guid>https://blog.youkuaiyun.com/weixin_38772011/article/details/90351602</guid><author>weixin_38772011</author><pubDate>Mon, 20 May 2019 00:10:35 +0800</pubDate><description><![CDATA[http://acm.hdu.edu.cn/showproblem.php?pid=6534



题意：l,r区间内差值小于k的对数




#include&lt;bits/stdc++.h&gt;
using namespace std;
typedef long long ll;
const int maxn = 27005;
int n,m,k;
int a[maxn];
int b...]]></description><category></category></item><item><title><![CDATA[华南理工大学“三七互娱杯” E (状压dp)]]></title><link>https://blog.youkuaiyun.com/weixin_38772011/article/details/90121558</link><guid>https://blog.youkuaiyun.com/weixin_38772011/article/details/90121558</guid><author>weixin_38772011</author><pubDate>Sun, 12 May 2019 00:29:30 +0800</pubDate><description><![CDATA[https://ac.nowcoder.com/acm/contest/874/E



n个球放在m个箱子中，花费为箱子倒出的球数。求最小花费




#include&lt;bits/stdc++.h&gt;
using namespace std;
const int maxx=20;
int a[maxx][maxx],c[maxx],f[1&lt;&lt;maxx][maxx];
i...]]></description><category></category></item><item><title><![CDATA[西北大学集训队选拔赛 B	饱和式救援(概率dp)]]></title><link>https://blog.youkuaiyun.com/weixin_38772011/article/details/90114969</link><guid>https://blog.youkuaiyun.com/weixin_38772011/article/details/90114969</guid><author>weixin_38772011</author><pubDate>Sat, 11 May 2019 21:03:22 +0800</pubDate><description><![CDATA[https://ac.nowcoder.com/acm/contest/892/B

n个选出至少k个的概率




#include&lt;bits/stdc++.h&gt;
using namespace std;
int n,m,k;
double dp[2005][2005];
double p[2005];
int l;
double r;
double sum=0;
int main...]]></description><category></category></item><item><title><![CDATA[HDU - 6482 组合数]]></title><link>https://blog.youkuaiyun.com/weixin_38772011/article/details/90112730</link><guid>https://blog.youkuaiyun.com/weixin_38772011/article/details/90112730</guid><author>weixin_38772011</author><pubDate>Sat, 11 May 2019 17:47:39 +0800</pubDate><description><![CDATA[http://acm.hdu.edu.cn/showproblem.php?pid=6482



从y轴上的点走（只能往右和往下）到x轴上的点的方案数是Cx+y,y。

很好理解,x+y段中选x段往右走。

题解: 总方案减去交叉方案

队友ac代码：


#include&lt;bits/stdc++.h&gt;
using namespace std;
typedef long long ...]]></description><category></category></item><item><title><![CDATA[HDU-6470 矩阵快速幂]]></title><link>https://blog.youkuaiyun.com/weixin_38772011/article/details/90111279</link><guid>https://blog.youkuaiyun.com/weixin_38772011/article/details/90111279</guid><author>weixin_38772011</author><pubDate>Sat, 11 May 2019 16:20:50 +0800</pubDate><description><![CDATA[#include &lt;bits/stdc++.h&gt;
using namespace std;

#define MOD 123456789
#define N 6
typedef long long LL;
typedef pair&lt;int, int&gt; Author;
vector&lt;pair&lt;string, int&gt; &gt; VP; 

struct M...]]></description><category></category></item><item><title><![CDATA[消息的传递 （tarjan）]]></title><link>https://blog.youkuaiyun.com/weixin_38772011/article/details/90109829</link><guid>https://blog.youkuaiyun.com/weixin_38772011/article/details/90109829</guid><author>weixin_38772011</author><pubDate>Sat, 11 May 2019 15:06:05 +0800</pubDate><description><![CDATA[http://lpoj.cn/problemdetail?problemID=104

tarjan 缩点 求入度为0的联通分量




#include&lt;bits/stdc++.h&gt;
#define MAXN 1000010
using namespace std;
int n,a;
int ans,num;
int tnt,head[MAXN],to[MAXN],next1[MA...]]></description><category></category></item><item><title><![CDATA[C1C2. Power Transmission (gcd离散化边)]]></title><link>https://blog.youkuaiyun.com/weixin_38772011/article/details/90109733</link><guid>https://blog.youkuaiyun.com/weixin_38772011/article/details/90109733</guid><author>weixin_38772011</author><pubDate>Sat, 11 May 2019 15:00:26 +0800</pubDate><description><![CDATA[https://codeforces.com/contest/1163/problem/C1

题意：给出二维平面上的至多1000个点，求两两连边后，相互相交的边的对数。




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

using ll = long long;
using pii = pair&lt;int, int&gt;;
...]]></description><category></category></item><item><title><![CDATA[HDU - 6485 模糊匹配（尺取法）]]></title><link>https://blog.youkuaiyun.com/weixin_38772011/article/details/90109657</link><guid>https://blog.youkuaiyun.com/weixin_38772011/article/details/90109657</guid><author>weixin_38772011</author><pubDate>Sat, 11 May 2019 14:56:05 +0800</pubDate><description><![CDATA[#include&lt;bits/stdc++.h&gt;
using namespace std;
int t;
int n;
int ans;
char s[5005],v[5005];
void solve(char *x, char *y)
{
	int l = 0,r = -1,cnt = 0;
	int lx = strlen(x);
	int ly = strlen(y);
	wh...]]></description><category></category></item><item><title><![CDATA[Transformer]]></title><link>https://blog.youkuaiyun.com/weixin_38772011/article/details/89815790</link><guid>https://blog.youkuaiyun.com/weixin_38772011/article/details/89815790</guid><author>weixin_38772011</author><pubDate>Sat, 04 May 2019 17:38:00 +0800</pubDate><description><![CDATA[论文：Attention Is All You Need

https://arxiv.org/pdf/1706.03762.pdf





解读：

https://blog.youkuaiyun.com/qq_41664845/article/details/84969266

https://blog.youkuaiyun.com/songbinxu/article/details/80332992

http...]]></description><category></category></item><item><title><![CDATA[激活函数 sigmoid，tanh，relu]]></title><link>https://blog.youkuaiyun.com/weixin_38772011/article/details/89815719</link><guid>https://blog.youkuaiyun.com/weixin_38772011/article/details/89815719</guid><author>weixin_38772011</author><pubDate>Sat, 04 May 2019 17:30:14 +0800</pubDate><description><![CDATA[https://blog.youkuaiyun.com/u011684265/article/details/78039280
]]></description><category></category></item></channel></rss>