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原创淘宝登陆界面

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2021-10-06 22:02:55 1655

原创广州大学招聘会总结（IT岗位）

今天无意中与朋友去了一波广州大学商业中心的招聘会，体验了一下招聘的气氛面了几家互联网的公司吧，题目总结如下 :TCP与UDP的协议的区别举具体例子，什么时候用TCP协议什么时候用UDP协议C++数组跟指针创建数组（动态数组）的区别二叉树可以用来干什么列举使用过的容器map内部是用什么数据结构实现的请讲解一下你对红黑树的理解define与const的区别面向对象有哪...

2019-03-30 18:10:16 382

原创 C++ 标准库中的numberic_limits

先来看一组example: #include <iostream> // std::cout#include <limits> // std::numeric_limitsint main () { std::cout << std::boolalpha; std::cout << "Minimum value ...

2018-08-12 10:58:49 418

原创 C++面试题汇总

一、求一个无序数组中的第k大元素（一）分治（二）最小堆（三）桶排序二、关于排序的稳定性概念：排序的稳定性是指如果在排序的序列中，存在前后相同的两个元素的话，排序前和排序后他们的相对位置不发生变化分类：堆排序、快速排序、希尔排序、直接选择排序不是稳定的排序算法，而基数排序、冒泡排序、直接插入排序、折半插入排序、归并排序是稳定的排序算法。三、TCP和UDP的区别和应用场景...

2018-05-29 10:44:14 1633 1

原创 UVA11426 FZU1969 51NOD1188 线性欧拉筛法+筛法

题目大意：计算：G=∑i<Ni=1∑j<=Nj=i+1GCD(i,j)G=\sum_{i=1}^{i<N}\sum_{j=i+1}^{j<=N}{GCD(i,j)}难度系数(往下递增):UVA11426FZU196951NOD1188题目思路一：G=∑ng=1g(∑phi(⌊gd⌋−1) G=\sum_{g=1}^{n}{g(\sum{phi(\lfloor{\frac{g}{d}}\rfloor-

2017-11-02 02:00:28 334

原创 hdu5618Jam's problem again (CDQ分治)

Jam's problem againTime Limit: 5000/2500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1417 Accepted Submission(s): 504Problem DescriptionJam lik

2017-10-12 08:59:39 329

原创 Codeforces Beta Round #86 (Div. 2 Only) E Double Happiness 素数筛法，费马平方和定理

题目链接E. Double Happinesstime limit per test3 secondsmemory limit per test128 megabytesinputstandard inputoutputstandard outputOn the math lesson

2017-10-08 19:40:13 508

原创 Codeforces Round #439 (Div. 2) C. The Intriguing Obsession dp

C. The Intriguing Obsessiontime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output— This is not playing but duty a

2017-10-08 14:19:02 493

原创 Codeforces Round #438 by Sberbank and Barcelona Bootcamp (Div. 1 + Div. 2 combined)C - Qualification

C. Qualification Roundstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputSnark and Philip are preparing the

2017-10-05 21:13:37 332

原创 51nod1228 序列求和（伯努利数）

1228 序列求和题目来源： HackerRank基准时间限制：3 秒空间限制：131072 KB 分值: 160 难度：6级算法题收藏关注T(n) = n^k，S(n) = T(1) + T(2) + ...... T(n)。给出n和k，求S(n)。例如k = 2，n = 5，S(n) = 1^2 + 2^2 +

2017-10-05 15:48:47 430

原创 hdu4777 Rabbit Kingdom （离线处理、互斥）

Online JudgeOnline ExerciseOnline TeachingOnline ContestsExercise AuthorF.A.QHand In HandOnline AcmersForum | DiscussStatistical ChartsProblem ArchiveRealtime Judge Statu

2017-09-28 08:49:23 436

原创 2017 ACM-ICPC 亚洲区（西安赛区）网络赛 G Xor lca+暴力

题目链接：https://nanti.jisuanke.com/t/17120题意：给你一棵树，每个点都有权值，q次询问，每次询问给你a,b,k，问从点a到点b每隔k步的异或和是多少。思路：现场没敢打，复杂度怎么算都不对。。。后来别人说可以1的情况lca其他暴力，我就试了一下，一开始是TLE，然后我各种迷之优化确实可以过。。。。走k步这里我是用了倍增的数组然后用logk的复杂度走k步。。。

2017-09-16 22:28:41 1026

原创 hdu 6194 string string string 后缀数组+rmq+容斥

string string stringTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1531 Accepted Submission(s): 440Problem DescriptionUncle

2017-09-15 19:31:36 291

转载后缀数组模板

来源：http://blog.csdn.net/yxuanwkeith/article/details/50636898。详情看上面，下面只写上我需要的部分。Str ：需要处理的字符串(长度为Len) Suffix[i] ：Str下标为i ~ Len的连续子串(即后缀) Rank[i] : Suffix[i]在所有后缀中的排名 SA[i] : 满足Suffix[SA[1]]

2017-09-15 15:56:04 289

原创 2017 icpc广西邀请赛 K.Query on A Tree (hdu 6191)可持久化字典树

Query on A TreeTime Limit: 20000/10000 MS (Java/Others) Memory Limit: 132768/132768 K (Java/Others)Total Submission(s): 0 Accepted Submission(s): 0Problem DescriptionMonkey A liv

2017-08-31 12:17:41 1154 2

转载高斯消元模板

#includeusing namespace std;#define eps 1e-9const int MAXN=220;const int N=3;double a[MAXN][MAXN],x[MAXN];//方程的左边的矩阵和等式右边的值，求解之后x存的就是结果int equ,var;//方程数和未知数个数/**返回0表示无解，1表示有解*/int Gauss()//高

2017-08-25 00:05:53 250

原创 hdu 6171 Admiral 双向bfs+hash

AdmiralTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 153428/153428 K (Java/Others)Total Submission(s): 206 Accepted Submission(s): 54Problem DescriptionSuppose that you ar

2017-08-24 21:48:54 810

原创 hdu 6166 Senior Pan 最短路

Senior PanTime Limit: 12000/6000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 179 Accepted Submission(s): 45Problem DescriptionSenior Pan fail

2017-08-22 20:32:50 1637 3

原创 hdu 6148 Valley Numer 数位dp

Valley NumerTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 0 Accepted Submission(s): 0Problem Description众所周知，度度熊非常喜欢数字。

2017-08-18 16:39:55 1345 4

原创 hdu 6118 度度熊的交易计划费用流

度度熊的交易计划Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 647 Accepted Submission(s): 226Problem Description度度熊参与了喵哈哈村的商业大会，但是

2017-08-14 11:38:20 512

原创 hdu 6119 小小粉丝度度熊

小小粉丝度度熊Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 14 Accepted Submission(s): 7Problem Description度度熊喜欢着喵哈哈村的大明星——星星小姐。

2017-08-13 17:59:12 319

原创 hdu 6109 数据分割并查集缩点建边

数据分割Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 476 Accepted Submission(s): 103Problem Description小w来到百度之星的赛场上，准备开始实现一个程序

2017-08-12 20:33:21 283

原创 hdu 6102 GCDispower 树状数组+容斥

GCDispowerTime Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 175 Accepted Submission(s): 87Problem DescriptionThere is a myster

2017-08-12 14:52:05 564

原创 hdu 6097 Mindis (反演点)

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6097题目大意:给你圆内两个点P、Q，求圆上一点D，使得 |QD|+|PD|最短在此之前我们先了解一下反演点的知识，参照百度百科:一般指二维反演中的点。二维上反演以一个特定的反演圆为基础:圆心O为反演中心,圆半径为常数k,把点P反演为点P'就是使得OP×OP'=r^2.

2017-08-11 17:13:13 406

原创 hdu 6086 Rikka with String ac自动机+dp

Rikka with StringTime Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 210 Accepted Submission(s): 58Problem DescriptionAs we know,

2017-08-09 12:29:15 1078

原创边双连通分量模板

#include #define MAXN 10005using namespace std;struct Tarjan{ struct edge { int u,v; edge(int uu=0,int vv=0):u(uu),v(vv){} }; int n; //µãµÄ¸öÊý vector e[MAXN]; //Í¼ int DFN[MAXN

2017-08-07 14:44:58 485

原创点双联通分量模板

#include #define MAXN 10005using namespace std;struct Tarjan{struct edge{int u,v;edge(int uu=0,int vv=0):u(uu),v(vv){}bool operator ==(const edge &p)const{return (u==

2017-08-07 14:23:24 292

原创 SW算法求全局最小割

//O(n^3)实现#include using namespace std;#define sfi(a) scanf("%d",&a)#define sfd(a) scanf("%lf",&a)#define sfl(a) scanf("%lld",&a)#define sfs(a) scanf("%s",a)#define rep(i,a,b) for(int i=int(

2017-08-07 14:12:26 573

原创字符串hash模板

const int seed = 31;ull h[maxn];ull base[maxn];ull string_hash(int l, int r){ return h[r] - h[l-1]*base[r-l+1];}void init(){ for(int i = 1; i <= len; ++i) base[i] = base[i-1]*seed; h

2017-08-05 16:14:52 253

转载大素数判断和素因子分解（miller-rabin，Pollard_rho算法）

转载自：http://www.cnblogs.com/kuangbin/archive/2012/08/19/2646396.html传说中的随机算法。效率极高。可以对一个2^63的素数进行判断。可以分解比较大的数的因子。素数判定的时间是O(log(n))Pollard_rho算法的时间复杂度是O(n^1/4)--->用于求一个大数的其中一个质因子#includ

2017-08-04 22:36:17 387

原创 Codeforces Round #426 (Div. 2)

D. The Bakerytime limit per test2.5 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputSome time ago Slastyona the Sweetmaid dec

2017-08-04 14:42:21 205

原创 hdu 6070 Dirt Ratio

Dirt RatioTime Limit: 18000/9000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)Total Submission(s): 384 Accepted Submission(s): 120Special JudgeProblem DescriptionI

2017-08-03 21:25:33 643

原创 hdu 6058 Kanade's sum

Kanade's sumTime Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 516 Accepted Submission(s): 182Problem DescriptionGive you an arr

2017-08-01 21:40:50 1016 2

原创 hdu 6035 Colorful Tree

Colorful TreeTime Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 182 Accepted Submission(s): 46Problem DescriptionThere is a tr

2017-07-25 21:05:37 1844 4

原创 Codeforces Round #384 (Div. 2) E. Vladik and cards

E. Vladik and cardstime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputVladik was bored on his way home and de

2017-07-21 17:32:14 219

原创最大团模板

int N=60;//1~n的节点bool G[N][N];//邻接矩阵int n,Max[N],Alt[N][N],ans;bool DFS(int cur,int tot){ if(!cur){ if(tot>ans){ans=tot;return 1;} return 0; } for(int i=1;i<=cur;i++){

2017-07-20 12:00:45 197

原创 hdu 5556 Land of Farms

Land of FarmsTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 675 Accepted Submission(s): 220Problem DescriptionFarmer John an

2017-07-20 11:53:04 315

原创 hdu 5514 Frogs

FrogsTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2313 Accepted Submission(s): 760Problem DescriptionThere are m stones

2017-07-19 12:27:01 239

原创 Codeforces Round #422 (Div. 2) C. Hacker, pack your bags!

C. Hacker, pack your bags!time limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputIt's well known that the best wa

2017-07-10 16:03:28 314

原创倍增LCA模板

const int M = 20;struct node{ int v, value, next;}p[maxn << 1];int len, head[maxn], dis[maxn], depth[maxn], n, m, father[maxn][25];void addedge(int u, int v, int value){ p[len].v = v; p[len].va

2017-06-03 16:42:25 210

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