其它 - 随笔分类 - Ramanujan

暴力 C++

摘要：一个面试题。主要复习C++语法与STL和list sort 二元谓语函数编写。阅读全文

posted @ 2016-09-08 23:17 Ramanujan 阅读(729) 评论(0) 推荐(0)

摘要：package test; import java.io.BufferedReader; import java.io.File; import java.io.FileOutputStream; import java.io.FileReader; import java.io.IOExcepti 阅读全文

posted @ 2016-09-05 22:59 Ramanujan 阅读(357) 评论(0) 推荐(0)

bfs

摘要：import java.util.LinkedList;import java.util.Queue;import java.util.Scanner;/** * * @author ramanu_jan * *//** * 四维点 */class Point{ int x, y, z, w; ... 阅读全文

posted @ 2014-10-23 19:20 Ramanujan 阅读(300) 评论(0) 推荐(0)

poj 2127 lcis wa

摘要：#include#include#include#include#includeusing namespace std;typedef long long ll;const int maxn = 505;const ll one = 1;const ll inf = one l[j-1][x]) l... 阅读全文

posted @ 2014-06-18 18:28 Ramanujan 阅读(324) 评论(0) 推荐(0)

Oracle笔记

摘要：一、概念1.表空间：段->分区->块。表空间分类：系统、临时表、永久表、大小文件表。段分类：数据段、临时段、索引段、回滚段。2.该关心的11G新特性：=>符号指定函数特定参数、序列、正则表达式、触发器、continue可以再循环使用。二、sql 基础（不熟部分）DDL：1.索引2.修改表：alter... 阅读全文

posted @ 2014-06-14 21:43 Ramanujan 阅读(223) 评论(0) 推荐(0)

三个可能要考的串匹配算法XD.

摘要：部分抄0.暴力算法//伪代码maybeint StrMatch(SString S, SString P)｛ i = 1； j = 1； while(i P[0]) return i – P[0]； return 0; ｝1.KMP相对暴力算法，KMP不同在于当两字符不... 阅读全文

posted @ 2014-06-11 23:21 Ramanujan 阅读(184) 评论(0) 推荐(0)

oracle pl / sql 触发器

摘要：好了，现在学习触发器：（Oracle 从入门到精通，抄的）1. 触发器出现的原因？完成其它约束难以/不能完成的约束。2.原理类似于过程，不过由oracle 自动执行。3.分类：DML触发器； instead of 触发器；系统触发器。触发：触发对象上，触发事件happen && 条件=TRUE... 阅读全文

posted @ 2014-06-10 22:48 Ramanujan 阅读(192) 评论(0) 推荐(0)

oracle 修改 system 密码，创建新用户

摘要：登陆sqlplus ：用户名： sqlplus/as sysdba密码：回车alter user identified by password;创建用户：create user identified by password. 阅读全文

posted @ 2014-06-10 19:31 Ramanujan 阅读(248) 评论(0) 推荐(0)

hdu 4460

摘要：break point 在边数很少，o(n*m)暴力出奇迹~~~#include#include#include#include#include#include#includeusing namespace std;const int maxn = 1005, maxm = 22222, inf=... 阅读全文

posted @ 2014-06-07 17:47 Ramanujan 阅读(218) 评论(0) 推荐(0)

hdu 2594 kmp

摘要：这个题和kmp算法的共同点，也就是可以用kmp解的原因，在于当前缀所在串（kmp中的模式串）字符pj≠后缀所在串(kmp中文本串)字符tj时，应使前缀串(kmp中模式串)尽量往右移动最大位移，而暴力算法则是每次移动位移为1。#include#include#include#includeusing ... 阅读全文

posted @ 2014-05-19 21:30 Ramanujan 阅读(220) 评论(0) 推荐(0)

hdu 1047

摘要：Integer InquiryTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10588Accepted Submission(s): 2737Problem DescriptionOne of the first users of BIT's new supercomputer was Chip Diller. He extended his exploration of powers of 3 to go from 0 to 333 阅读全文

posted @ 2014-03-07 00:03 Ramanujan 阅读(310) 评论(0) 推荐(0)

poj 3278 bfs

摘要：Catch That CowTime Limit: 2000MSMemory Limit: 65536KTotal Submissions: 41703Accepted: 13005DescriptionFarmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100 阅读全文

posted @ 2014-03-05 16:52 Ramanujan 阅读(222) 评论(0) 推荐(0)

hdu 1869 floyd 模板

摘要：六度分离Time Limit: 5000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3562Accepted Submission(s): 1412Problem Description1967年，美国著名的社会学家斯坦利·米尔格兰姆提出了一个名为“小世界现象(small world phenomenon)”的著名假说，大意是说，任何2个素不相识的人中间最多只隔着6个人，即只用6个人就可以将他们联系在一起，因此他的理论也被称为“六度分离”理论(six degre 阅读全文

posted @ 2014-03-05 16:13 Ramanujan 阅读(253) 评论(0) 推荐(0)

hdu 1004 shui

摘要：Let the Balloon RiseTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 65716Accepted Submission(s): 24329Problem DescriptionContest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges' favorite time 阅读全文

posted @ 2014-03-05 14:31 Ramanujan 阅读(164) 评论(0) 推荐(0)

#notes# database chapter 5

摘要：not clear enough:1.haven't mastered definition of : a.tuple relational calculus b.domain relational calculus c.datalog && can't use these methods to solve pros smoothly2.spent much time in understanding recursive programs && semantics of program. 3.still not get the meaning o 阅读全文

posted @ 2013-11-12 01:40 Ramanujan 阅读(222) 评论(0) 推荐(0)

训练 / 校内赛题目总结

摘要：虽然regional都没去过，就要退役了，虽然没有认识很多人，但在多个人身上验证了‘山外有山，人外有人’这句话，也算扩展了自己的视野吧。另外，虽然ACM成绩确实不咋地，但还是学到不少东西的，勇气、自信、耐心、代码能力、算法YY、决心。1.UESTC 1855 http://www.cnblogs.com/ramanujan/p/3412843.html2.uestc 1847 http://www.cnblogs.com/ramanujan/p/3412848.html3.poj 2632 http://www.cnblogs.com/ramanujan/p/3412860.html4.hd 阅读全文

posted @ 2013-11-07 19:10 Ramanujan 阅读(181) 评论(0) 推荐(0)

poj 1330 , LCA, 乱搞

摘要：K - Nearest Common Ancestors Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u Submit Status Practice POJ 1330Description A rooted tree is a well-known data structure in computer science and engineering. An example is shown below: In the figure, each node is labeled with an in 阅读全文

posted @ 2013-11-07 18:55 Ramanujan 阅读(230) 评论(0) 推荐(0)

poj 2001, 乱搞

摘要：A - Shortest Prefixes Time Limit:1000MS Memory Limit:30000KB 64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2001Description A prefix of a string is a substring starting at the beginning of the given string. The prefixes of "carbon" are: "c", "ca", "ca 阅读全文

posted @ 2013-11-07 18:53 Ramanujan 阅读(157) 评论(0) 推荐(0)

poj 2182,乱搞

摘要：D - Lost Cows Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2182DescriptionN (2 #define N 8005using namespace std;struct NODE{ int l,r,num;}node[Nleft){ int mid=(right+left)>>1; i=i=c) findBrand(i,c,j); else{ c-=node[i].num... 阅读全文

posted @ 2013-11-07 18:51 Ramanujan 阅读(179) 评论(0) 推荐(0)

hust 1606，乱搞

摘要：G - Naive Time Limit:3000MS Memory Limit:131072KB 64bit IO Format:%lld & %llu Submit Status Practice HUST 1606DescriptionGive you a positive integer x, determine whether it is the sum of three positive cubic numbers.InputThere’re several test cases. For each case: Only one line containing an int 阅读全文

posted @ 2013-11-07 18:42 Ramanujan 阅读(220) 评论(0) 推荐(0)

Ramanujan

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