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LeetCode-187 Repeated DNA Sequences
All DNA is composed of a series of nucleotides abbreviated as A, C, G, and T, for example: "ACGAATTCCG". When studying DNA, it is sometimes useful to ...
分类:其他好文   时间:2015-04-21 00:25:31    阅读次数:138
protobuf的配置原始文件.proto中,使用修饰符repeated、required、optional生成的java文件的区别
最近在做一个基于Netty框架的项目。 该项目使用了Google的Protobuf作为编解码的框架。(没有使用XML或者json是为了更好的提高效率)。 经过了一段时间的使用我对Protobuf的原始文件进行一个简单的使用方法总结。 我主要在这里说明一下,原始文件.proto中,修饰符repeated、required、optional的区别。 0.windows环境,java工程 1....
分类:编程语言   时间:2015-04-20 15:08:27    阅读次数:216
Repeated DNA Sequences @leetcode
早上到公司第一件事变成了刷leetcode,发现各种题目的各种解法,真心是个挺有趣的过程。比如今天早上碰到的这个DNA序列的问题,一开始完全没有头绪,但是后来看了些文章发现,真的是二进制大法好啊!会了二进制,走遍天下都不怕啊。原题如下: All DNA is composed of a series of nucleotides abbreviated as A, C, G, and T,...
分类:其他好文   时间:2015-04-20 13:14:15    阅读次数:159
LeetCode Repeated DNA Sequences
1 class Solution { 2 private: 3 char tbl[256]; 4 public: 5 vector findRepeatedDnaSequences(string s) { 6 vector res; 7 8 ...
分类:其他好文   时间:2015-04-19 22:46:06    阅读次数:224
2764: [JLOI2011]基因补全
2764: [JLOI2011]基因补全Time Limit:10 SecMemory Limit:128 MBSubmit:570Solved:187[Submit][Status][Discuss]Description在生物课中我们学过,碱基组成了DNA(脱氧核糖核酸),他们分别可以用大写字母...
分类:其他好文   时间:2015-04-17 23:37:08    阅读次数:171
杭电1379(DNA Sorting)java面向对象编程
点击打开链杭电1379 Problem Description One measure of ``unsortedness'' in a sequence is the number of pairs of entries that are out of order with respect to each other. For instance, in the letter sequen...
分类:编程语言   时间:2015-04-16 21:58:10    阅读次数:183
Bash:常用命令工具-uniq
NAME uniq - report or omit repeated linesSYNOPSIS uniq [OPTION]... [INPUT [OUTPUT]]DESCRIPTION Filter adjacent matching lines from I...
分类:其他好文   时间:2015-04-16 19:07:07    阅读次数:138
Gym - 100548C The Problem Needs 3D Arrays
Problem C. The Problem Needs 3D Arrays Time Limit: 6000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64uDescriptionA permutation is a sequenc.....
分类:其他好文   时间:2015-04-14 14:32:41    阅读次数:96
POJ1007--DNA Sorting
#include #include #include using namespace std; int n,m; struct e { char a[200]; int id; int px; }s[200]; int cmp(e s1,e s2) { if(s1.px==s2.px) return s1.id<s2.id; else return s1....
分类:其他好文   时间:2015-04-14 08:35:35    阅读次数:171
poj1141Brackets Sequence
Brackets Sequence Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 26752   Accepted: 7553   Special Judge Description Let us define a regular brackets sequenc...
分类:其他好文   时间:2015-04-13 22:58:46    阅读次数:171
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