搜索资源列表
maze1
- 从TS流中读取各表信息,从pat表中读取TS_ID,PMT_PID,SERVICE_ID,再去确定pmt和sdt,以及nit的信息。-get the information of transport stream.you can get the infomation of program access table,program map table,network information table and service descr iption table
PATandPMT
- 遍历各TS包,输出包长及包数目,从中找PAT包,读节目号及对应PMT的PID,读PMT的视、音频PID。-Crawl ts video stream packet
paipai
- 采集拍拍卖家QQ号,按查询内容页搜索采集,使用正则表达式-Acquisition pat seller QQ number, acquisition, according to the query page search using regular expressions
pat
- ZJU-pat 1042~1045题目源码-The title source code of ZJU-pat 1042 ~ 1045
pat1017
- 浙江大学的PAT程序能力测试中的第1017题。 -failed to translate
pat1018
- 浙江大学的PAT程序能力测试中的第1018题。涉及到最短路径 -failed to translate
no_1014
- 浙大PAT no_1014福尔摩斯的约会。 大侦探福尔摩斯接到一张奇怪的字条:“我们约会吧! 3485djDkxh4hhGE 2984akDfkkkkggEdsb s&hgsfdk d&Hyscvnm”。大侦探很快就明白了,字条上奇怪的乱码实际上就是约会的时间“星期四 14:04”,因为前面两字符串中第1对相同的大写英文字母(大小写有区分)是第4个字母 D ,代表星期四;第2对相同的字符是 E ,那是第5个英文字母,代表一天里的第14个钟头(于是一天的0点到23点由数字0到9、以及大写字母A
1073
- PAT1073可以使用,今年考试题,关于数字的表示的题-PAT1073,pat.zju.edu.cn,you can get exercise at here
1001
- PAT基础题1001.害死人不偿命的(3n+1)猜想 简洁易懂 可以给大家提供一些解题思路 扩展思维-PAT basic question 1001. Killed attractiveness of (3n+1) guess concise and can give you some idea of solving the expansion of thinking
1002
- PAT基础题1002. 写出这个数 简洁易懂 可以给大家提供一些解题思路 扩展思维 -PAT basic question 1002. Write this number can be simple and easy to understand solving ideas to provide some extended thinking
1003
- PAT基础题1003. 我要通过! 简洁易懂 可以给大家提供一些解题思路 扩展思维 -PAT basic question 1003 I want to pass! Easy to read you can give us some idea of solving the expansion of thinking
1004
- PAT基础题1004. 成绩排名 简洁易懂 可以给大家提供一些解题思路 扩展思维 -PAT basic question 1004. Standings concise and can give you some idea of solving the expansion of thinking
1005
- PAT基础题1005. 继续(3n+1)猜想 简洁易懂 可以给大家提供一些解题思路 扩展思维 -PAT basic question 1005. Continue (3n+1) guess concise and can give you some idea of solving the expansion of thinking
PAT_Advanced-Level1001~1020
- 浙大PAT甲级1001~1020源代码,所有都经过测试-Zhejiang PAT Grade 1001 to 1020 source code, all are tested
pat
- pat.zju.edu.cn 计算机编程训练平台上基础题和部分提高题的源码,C语言,gcc编译通过。该平台上练练手不错,但是一些测试点不告诉你错在哪里十分纠结。-Source-based questions and some questions on pat.zju.edu.cn improve computer programming training platform, C language, gcc compiler. To practice good hand on the platfo
pat1044
- pat1044 题目大意:给出长度为n的一个整数序列,找出其中所有和为m的子序列(连续的),如果没有任何子序列的和等于m,那么找出大于m的最小和。如长度为8的序列3,2,1,5,4,6,8,7,找出和为15的子序列有三个:1~5,4~6,7~8。 这道题思路并不难,一般都能想到通过遍历来解决,但是如果不注意细节是很难拿到满分的。 -pat 1044
Pat
- Pat部分真题的题目和解答,供求职需要刷题的同学参考-code for some tests in pat exam
A1001
- pat advance level 1001. A+B Format-Calculate a+ b and output the sum in standard format that is, the digits must be separated into groups of three by commas (unless there are less than four digits). Input Each input file contains one test
A1002
- pat advance level 1002 A+B for Polynomials-This time, you are supposed to find A+B where A and B are two polynomials. Input Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a poly
PAT
- 刷PAT必备,包含PAT题集,英文单词解释,部分答案。(Brush PAT must include the PAT problem set, English word explanation, partial answer.)