搜索资源列表
1001
- POJ1001 PASCAL源代码.poj.org Problems 1001-POJ1001 PASCAL Source Code.poj.org Problems 1001
zhby1990
- 北大poj 上的题集 acm 我做过的题-Acm question I have ever done on the problem sets in Beijing University poj
questoes1
- Question 1068 on poj (PKU)
ACMPedia
- ACM资料大全: 含 北大ACM配套教程.pdf: 北大百练OJ配套用; ACM程序设计.pdf : 用STL进行ACM编程,北大OJ用 ACM的算法分类(北邮OJ).htm 题目分类(文件夹): 含POJ,北邮OJ及ZJU的部分OJ题分类 ACM算法模板(吉林大学): 久负盛名的ACM算法模板; C++函数手册+(LibraryFunctions).chm : C++快速参考,尤其是STL参考很有用;-ACM Sourcebook: Contain Bei
communication-systems-
- poj通信系统 题库中的一道题,可供不会写的同学们下载使用-title the poj communication systems the questions in the one, for not write students download
poj_2787
- poj 2787算24的答案,供想要参考的同学们下载使用-poj 2787 count of 24 the answer for students who want to refer to Download
poj_2775
- poj 2775文件结构的答案,供想要参考的同学们下载使用-poj 2775 file structure of the answer for students who want to refer to Download
poj_2738
- poj 2738 浮点数加法的答案,供想要参考的同学们下载使用-answer to poj 2738 floating-point addition, for the students want to reference download
poj1141
- POJ 1141题目源码+源码 题意: 给你一段 括号序列 让你输出 一个最小的让括号合法的序列 黑书上 有将 思路 : dp[a][b] 代表 a b 之间 最小的需要填加序列数目 a==b时 dp[a][b]=1 a>b时 dp[a][b]=0 s[a] 与s[b]配对时 dp[a][b]=dp[a+1][b-1] 否则 dp[a][b]=min(dp[a][k],dp[k+1][b]) a<=k<b 记录路径采用 一般DP记录路径
487-3279
- poj 487-3279,用c语言AC的代码-poj 487-3279,accepted in C language
Flip-game
- poj Flig game,have been accepted,please use it。
The-Buses
- POJ 1167 The Buses源码。其实也是IOI 1994的题。深搜+剪枝-POJ 1167 The Buses
Finding-Liars
- POJ 1332 Finding Liars. 此题较难-POJ 1332 Finding Liars
3032
- 一个经典纸牌游戏,第一次将一张纸牌从最上面移到最先面,从最上面抽一张放到桌子上,第二次两张,从最上面抽一张放到桌子上,第三次三张.......一次类推。使放到桌上的牌的顺序为1、2....n。求开始的顺序。-POJ 3032 The magician shuffles a small pack of cards, holds it face down and performs the following procedure: The top card is moved to
Poj
- 关于C的基础一百道练习题,西工大的日常训练题,对于基础的C友特别哟哟用-The basis of C, one hundred exercises, daily training title of NPU-based C-friends in particular yo yo with
SolvedHZIEE
- 杭州电子科大hdu poj 上面水题的答案,供不辞辛劳、努力刷题的同学们参考学习。-The above Hangzhou the UESTC hdu poj, water the answer for tirelessly to brush the question of the students for reference.
chessboard-distance
- 一道poj上的问题的解答,棋盘上的距离问题。即实现棋盘上从一个点移动到王,后,兵之间最短步数的解答。-A poj answers to your questions on the chessboard distance problem. That is, move from one point to the king on the board, after the shortest number of steps between the soldiers answer.
Lazy-Cows
- poj的一道题目,lazy cows,很经典,代码有注释,可读性强-lazy cows from poj, classic, readable
1001
- c++算法源代码,主要是POJ上面的算法,根据标号可以找到。-c++ algorithm source code, POJ above algorithm, based on the label can be found.
1002
- c++算法源代码,主要是POJ上面的算法,根据标号可以找到。-c++ algorithm source code, POJ above algorithm, based on the label can be found.