手机版的贪吃蛇相信已经玩腻了,来玩玩vc 做的吃苹果的贪吃蛇,
Mobile version of the greedy snake believe to be tired of playing, to play with the vc to eat the apple snake, (2018-12-01, Dev C++, 6KB, 下载0次)
在玩‘连连看’,由于时间限制得太紧,总是玩得不爽,于是费半天时间
In 'Lianliankan', because the time limit is too tight, always play hard, so spend half a day (2018-11-29, Dev C++, 63KB, 下载0次)
一个内存修改器,可作为游戏玩家修改装备的好工具,也可为编程者提供参考
A memory modifier can be used as a good tool for players to modify equipment, but also for programmers to provide a reference (2018-11-27, Dev C++, 51KB, 下载0次)
算法简单描述如下： 如果当前棋局为终局状态,则返回状态分 从当前棋局的状态出发,找出一个可走的步数,试走此部,新状态扩展
The algorithm is described as follows: if the current game is the final state, then the return state is divided from the current state of the chess game, find out a walkable number of steps, try this part, the new state extension (2018-11-21, Dev C++, 31KB, 下载0次)
msp430仿真工具的驱动程序，如果你的电脑不能识别仿真硬件，试一试这个吧
msp430 driver simulation tools, if your computer does not recognize the hardware emulation, try the bar.) (2018-08-20, Dev C++, 3297KB, 下载0次)
若无向图表示高速公路网，其中顶点表示城市，边表示城市之间的高速公路。试设计一个找路程序，获取两个城市之间的所有简单路径
If an undirected graph is used to express a highway network, the vertices represent the city, and the side represents the expressway between the cities. Try to design a path finding program to get all the simple paths between two cities (2018-01-12, Dev C++, 1KB, 下载3次)
描述
某公司在对应聘者做过一轮笔试之后，从中选出n 人继续进行面试，每位应聘者被分配了一个整数ID。
为公平起见，组织者决定利用会议室外的圆桌，按以下方法“随机”确定面试顺序：第一个到达的应聘者在圆桌周围任意选择一个位置坐下；此后到达的每位应聘者都从前一应聘者出发，沿逆时针方向围圆桌走过m 人（前一应聘者算作走过的第1 人，同一人可能经过多次），并紧邻第m 人右侧就座；所有应聘者到齐后，从最后到达者出发，绕圆桌以顺时针方向为序进行面试。
这里假定应聘者到达的时刻互异，且相对的就坐位置确定后，左、右两人之间总能插入一把椅子。
试编写一个程序，确定面试顺序。
输入
共2行。
第1行包含两个整数， n和m。
第2行包含n个整数，表示先后到达的n个应聘者的ID。
输出
共1行。以空格分隔的n个整数，分别表示顺次进行面试的应聘者的ID。
After a company has done a written test of the candidate, n people are selected to continue the interview. Each candidate is assigned an integer ID.
To be fair, the organizers decided to use the round table outside the conference room to "randomly" determine the interview sequence as follows: The first candidate who arrives at a random place around the round table sits down; each candidate arrives thereafter Candidates starting from the round table counterclockwise around m people (the previous candidate counted as walking through the first person, the same person may go through many times), and the m-th right seat; all candidates to Qi, starting from the last arrivals, around the round table in order to conduct an interview in the clockwise direction.
Here assume that candidates arrive at different times, and the relative seating position is determined, the left and right can always insert a chair between the two. (2017-12-16, Dev C++, 450KB, 下载3次)
问题描述
若用无向图表示高速公路网，其中顶点表示城市，边表示城市之间的高速公路。试设计一个找路程序，获取两个城市之间的所有简单路径。
基本要求
（1） 输入参数：结点总数，结点的城市编号（4位长的数字，例如电话区号，长沙是0731），连接城市的高速公路（用高速公路连接的两个城市编号标记）。
（2） 输入 要求取所有简单路径的两个城市编号。
（3） 将所有路径（有城市编号组成）输出到用户指定的文件中。
Problem descriptionIf use undirected graph said freeways, with vertex said city, edge said highway between the two cities.Try to design a route program, obtain all simple paths between the two cities. The basic requirements(1) input parameters: the total number of nodes, the nodes of the city number (4 digits long Numbers, telephone area code, for example, changsha is 0731), connected to the city expressway (with high-speed highway connects the two cities of number tags).(2) input requirements take number all simple paths of the two cities.(3) will all paths (a number of city) output to the user specified file. (2017-01-02, Dev C++, 3KB, 下载4次)
问题描述
设Pn(x)和Qm(x)分别两个一元多项式。试编写程序实现一元多项式的加法运算。
基本要求
需要基于线性表的基本操作来实现一元多项式的加法运算
需要利用有序链表来实现线性表。
Problem descriptionPn (x) and Qm (x), respectively, two unary polynomial.Try to write programs achieve one yuan polynomial addition operation. The basic requirementsBased on the basic operation of the linear table is needed to achieve one yuan polynomial addition operationNeed to use an orderly linked list to achieve linear table. (2017-01-02, Dev C++, 1KB, 下载6次)
词法分析器一枚，可以试下多种语言的词法分析功能。
A lexical analyzer, you can try multilingual lexical analysis. (2016-08-01, Dev C++, 9KB, 下载2次)
中国罗斯福互助基金程序-仿MMM中国罗斯福互助基金程序-仿MMM
Chinese Roosevelt mutual fund program- Imitation MMMChinese Roosevelt mutual fund program- Imitation MMM (2016-03-04, Dev C++, 22186KB, 下载59次)
关于直流电机的NIOS源码程序，已调试，建议在EL-ARM-DSP-IV实验箱上调试。
NIOS source program for DC motor，debugging on EL-ARM-DSP-IV experiment box. (2015-10-06, Dev C++, 16456KB, 下载3次)
关于DA转换的NIOS源码程序，已调试，建议在EL-ARM-DSP-IV实验箱上调试。
NIOS source program for DA conversion，debugging on EL-ARM-DSP-IV experiment box. (2015-10-06, Dev C++, 16363KB, 下载1次)
此程序为21点扑克牌游戏，玩家一共可以要五张牌，但如果牌的点数超过21，则自动出局；在不超过21点的情况下，玩家与庄家比牌的点数大小，大者为赢，相同则为平局。
程序说明：
1） 将所有的函数与相应的数据封装到类中，并改写主程序。使程序简化。
2） 程序中一共有13张扑克牌可以随时抽取，大于10的点数为0.5。
3） 超级玩家永远不会输掉，即超级玩家可以查看下一张牌，若大于21点，则可以拒绝，当然游戏规则上不能泄露这点秘密。
4） 超级玩家可以查看下一张牌，即输入指定的字符串或字符，然后按要求输入正确密码后才能看自己的和计算机的牌，并指定下一级牌的点数。
5） 每次要牌后可以设定赔率，即如果开始下的是10元钱的注，如果牌好，你可以要求继续下注，当然不能超过你所有的总钱数。
6） 将界面的提示说明改成中文界面，界面的解释详细友好，可以吸引更多的玩家。
This procedure for the 21 points of the game, the player can be a total of five cards, but if the number of points more than 21, then automatically out in no more than 21 points, the players and the dealer than the number of points, the big winner, the same as a draw.
Program description:
1) all the functions and the corresponding data are encapsulated in the class, and the main program is rewritten. Simplify procedures.
2) a total of 13 cards can be drawn at any time, more than 10 points to 0.5.
3) the super players will never lose, that is, the super players can look at the next card, if more than 21 points, you can refuse, of course, the game can not reveal this secret.
4) a super player can look at the next card, that is, enter the specified string or character, and then press the request to enter the correct password to see their own and the computer s card, and specify the next level.
5) every time you have to sign the odds, that if you start with a 10 yuan not (2015-09-02, Dev C++, 2KB, 下载5次)