其中，A为入口，B为出口，S为中转盲端。所有铁道均为单轨单向式：列车行驶的方向只能是从A到S，再从S到B；也可以不在S中驻留，直接从A驶向B；另外，由于S较为窄小，车厢在S中不能调头或超车，这意味着任何一个在S中驻留的车厢能从S中驶出，当且仅当所有在它之后驶入S的车厢都已经从S中驶出（或者没有车厢在它之后驶入S）。因为车厢可在S中驻留，所以它们从B端驶出的次序，可能与从A端驶入的次序不同。不过S的容量有限，同时驻留的车厢不得超过m节。
设某列车由编号依次为{a1, a2, ..., an}的n节车厢组成。调度员希望知道，按照以上交通规则，这些车厢能否以{1, 2, ..., n}的次序，重新排列后从B端驶出。
Among them, a is the inlet, B is the outlet, and S is the transfer blind end. All railways are monorail unidirectional: the train can only travel from a to s, and then from s to B; it can also run directly from a to B without staying in S; in addition, due to the narrow size of S, cars can not turn around or overtake in s, which means that any car staying in s can move out of s if and only if all the cars that enter s after it have moved out of S (or No cars enter s after it. Because the cars can stay in s, the order in which they leave from end B may be different from that in which they enter from end a. However, the capacity of S is limited and no more than m cars can stay at the same time. (2020-11-23, Visual C++, 2KB, 下载0次)
教练员 A、B 和 C 将要从编号为1到n的队员中挑选自己的队员。为公平起见，每个教练都根据自己的喜好程度将队员排序；你负责根据以下规则为他们分配队员。
你拿到的数据是 a、b、c 三个数组，表示三个教练对队员的喜好程度排序，每个数组都是数字 1 到 n 的一个排列，下标越小表示教练越喜欢该队员（例如教练 A 最喜欢编号为 a[0] 的队员，其次是编号为 a[1] 的队员）。你的分组规则是，从还未被分配的队员中找一个教练A最喜欢的队员分到 A 组；然后，在未分配的队员中分配教练B最喜欢的队员到 B 组；然后是教练 C；再是教练 A、B......依次类推直到所有队员分配完毕。
现在队员 k 希望知道自己被分配给哪位教练，请你来告诉他。
Among them, a is the inlet, B is the outlet, and S is the transfer blind end. All railways are monorail unidirectional: the train can only travel from a to s, and then from s to B; it can also run directly from a to B without staying in S; in addition, due to the narrow size of S, cars can not turn around or overtake in s, which means that any car staying in s can move out of s if and only if all the cars that enter s after it have moved out of S (or No cars enter s after it. Because the cars can stay in s, the order in which they leave from end B may be different from that in which they enter from end a. However, the capacity of S is limited and no more than m cars can stay at the same time.
Suppose that a train is composed of N cars numbered {A1, A2,..., an}. The dispatcher would like to know whether, according to the above traffic rules, these cars can be rearranged in the order of {1,2,..., n} and then drive out from end B. (2020-11-23, Visual C++, 1KB, 下载0次)
Main 97d Fix Cam 3d+otros
Main 97d Fix Cam 3d+otros (2020-08-21, Visual C++, 8386KB, 下载0次)
Munonlineeeeeee free
Munonlineeeeeee free (2020-06-08, Visual C++, 239KB, 下载2次)
BF算法实现,c++算法，数据结构课程作业
Data structure code,for reference only,BF algorithm (2019-11-20, Visual C++, 1079KB, 下载0次)
某列车调度站的铁道联接结构如图所示。
其中，A为入口，B为出口，S为中转盲端。所有铁道均为单轨单向式：列车行驶的方向只能是从A到S，再从S到B；也可以不在S中驻留，直接从A驶向B；另外，不允许超车。因为车厢可在S中驻留，所以它们从B端驶出的次序，可能与从A端驶入的次序不同。不过S的容量有限，同时驻留的车厢不得超过m节。
设某列车由编号依次为{a1, a2, ..., an}的n节车厢组成。调度员希望知道，按照以上交通规则，这些车厢能否以{1, 2, ..., n}的次序，重新排列后从B端驶出。
In this station, A is the entrance for each train and B is the exit. S is the transfer end. All single tracks are one-way, which means that the train can enter the station from A to S, and pull out from S to B. Note that the overtaking is not allowed. Because the compartments can reside in S, the order that they pull out at B may differ from that they enter at A. However, because of the limited capacity of S, no more that m compartments can reside at S simultaneously.
Assume that a train consist of n compartments labeled {1, 2, , n}. A dispatcher wants to know whether these compartments can pull out at B in the order of {a1, a2, , an} (a sequence). If can, in what order he should operate it? (2019-04-21, Visual C++, 1106KB, 下载1次)
某列车调度站的铁道联接结构如图所示。其中，A为入口，B为出口，S为中转盲端。所有铁道均为单轨单向式：列车行驶的方向只能是从A到S，再从S到B；也可以不在S中驻留，直接从A驶向B；另外，不允许超车。因为车厢可在S中驻留，所以它们从B端驶出的次序，可能与从A端驶入的次序不同。不过S的容量有限，同时驻留的车厢不得超过m节。设某列车由编号依次为{a1, a2, ..., an}的n节车厢组成。调度员希望知道，按照以上交通规则，这些车厢能否以{1, 2, ..., n}的次序，重新排列后从B端驶出。
In this station A is the entrance for each train and B is the exit S is the transfer end. All single tracks are one-way which means that the train can enter the station from A to S and pull out from S to B Note that the overtaking is not allowed Because the compartments can reside in S the order that they pull out at B may differ from that they enter at A However because of the limited capacity of S no more that m compartments can reside at S simultaneously Assume that a train consist of n compartments labeled 1 2 n A dispatcher wants to know whether these compartments can pull out at B in the order of a1 a2 an a sequence If can in what order he should operate it (2019-04-21, Visual C++, 1106KB, 下载0次)
数据结构课程设计 电梯调度算法的实现，实现简单，可直接使用
Elevator is defined in this program.You can use it directly; (2018-12-23, Visual C++, 27481KB, 下载3次)
官方大灰狼远程管理V8.97，免杀源码可以自己编译
The official wolf remote management V8.97 source code, you can compile free (2018-02-28, Visual C++, 4564KB, 下载29次)
这是一个地方的话可以吗？HK搜狐视频聊天的吗？？？
this is a jxbsjxbsjdbxjsbxbjdbxvhexj (2018-02-02, Visual C++, 556KB, 下载1次)
折半查找法是效率较高的一种查找方法。假设有已经按照从小到大的顺序排列好的五个整数a0~a4，要查找的数是X，其基本思想是: 设查找数据的范围下限为l=1，上限为h=5，求中点m=(l+h)/2，用X与中点元素am比较，若X等于am，即找到，停止查找;否则，若X大于am，替换下限l=m+1，到下半段继续查找;若X小于am，换上限h=m-1，到上半段继续查找;如此重复前面的过程直到找到或者l>h为止。如果l>h，说明没有此数，打印找不到信息，程序结束。
The method of binary search is one of the most efficient methods. Hypothesis has already ranked in accordance with the order of the growing up five integers a0 ~ a4, to find the number is X, its basic idea is: the scope of the data set to find the lower limit for l = 1, up to h = 5, strives for the midpoint m = (l + h) / 2, with X compared with neutral element am, if X is equal to the am, is found, stop looking for; Otherwise, if X is greater than am, replace the lower bound l=m+ 1, and continue to find it in the lower half; If X is less than am, change the upper limit h is equal to m minus 1, and continue to find it in the first half; Repeat the previous process until you find or l> h. If l> h, you don't have this number, print can't find the information, the program ends. (2018-01-03, Visual C++, 36KB, 下载1次)
CPU硬改需要刷新bios，用它来改你的bios，管理通过吧
bios changer refresher speed stable (2017-12-26, Visual C++, 394KB, 下载4次)
摁1输出矩键，摁2stm32单片机上的两个灯亮，摁3的蜂鸣器一直响，摁4 4个led灯隔一秒闪烁，摁5，蜂鸣器响三秒，摁6，数码管显示数字六
1 矩键 C.01234567 +
2 亮灯 B5.E5 +
3 蜂鸣器 B8 +
4 4个灯走马灯 1秒 F.3579 共阴极 +
5 蜂鸣器响三秒 B8 +
6 数码管显示6 A.01234567 共阴极 +
Press the 1 key moment output two lights, press the 2stm32 microcontroller on the light, press the 3 buzzer ringing, press the 44 LED lights every seconds flashing, press the buzzer 5, three seconds, press 6, six digital tube display digital (2017-11-21, Visual C++, 4929KB, 下载1次)
Train
描述
某列车调度站的铁道联接结构如图所示。
其中，A为入口，B为出口，S为中转盲端。所有铁道均为单轨单向式：列车行驶的方向只能是从A到S，再从S到B；也可以不在S中驻留，直接从A驶向B；另外，不允许超车。因为车厢可在S中驻留，所以它们从B端驶出的次序，可能与从A端驶入的次序不同。不过S的容量有限，同时驻留的车厢不得超过m节。
设某列车由编号依次为{a1, a2, ..., an}的n节车厢组成。调度员希望知道，按照以上交通规则，这些车厢能否以{1, 2, ..., n}的次序，重新排列后从B端驶出。
THU2017 2-1 Train
describe
The railway connection structure of a train dispatching station is shown in the diagram.
Among them, A is the entrance, B is the exit, and S is the blind end of the transfer. All tracks are monorail one-way: the direction of the train is only from A to S, and then from S to B; also can not stay in the S, directly from A to B; in addition, not allowed overtaking. Because the carriage can reside in the S, the order that they exit from the B end may be different from the order from the A end. However, the capacity of S is limited, and the carriage shall not be longer than m. (2017-11-11, Visual C++, 42KB, 下载3次)
针对ADI公司的cces操作平台， 上电测试程序，测试上电后bf-706开发板的功能
power on self-test for cces (2017-10-20, Visual C++, 79KB, 下载4次)
官方大灰狼远程管理V8.97，免杀源码可以自己编译
You can compile free source code (2017-09-12, Visual C++, 12628KB, 下载142次)
PIC单片机系统隐藏的延时方法,该函数为编译器自带延时函数
Hidden delay method of PIC single chip microcomputer system (2017-08-18, Visual C++, 6KB, 下载4次)
实现二进97小波变换的一个例子,对深入理解小波变换有帮助,
Realization of binary wavelet transform is an example of 97 has help to the deep understanding of wavelet transform, (2017-08-12, Visual C++, 2KB, 下载1次)
可以实现二进97小波提升格式的小波变换,希望对大家学习和研究提
Can achieve binary wavelet ascend 97 format of wavelet transform, hope to study and research (2017-08-12, Visual C++, 2KB, 下载3次)
反向阵列（multiblock）:
给定指针d_a中的输入数组{a0，a1，...，an-1}，将反向数组{an-1，an-2，...，a0}存储在指针d_b中
A: 从“reverseArray_multiblock”模板开始
B:多个256线程块启动,要颠倒大小为N，N / 256块的数组
第1部分：计算要启动的块数
第2部分：实现内核reverseArrayBlock
请注意，现在您必须同时进行计算
块内的相反位置
反向偏移到块的开始
Reverse array (multiblock):
Given the pointer d_a in the input array {a0, A1,..., an-1}, the reverse array {an-1, An-2,..., a0} stored in the pointer d_b
A: starts with the "reverseArray_multiblock" template
B: more than 256 thread block start, to reverse the size of N, N / 256 block array
The first part: the number of calculation to start
The second part: implementation of kernel reverseArrayBlock
Attention, now you must calculate at the same time
The reverse position in a block
Reverse offset to block start (2017-06-24, Visual C++, 6KB, 下载1次)