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44
说明: 给定程序中,函数fun的功能是:将a所指3×5矩阵中第k列的元素左移到第0列,第k列以后的每列元素依次左移,原来左边的各列依次绕道右边。
例如,有以下矩阵:
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
若k结果为2,程序执行结果为:
3 4 5 1 2
3 4 5 1 2
3 4 5 1 2
请在程序的下划线处填入正确的内容并把下划线删除,使程序得出正确的结果。
注意:源文件存放在考生文件夹下的BLANK1.C中
不得增行或删行,也不得更改程序的结构!
(A given program, function fun feature is: a 3 × 5 matrix referred to in the first k elements of the left column to move Section 0, the k elements of each column after column followed by the left, the original columns in the left turn Bypass on the right. For example, the following matrix: 123,451,234,512,345 If k is 2, the program execution results: 345,123,451,234,512 in the program underscore the right content at the fill and to remove the underscore, so that the program reach the right result. Note: The source files in the folder of BLANK1.C candidates are not allowed by the line or delete line, nor change the structure of the program!)
- 2011-08-11 07:53:10下载
- 积分:1
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Numbersequence
说明: 简单的数据结构算法 在POJ上的题目 利用C++实现(Simple data structure algorithm in POJ on the subject using C++ Implementation)
- 2010-05-04 19:03:45下载
- 积分:1
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5100309203_5_8
二叉树类的前序,后序,层次遍历,其以链表方式存储(Binary tree class pre-order, post-order, hierarchy traversal, the linked list storage)
- 2012-12-16 14:16:41下载
- 积分:1
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arrangement
假设要在足够多的会场里安排一批活动,并希望使用尽可能少的会场。设计一个有效的贪心算法进行安排。(这个问题实际上是著名的图着色问题。若将每一个活动作为图的一个顶点,不相容活动间用边相连。使相邻顶点着有不同颜色的最小着色数,相应于要找的最小会场数。)
对于给定的k 个待安排的活动,编程计算使用最少会场的时间表。
(Suppose you want to arrange a number of activities in the hall more than enough, and want to use as little as possible venue. Design an effective greedy algorithm arrangements. (This problem is actually well-known graph coloring problem. If each activity as a vertex map, with the side connected between incompatible activities. Makes coloring adjacent vertices with minimum number of different colors, corresponding to looking The minimum number of venue.)
K to be arranged for a given activity, calculated using a minimum of venue programming schedule.)
- 2014-10-22 11:00:45下载
- 积分:1
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Online_RandMarkingAlgorithm
本程序解决的是:模拟类似计算机主、缓存的存储结构,当用户请求某个页面时,若某个页面已在缓存中,则直接从缓存中去取,不会产生失误(即未在缓存中命中),否则,就到主存中去取,代价变大。
本程序采用随机标记算法,即在在线标记算法的基础上,对于替换页面时,采用随机从未标记的缓存项中选出一项进行替换。
代码、文档详尽(Addressed in this program are: analog computer-like main, cache storage structure, when the user requests a page, a page in the cache directly from the cache to fetch, does not produce errors (that is, not in the cache in the hit), otherwise, to the main memory to fetch a consideration of larger. This program uses a random marking algorithm, online marking algorithm based on random never mark the cache entry selected a replacement for the replacement page. Code, documentation detailed)
- 2012-09-19 13:01:47下载
- 积分:1
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tree
使用非递归的方法遍历二叉树,可以直接运行。(
Non-recursive binary tree traversal.)
- 2013-12-04 16:08:43下载
- 积分:1
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StringClass
这个代码是STRING类的声明及实现,因为我认为C++类库中的string类比较庞大,找一个函数需要很长时间,所以用一个简单明了的代替之.(STRING kind of statement and realized, because I think the C library string of relatively large, a function for a very long time. Therefore, the use of a simple and straightforward replaced it.)
- 2006-05-17 17:46:58下载
- 积分:1
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LINK
建立链表的算法:
新节点链入表尾建立链表:
动态分配第一个节点变量;
用指针head 妥善保存链表第一个节点的地址值;
依次动态分配一个新节点变量;
将上一个节点的指针域指向新生成的节点;
循环3),4)步骤,直到所有节点均连入链表;
将最后一个节点的指针域赋值为0;
(Establish the list of algorithms:
The new node links footer build list:
Dynamic allocation of the first node variable
Keep the list head pointer of the first node address value
In order to dynamically allocate a new node variable
The domain of a node pointer to the new generation of nodes
Cycle 3), 4) step, until all nodes are connected to the chain
Pointer field will last node assigned to 0 )
- 2015-01-05 16:03:48下载
- 积分:1
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kalman_C
说明: 离散随机线性系统的卡尔曼滤波。
其中13lman.c是卡尔曼滤波函数,4rinv.c是滤波函数中用到的矩阵求逆函数,13lman0.c是主程序。(discrete stochastic linear Kalman filtering system. 13lman.c which is the Kalman filter function, 4rinv.c filtering function is used in the matrix inversion function, is the main program 13lman0.c.)
- 2006-03-01 19:20:13下载
- 积分:1
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xuanzexingjiegou
选择型结构c语言编程 (Choose the type of structure of C language programming)
- 2013-09-24 22:11:37下载
- 积分:1