字符串匹配算法-BF算法，它是字符串匹配的最简单算法，适合入门时参考。
String matching algorithm-BF algorithm, which is the most simple string matching algorithm, suitable for entry-reference. (2008-06-02, Visual C++, 1KB, 下载40次)
卡耐基网站SSD6课程练习5作业.绝对好用，欢迎大家下载使用
Exercise 5 Carnegie website SSD6 course work. Absolutely easy to use, welcome to download (2011-12-21, Visual C++, 643KB, 下载20次)
问题描述与实验目的
给定8*8方格棋盘，求棋盘上一只马从一个位置到达另一位置的最短路径长。
注意马是走“日”形的。
输入
输入有若干测试数据。
每组测试数据仅1行，每行上有2个方格pos1、pos2，之间用一个空格隔开，每格方格表示棋盘上的一个位置，该位置由表示列的1个字母（a-h）及表示行的一个数字（1-8）构成，如“d7”表示第4列第7行。
输出
对输入中每行上的2个方格pos1、pos2，输出马从位置pos1跳到pos2所需的最短路径长。如“a1==>a2: 3 moves”表示从位置a1跳到a2所需的最少步数是3。
注意：按输出样例所示格式输出，如“a1==>a2: 3 moves”中冒号后有一个空格，再跟着所需的最少步数。
实验结果
输入样例
a1 a2
a1 a3
a1 h8
g2 b8
输出样例
a1==>a2: 3 moves
a1==>a3: 2 moves
a1==>h8: 6 moves
g2==>b8: 5 moves
Description and purpose of the experiment
8* 8 grid for a given board, find the board a horse from one location to another location to reach the shortest path length.
Note that horse is to take the "day" shape.
Input
Enter a number of test data.
Only 1 line of each test data, each line has 2 squares pos1, pos2, separated by a space, each grid square, said a board position, the position of column 1 by the said letters ( ah), and that the line of a number (1-8) form, such as "d7" said the line 7 column 4.
Output
Each line in the input box on the two pos1, pos2, output horse to jump from position pos1 pos2 shortest path length required. Such as "a1 ==> a2: 3 moves" that jump from position a1 a2 minimum required number of steps is 3.
Note: The sample shown by the output format of the output, such as "a1 ==> a2: 3 moves" in a space after the colon, and then follow the required minimum number of steps.
Experimental results
Input sample
a1 a2
a1 a3
a1 h8
g2 b8
Sample Ou (2010-11-08, Visual C++, 850KB, 下载18次)
实现字符串的查找功能：BF算法，KMP算法，对上述两个算法进行时间复杂性分析并设计实验程序验证分析结果
Find function for string: BF algorithms, KMP algorithm, the two algorithms of time complexity analysis and experimental design analysis of program verification (2011-01-11, Visual C++, 25KB, 下载16次)
最优车皮排序问题
.问题描述：
在一个列车调度站中，k 条轨道连接到k 条侧轨处，形成k 个铁路转轨栈，从左到右依
次记为H1，H2，…，Hk 如下图所示。其中左边轨道为车皮入口（记为H0），右边轨道为出
口（记为Hk+1），编号为a1, a2,.,an 的n 个车皮从入口依次进入转轨栈，由调度室安排车
皮进出栈次序，并对车皮按其出栈次序重新排序为1，2，…，n。
调度室在安排车皮进出栈次序时，遵循以下规则：
（1）
车皮入口H0 处的车皮可以进入转轨栈H1，H2，…，Hk 之一，或直接进入车
皮出口Hk+1 。
（2）
转轨栈H1，H2，…，Hk 处的车皮可以进入车皮出口Hk+1 。
.编程任务：
给定正整数n，和n 个车皮的初始编号a1, a2,.,an ，编程计算最少需要多少个转轨栈
才能按照规则（1）和（2）调度车皮，使车皮在车皮出口按照1，2，…，n 的顺序输出。
.数据输入：
由文件input.txt 给出输入数据。第1 行有1 个正整数n，表示有n 个车皮。第2 行是n
个车皮的初始编号a1, a2,.,an 。
wagon optimal scheduling problem. Problem description : In a train station, k, k track connected to the side of the tracks, forming k stack rail transition from left to right followed credited to H1, H2, ..., Hk as in the figure below. Which left orbit for imported cars (credited to H0), the right track for exports (credited as a Hk), the No. a1, a2,. , An n-wagon from the entrance into the transition stack followed by the dispatch room arrangements wagon Stack access priorities, as well as cars stack up according to the order reordering of 1,2, ..., n. The dispatch room in arranging access stack wagon order, to abide by the following rules : (1) H0 wagon entrance of cars can enter the transition stack H1, H2, ..., Hk, or directly into a wagon export Hk. (2) Transition stack H1, H2, ..., H (2006-03-29, Visual C++, 29KB, 下载12次)
学生成绩管理系统 数据结构二叉树实现。C++编程语言
the manage of the student s grade. (2012-03-17, Visual C++, 935KB, 下载10次)
旅游背包问题，采用近似算法BF、CF算法。题目：第i个物体大小为 Si, 箱子尺寸为c（maxSi<c). 现在有n个物体和m个箱子。请问，我们最少需要多少个箱子来储存这些物体？
要求：
输入 文件名
输出：最少箱子数量
文件名对应的文件内容是
第一行 物体数量n 箱子尺寸C
第二行 用空格分隔的i个物体的大小
其中的数字都是整数，物体和箱子均不能分隔
Travel knapsack problem, the use of approximation algorithms BF, CF algorithms. Title: The size of the i-th object Si, box size c (maxSi <c). 现在有n个物体和m个箱子。请问，我们最少需要多少个箱子来储存这些物体？
要求：
输入 文件名
输出：最少箱子数量
文件名对应的文件内容是
第一行 物体数量n 箱子尺寸C
第二行 用空格分隔的i个物体的大小
其中的数字都是整数，物体和箱子均不能分隔 (2013-07-03, Visual C++, 1KB, 下载7次)
设有n种不同面值a1, a2,…, an的邮票，规定每封信最多贴m张邮票。对于给定的m，n，求出最大的邮资连续区间。
with different denominations a1, a2, ..., an stamps, each up to the letter m stamp affixed. For a given set of m, n, obtained the largest postage continuous interval. (2007-06-12, Visual C++, 403KB, 下载7次)
对单链表实现就地逆置,即利用原表的存储空间将线性表（a1,a2,…,an）逆置为（an,an-1,…,a1）。输入数据：文本文件input.txt
文件分2行：
第1行是单链表的数据元素个数n，n<20。如果n<1或n>19则视为输入数据出错。
第2行是单链表的n个数据元素，每个元素都是-999～999之间的十进制整数，元素之间用空格分开。
输出数据：文件output.txt
文件只有1行：
第1行是完成逆置之后的单链表n个元素的十进制整数，每个元素之间用空格分开。
如果输入数据出错，或者运行过程出错、问题无解等，则第1行输出字符串“ERROR”。
Single linked list implementation situ reverse position, namely the use of the storage space of the original tables linear table (a1, a2, ..., an) inverse set (an, an-1, ..., a1). Input data: text file input.txt file sub 2 lines: Line 1 is the number of a single list of data elements n, n < 20. If n < 1 or n> 19 is considered to input data error. The second line is a single-chain n data elements, and each element is a decimal integer between-999 to 999, separated by a space between the elements. Output data: file output.txt file has only one line: Line 1 is set after the completion of the reverse single linked list of n decimal integer elements, separated by a space between each element. If the input data error, or an error during the operation, the problem has no solution, etc., then the first line of the output string " ERROR" . (2014-05-08, Visual C++, 1KB, 下载6次)
设S=(x1,x2,…,xn)是有序集，且x1<x2<…<xn ,表示有序集S的二叉搜索树， 利用二叉树的结点存储有序集中的元素。且已知其存取概率分布为(a1,a2,…,an)，求在所有表示有序集的二叉树中找出一棵具有最小平均路长的二叉搜索树问题。
Set S = (x1, x2, ..., xn) is an ordered set, and x1 <x2<…<xn ,表示有序集S的二叉搜索树， 利用二叉树的结点存储有序集中的元素。且已知其存取概率分布为(a1,a2,…,an)，求在所有表示有序集的二叉树中找出一棵具有最小平均路长的二叉搜索树问题。 (2010-05-15, Visual C++, 1KB, 下载6次)
最大递增子序列，动态规划经典算法
设L=<a1,a2,…,an>是n个不同的实数的序列，L的递增子序列是这样一个子序列Lin=<aK1,ak2,…,akm>，其中k1<k2<…<km且aK1<ak2<…<akm。求最大的m值。
Maximum increment sequence, the classic dynamic programming algorithm set L = <a1,a2,…,an> Are n different sequences of real numbers, L is the sequence of increasing sub-sequence of Lin = a <aK1,ak2,…,akm> , Where k1 <k2<…<km且aK1<ak2<…<akm。求最大的m值。 (2011-11-10, Visual C++, 1KB, 下载6次)
从文件读入30个无序整数，建立一个双向循环链表并输出，调整链表顺序为
（a1,a3.a5…,a2,a4,a6…）并输出。
Read from the document 30 disorderly integer, a bidirectional circular list and output adjustment linked list of the order of (a1, a3.a5 ..., a2, a4, a6 ...) and outputs. (2012-12-16, Visual C++, 1KB, 下载5次)
一个给定的S={a1,a2,a3,…,an},若有P={a_x1,a_x2,a_x3,…,a_xm},满足(x1 < x2 < … <
xm)且( ax1 < ax2 < … < axm),就称P 为S 的一个上升序列。如果有多个P 满足条件，那么
我们想求字典序最小的。
A given S = {a1, a2, a3, ..., an}, if P = {a_x1, a_x2, a_x3, ..., a_xm}, satisfy (x1 < x2 < ... < xm) and (ax1 < ax2 < ... < axm), called P S in a rising sequence. If you have multiple P satisfy the condition, then we would like to seek the lexicographical smallest. (2012-10-26, Visual C++, 65KB, 下载5次)
给定n个石子，其重量分别为a1，a2，a3，……，an，要示将其划分成m份，每一份的划分费用定义为这份石子中最大重量与最小重量的差的平方。总划分费用等于m份划分费用之和。对于给定的n个石子，求一种划分方案，使得总划分费用最小。
Given an n stone, its weight, respectively a1, a2, a3, ......, an, to be shown to be divided into m parts, each a division of costs defined in this stone maximum weight and minimum weight square. The total of expenses equal to the sum of m copies of the division of costs and For a given n stones, seeking a division of the program, making the total division of the minimum cost.
[auto translate] (2012-06-25, Visual C++, 14KB, 下载4次)
Description:
计算机似乎生来就是干苦力的，不是吗？统计就是这种苦力中的一种，而如果你运气不好，刚好碰到有很多重复数据的时候，就更加令人感到乏味。那么，你的任务来了。现在，给你一列数 A1, A2, … An 。然后要你回答，对于给定的区间，里面到底有多少个不重复的数。
Input:
一个整数 T (T <= 10) ，表示有 T 组测试数据。 每组数据第一行是数 N (1 <= N <= 30000) 。 第二行有 N 个数 A1 A2 … An (0 <= Ai <= 2^31 - 1) 。 第三行是数 Q (1 <= Q <= 100000) 。 接下来 Q 行是 Q 个询问 X Y (1 <= x <= y <= N) 。
Output:
对于每个询问，输出一个答案，一个答案占一行。
Sample Input:
2
3
1 1 4
2
1 2
2 3
5
1 1 2 1 3
3
1 5
2 4
3 5
Sample Output:
1
2
3
2
3
Source:
ycc
Think:
The computer, which seems to do the coolie was born, isn t it? Statistics is one of the the workhorses, and if you are unlucky, just come up against many repeating data, more people feel boring. So, your task. Now, give you a list of several A1, A2,... An. Then will you answer, for a given interval, exactly how many don t repeat number.
Input:
An integer T (T < = 10), said a T test data. Each group of data is the first line number N (1 < = N < = 30000). The second line have N number A1 A2... An (0 < = Ai < = 2 ^ 31-1).The third line is a number (1 Q < = Q < = 100000). The next line is a Q Q ask X Y (1 < = X < = Y < = N).
Output:
For each inquiry, the output one answer, an answer of a line.
Sample Input:
2
3
1 1 4
2
1 2
2 3
5
1 1 2 1 3
3
1 of 5
2 4
3 5
Sample Output:
1
2
3
2
3
Returned:
ycc
(2011-08-26, Visual C++, 9KB, 下载4次)
设A＝（a1， ……，am）和B＝（b1，……，bn）均为顺序表，A’和B’分别为A和B中除去最大共同前缀后的子表
试写一个比较A，B大小的算法
Set A = (a1,......, am) and B = (b1,......, bn) are all order list, A "and" B for A and B were removed after the son of the most common prefix table
Try to write A more A, B the size of the algorithm
(2011-11-27, Visual C++, 1KB, 下载4次)
本实训是有关线性表的顺序存储结构的应用，在本实训的实例程序中，通过C语言中提供的数组来存储两个已知的线性表，然后利用数组元素的下标来对线性表进行比较。通过对本实训的学习，可以理解线性表在顺序存储结构下的操作方法。
在实训中，我们设A=（a1,a2,…,an）和B=（b1,b2,…,bm）是两个线性表，其数据元素的类型是整型。若n=m,且ai=bi,则称A=B 若ai=bi，而aj<bj,则称A<B；除此以外，均称A>B。设计一比较大小的程序。
Training is related to the linear form the structure of the order of storage applications, in this example of practical training program, through the C language to provide an array to store the two known linear form, and then use the array element subscript to comparison of the linear form. Through the Training of the study, the linear form can be understood in the order of the storage structure of the operation. In Training, we set A = (a1, a2, ..., an) and B = (b1, b2, ..., bm) are two linear forms, the data elements of type integer. If n = m, and ai = bi, say A = B if ai = bi, and aj (2008-05-15, Visual C++, 4KB, 下载4次)
利用线性表实现一个一元多项式Polynomial
f(x) = a0 + a1x + a2x2 + a3x3 + … + anxn
The realization of a linear form of Multinomial Polynomial f (x) = a0+ a1x+ a2x2+ a3x3+ ...+ anxn (2010-08-19, Visual C++, 974KB, 下载3次)
问题1：输入整数n，输出1~n的全排列
问题2：输入n个整数：a1,a2……an(可能重复)，输出它们的全排列。
问题3：输入两个整数n，k（n>k），输出{1,2……n}的所有（n，k）组合。
Question 1: input integer n, output 1 ~ n the arrangement
Question 2: input n integer: a1, a2...... An (may repeat), they are all output.
Question 3: input two integer n, k (n > k), output {1, 2... N} all (n, k) combination.
(2012-06-24, Visual C++, 12KB, 下载3次)
已知线性表（a1 a2 a3 …an）按顺序存于内存，每个元素都是整数，试设计用最少时间把所有值为负数的元素移到全部正数值元素前边的算法
Order known linear form (a1 a2 a3 ... an) stored in the memory, each element is an integer, with the least time trial design value is negative elements over all positive numerical elements front algorithm (2012-10-21, Visual C++, 287KB, 下载3次)