 Combinations

Description
Computing the exact number of ways that N things can be taken M at a time can be a great challenge when N and/or M become very large. Challenges are the stuff of contests. Therefore, you are to make just such a computation given the following:
GIVEN: 5 <= N <= 100; 5 <= M <= 100; M <= N
Compute the EXACT value of: C = N! / (NM)!M!
You may assume that the final value of C will fit in a 32bit Pascal LongInt or a C long. For the record, the exact value of 100! is:
93,326,215,443,944,152,681,699,238,856,266,700,490,715,968,264,381,621, 468,592,963,895,217,599,993,229,915,608,941,463,976,156,518,286,253, 697,920,827,223,758,251,185,210,916,864,000,000,000,000,000,000,000,000
InputThe input to this program will be one or more lines each containing zero or more leading spaces, a value for N, one or more spaces, and a value for M. The last line of the input file will contain a dummy N, M pair with both values equal to zero. Your program should terminate when this line is read.
OutputThe output from this program should be in the form:
N things taken M at a time is C exactly.
Sample Input100 6
20 5
18 6
0 0
Sample Output100 things taken 6 at a time is 1192052400 exactly.
20 things taken 5 at a time is 15504 exactly.
18 things taken 6 at a time is 18564 exactly.
Identifying Legal Pascal Real Constants _course
20171017Description Pascal requires that real constants have either a decimal point, or an exponent (starting with the letter e or E, and officially called a scale factor), or both, in addition to the usual collection of decimal digits. If a decimal point is included it must have at least one decimal digit on each side of it. As expected, a sign (+ or ) may precede the entire number, or the exponent, or both. Exponents may not include fractional digits. Blanks may precede or follow the real constant, but they may not be embedded within it. Note that the Pascal syntax rules for real constants make no assumptions about the range of real values, and neither does this problem. Your task in this problem is to identify legal Pascal real constants. Input Each line of the input data contains a candidate which you are to classify. Output For each line of the input, display your finding as illustrated in the example shown below. The input terminates with a line that contains only an asterisk in column one. Sample Input 1.2 1. 1.0e55 e12 6.5E 1e12 +4.1234567890E99999 7.6e+12.5 99 * Sample Output 1.2 is legal. 1. is illegal. 1.0e55 is legal. e12 is illegal. 6.5E is illegal. 1e12 is legal. +4.1234567890E99999 is legal. 7.6e+12.5 is illegal. 99 is illegal.
Pascal Library _course
20161224Description Pascal University, one of the oldest in the country, needs to renovate its Library Building, because after all these centuries the building started to show the effects of supporting the weight of the enormous amount of books it houses. To help in the renovation, the Alumni Association of the University decided to organize a series of fundraising dinners, for which all alumni were invited. These events proved to be a huge success and several were organized during the past year. (One of the reasons for the success of this initiative seems to be the fact that students that went through the Pascal system of education have fond memories of that time and would love to see a renovated Pascal Library.) The organizers maintained a spreadsheet indicating which alumni participated in each dinner. Now they want your help to determine whether any alumnus or alumna took part in all of the dinners. Input The input contains several test cases. The first line of a test case contains two integers N and D indicating respectively the number of alumni and the number of dinners organized (1 <= N <= 100 and 1 <= D <= 500). Alumni are identified by integers from 1 to N. Each of the next D lines describes the attendees of a dinner, and contains N integers Xi indicating if the alumnus/alumna i attended that dinner (Xi = 1) or not (Xi = 0). The end of input is indicated by N = D = 0. Output For each test case in the input your program must produce one line of output, containing either the word `yes', in case there exists at least one alumnus/alumna that attended all dinners, or the word `no' otherwise. Sample Input 3 3 1 1 1 0 1 1 1 1 1 7 2 1 0 1 0 1 0 1 0 1 0 1 0 1 0 0 0 Sample Output yes no
字母猜谜问题，用动态规划算法怎么解决？_course
20181110Problem Description Anna Graham is a puzzle maker who prides herself in the quality and complexity of her work. She makes puzzles of all kinds  crosswords, logic problems, acrostics, and word search puzzles, to name but a few. For each puzzle, she has developed a set of rules which she constrains herself to follow. For word search puzzles, she insists not only that all the words be connected to one another (as in most word search puzzles), but also that removing any word from the word list will not cause one or more words to become disconnected from the rest. (Two words are connected if they contain a common letter in the grid.) The example word search puzzle on the left satisfies this condition, but the one on the right does not (removing the word Pascal from the word list disconnects Java from the other two words). ![](http://acm.hdu.edu.cn/forum/attachment/11_237_de4c3a5a3af1175.jpg) Your job is to write a program that checks to see if Anna’s word search problems are up to snuff. Input Input will consist of multiple test cases. The first line of each test case contains 3 integers n m l, where n and m are the number of rows and columns in the puzzle and l is the number of words. Following this are n lines containing m uppercase characters each (the puzzle) followed by l lines containing one word each (the word list, in mixed case). Each word in the word list will appear in the puzzle exactly once. There will be no more than 100 rows and 100 columns in the puzzle and no more than 100 words to search for. There will be no spaces in the input words. Output For each problem instance, output the word Yes or No depending on whether the puzzle satisfies Anna’s constraints. Sample Input 5 6 3 PBROGR PASCAL ASMMIN GIICON TCELST BASIC LISP Pascal 5 6 4 PBROJR PASCAL ASMMVN GIICAN TCELST BASIC Java LISP Pascal 0 0 0 Sample Output Yes No
Combinations _course
20171015Description Computing the exact number of ways that N things can be taken M at a time can be a great challenge when N and/or M become very large. Challenges are the stuff of contests. Therefore, you are to make just such a computation given the following: GIVEN: 5 <= N <= 100; 5 <= M <= 100; M <= N Compute the EXACT value of: C = N! / (NM)!M! You may assume that the final value of C will fit in a 32bit Pascal LongInt or a C long. For the record, the exact value of 100! is: 93,326,215,443,944,152,681,699,238,856,266,700,490,715,968,264,381,621, 468,592,963,895,217,599,993,229,915,608,941,463,976,156,518,286,253, 697,920,827,223,758,251,185,210,916,864,000,000,000,000,000,000,000,000 Input The input to this program will be one or more lines each containing zero or more leading spaces, a value for N, one or more spaces, and a value for M. The last line of the input file will contain a dummy N, M pair with both values equal to zero. Your program should terminate when this line is read. Output The output from this program should be in the form: N things taken M at a time is C exactly. Sample Input 100 6 20 5 18 6 0 0 Sample Output 100 things taken 6 at a time is 1192052400 exactly. 20 things taken 5 at a time is 15504 exactly. 18 things taken 6 at a time is 18564 exactly.
A Number from Yanghui Triangle _course
20170104Description Yanghui Triangle, as known as Pascal’s Triangle, is a number triangle with numbers arranged in staggered row such that an, 0 = 1 (n ≥ 0) an, n = 1 (n ≥ 0) an, r = an − 1, r − 1 + an − 1, r (0 < r < n) Consider a decimal number p with sequentially consists of kdigit decimal representations of the numbers in the nth row of Yanghui Triangle where log10 an, r < k for all 0 ≤ r ≤ n. Here are some examples: n k p 0 1 1 3 2 01030301 5 2 010510100501 The numerical value of p can be very large. Your task is to calculate p mod m for some integer m. Input The input consists of T test cases. The number of test cases T is given in the first line of the input file. Each test case consists of one line containing n, k, m (0 ≤ n ≤ 1 000 000, 1 ≤ k ≤ 1 000 000, 2 ≤ m ≤ 109) in order. Output The output should contain p mod m, one per line. Sample Input 3 0 2 7 2 2 12345 5 2 1000000 Sample Output 1 10201 100501
字谜的一个问题，怎么用的 C 语言_course
20190810Problem Description Anna Graham is a puzzle maker who prides herself in the quality and complexity of her work. She makes puzzles of all kinds  crosswords, logic problems, acrostics, and word search puzzles, to name but a few. For each puzzle, she has developed a set of rules which she constrains herself to follow. For word search puzzles, she insists not only that all the words be connected to one another (as in most word search puzzles), but also that removing any word from the word list will not cause one or more words to become disconnected from the rest. (Two words are connected if they contain a common letter in the grid.) The example word search puzzle on the left satisfies this condition, but the one on the right does not (removing the word Pascal from the word list disconnects Java from the other two words). Your job is to write a program that checks to see if Anna’s word search problems are up to snuff. Input Input will consist of multiple test cases. The first line of each test case contains 3 integers n m l, where n and m are the number of rows and columns in the puzzle and l is the number of words. Following this are n lines containing m uppercase characters each (the puzzle) followed by l lines containing one word each (the word list, in mixed case). Each word in the word list will appear in the puzzle exactly once. There will be no more than 100 rows and 100 columns in the puzzle and no more than 100 words to search for. There will be no spaces in the input words. Output For each problem instance, output the word Yes or No depending on whether the puzzle satisfies Anna’s constraints. Sample Input 5 6 3 PBROGR PASCAL ASMMIN GIICON TCELST BASIC LISP Pascal 5 6 4 PBROJR PASCAL ASMMVN GIICAN TCELST BASIC Java LISP Pascal 0 0 0 Sample Output Yes No
Who owns the Amiga? _course
20170622Problem Description In a corridor in a student dormitory, there are ve rooms numbered 1, 2, 3, 4 and 5; room number 1 is the leftmost room. The rooms have doors in different colours: blue, green, red, white and yellow, but not necessarily in that order. In these rooms live five students Anna, Bernhard, Chris, David and Ellen of five different nationalities Danish, Finnish, Icelandic, Norwegian and Swedish. (Both the names and the nationalities are given in alphabetical order, so it does not follow automatically that Anna is Danish.) These students have one computer each, and these computers are of different kinds: Amiga, Atari, Linux, Mac and Windows (given here in alphabetical order). They each have their own favourite programming language: C, C++, Java, Pascal and Perl (also listed in alphabetical order). You want to find out who owns the Amiga computer based on some facts about the students. Input The input consists of several scenarios. The rst input line contains a number 1?000 indicating how many scenarios there are. Each scenario starts with a line with a number 1?000 telling how many fact lines there are for that scenario. Then follow the fact lines which each contains three words separated by one or more spaces: －－－The first and third word is one of these names: 1 2 3 4 5 blue green red white yellow anna bernhard chris david ellen danish finnish icelandic norwegian swedish amiga atari linux mac windows c c++ java pascal perl (Note that no uppercase letters are used.) －－－The second word specifies a relationship; it is one of sameas leftof rightof nextto sameas tells that the first and third fact words apply to the same room; for instance blue sameas bernhard tells that Bernhard lives in the room with a blue door. leftof tells that the first fact word applies to the room immediately to the left of the one to which the third fact word applies. For example, chris leftof perl means that Chris lives in the room immediately to the left of the Perl programmer. rightof tells that the rst fact word applies to the room immediately to the right of the one to which the third fact word applies. nextto tells that the two fact words apply to rooms next to each other. For example, swedish nextto linux means that the Swedish student lives in the next room (either to the left or the right) of the owner of the Linux computer. You may assume that there are no inconsistencies in the input data. In other words, there will in every scenario be at least one person who may own the Amiga without violating the constraints. Output For each scenario, you should print a line starting with scenario #n : where n is the scenario number. If you can determine who (i.e., Anna, Bernhard, Chris, David or Ellen) owns the Amiga, you continue the line with xxxx owns the amiga. or, if you cannot name the Amiga owner, you print cannot identify the amiga owner. Sample Input 2 8 red sameas 1 danish sameas 1 perl sameas 5 atari sameas 2 linux sameas 3 mac sameas 4 windows sameas 5 anna sameas 1 8 chris leftof amiga chris leftof 4 c sameas 1 danish sameas 1 red sameas 1 linux sameas red windows nextto linux mac leftof swedish Sample Output scenario #1: anna owns the amiga. scenario #2: cannot identify the amiga owner.
带有斜线方向的单词的连线的问题，怎么使用C语言的程序的编写之后的计算方式来实现查找_course
20190531Problem Description Anna Graham is a puzzle maker who prides herself in the quality and complexity of her work. She makes puzzles of all kinds  crosswords, logic problems, acrostics, and word search puzzles, to name but a few. For each puzzle, she has developed a set of rules which she constrains herself to follow. For word search puzzles, she insists not only that all the words be connected to one another (as in most word search puzzles), but also that removing any word from the word list will not cause one or more words to become disconnected from the rest. (Two words are connected if they contain a common letter in the grid.) The example word search puzzle on the left satisfies this condition, but the one on the right does not (removing the word Pascal from the word list disconnects Java from the other two words). Your job is to write a program that checks to see if Anna’s word search problems are up to snuff. Input Input will consist of multiple test cases. The first line of each test case contains 3 integers n m l, where n and m are the number of rows and columns in the puzzle and l is the number of words. Following this are n lines containing m uppercase characters each (the puzzle) followed by l lines containing one word each (the word list, in mixed case). Each word in the word list will appear in the puzzle exactly once. There will be no more than 100 rows and 100 columns in the puzzle and no more than 100 words to search for. There will be no spaces in the input words. Output For each problem instance, output the word Yes or No depending on whether the puzzle satisfies Anna’s constraints. Sample Input 5 6 3 PBROGR PASCAL ASMMIN GIICON TCELST BASIC LISP Pascal 5 6 4 PBROJR PASCAL ASMMVN GIICAN TCELST BASIC Java LISP Pascal 0 0 0 Sample Output Yes No
常数的判断，Identifying Legal Pascal Real Constants_course
20191001Description Pascal requires that real constants have either a decimal point, or an exponent (starting with the letter e or E, and officially called a scale factor), or both, in addition to the usual collection of decimal digits. If a decimal point is included it must have at least one decimal digit on each side of it. As expected, a sign (+ or ) may precede the entire number, or the exponent, or both. Exponents may not include fractional digits. Blanks may precede or follow the real constant, but they may not be embedded within it. Note that the Pascal syntax rules for real constants make no assumptions about the range of real values, and neither does this problem. Your task in this problem is to identify legal Pascal real constants. Input Each line of the input data contains a candidate which you are to classify. Output For each line of the input, display your finding as illustrated in the example shown below. The input terminates with a line that contains only an asterisk in column one. Sample Input 1.2 1. 1.0e55 e12 6.5E 1e12 +4.1234567890E99999 7.6e+12.5 99 * Sample Output 1.2 is legal. 1. is illegal. 1.0e55 is legal. e12 is illegal. 6.5E is illegal. 1e12 is legal. +4.1234567890E99999 is legal. 7.6e+12.5 is illegal. 99 is illegal.
Pascal's Travels _course
20161221Description An n x n game board is populated with integers, one nonnegative integer per square. The goal is to travel along any legitimate path from the upper left corner to the lower right corner of the board. The integer in any one square dictates how large a step away from that location must be. If the step size would advance travel off the game board, then a step in that particular direction is forbidden. All steps must be either to the right or toward the bottom. Note that a 0 is a dead end which prevents any further progress. Consider the 4 x 4 board shown in Figure 1, where the solid circle identifies the start position and the dashed circle identifies the target. Figure 2 shows the three paths from the start to the target, with the irrelevant numbers in each removed. ![](http://poj.org/images/2704_2.jpg) Figure 1 Figure 2 Input The input contains data for one to thirty boards, followed by a final line containing only the integer 1. The data for a board starts with a line containing a single positive integer n, 4 <= n <= 34, which is the number of rows in this board. This is followed by n rows of data. Each row contains n single digits, 09, with no spaces between them. Output The output consists of one line for each board, containing a single integer, which is the number of paths from the upper left corner to the lower right corner. There will be fewer than 263 paths for any board. Sample Input 4 2331 1213 1231 3110 4 3332 1213 1232 2120 5 11101 01111 11111 11101 11101 1 Sample Output 3 0 7
一个字符串匹配的搜索的方式，按照字母顺序，采用C语言的实现_course
20190120Problem Description In a corridor in a student dormitory, there are ve rooms numbered 1, 2, 3, 4 and 5; room number 1 is the leftmost room. The rooms have doors in different colours: blue, green, red, white and yellow, but not necessarily in that order. In these rooms live five students Anna, Bernhard, Chris, David and Ellen of five different nationalities Danish, Finnish, Icelandic, Norwegian and Swedish. (Both the names and the nationalities are given in alphabetical order, so it does not follow automatically that Anna is Danish.) These students have one computer each, and these computers are of different kinds: Amiga, Atari, Linux, Mac and Windows (given here in alphabetical order). They each have their own favourite programming language: C, C++, Java, Pascal and Perl (also listed in alphabetical order). You want to find out who owns the Amiga computer based on some facts about the students. Input The input consists of several scenarios. The rst input line contains a number 1?000 indicating how many scenarios there are. Each scenario starts with a line with a number 1?000 telling how many fact lines there are for that scenario. Then follow the fact lines which each contains three words separated by one or more spaces: －－－The first and third word is one of these names: 1 2 3 4 5 blue green red white yellow anna bernhard chris david ellen danish finnish icelandic norwegian swedish amiga atari linux mac windows c c++ java pascal perl (Note that no uppercase letters are used.) －－－The second word specifies a relationship; it is one of sameas leftof rightof nextto sameas tells that the first and third fact words apply to the same room; for instance blue sameas bernhard tells that Bernhard lives in the room with a blue door. leftof tells that the first fact word applies to the room immediately to the left of the one to which the third fact word applies. For example, chris leftof perl means that Chris lives in the room immediately to the left of the Perl programmer. rightof tells that the rst fact word applies to the room immediately to the right of the one to which the third fact word applies. nextto tells that the two fact words apply to rooms next to each other. For example, swedish nextto linux means that the Swedish student lives in the next room (either to the left or the right) of the owner of the Linux computer. You may assume that there are no inconsistencies in the input data. In other words, there will in every scenario be at least one person who may own the Amiga without violating the constraints. Output For each scenario, you should print a line starting with scenario #n : where n is the scenario number. If you can determine who (i.e., Anna, Bernhard, Chris, David or Ellen) owns the Amiga, you continue the line with xxxx owns the amiga. or, if you cannot name the Amiga owner, you print cannot identify the amiga owner. Sample Input 2 8 red sameas 1 danish sameas 1 perl sameas 5 atari sameas 2 linux sameas 3 mac sameas 4 windows sameas 5 anna sameas 1 8 chris leftof amiga chris leftof 4 c sameas 1 danish sameas 1 red sameas 1 linux sameas red windows nextto linux mac leftof swedish Sample Output scenario #1: anna owns the amiga. scenario #2: cannot identify the amiga owner.
Faulty Odometer _course
20170822Description You are given a car odometer which displays the miles traveled as an integer. The odometer has a defect, however: it proceeds from the digit 3 to the digit 5, always skipping over the digit 4. This defect shows up in all positions (the one's, the ten's, the hundred's, etc.). For example, if the odometer displays 15339 and the car travels one mile, odometer reading changes to 15350 (instead of 15340). Input Each line of input contains a positive integer in the range 1..999999999 which represents an odometer reading. (Leading zeros will not appear in the input.) The end of input is indicated by a line containing a single 0. You may assume that no odometer reading will contain the digit 4. Output Each line of input will produce exactly one line of output, which will contain: the odometer reading from the input, a colon, one blank space, and the actual number of miles traveled by the car. Sample Input 13 15 2003 2005 239 250 1399 1500 999999 0 Sample Output 13: 12 15: 13 2003: 1461 2005: 1462 239: 197 250: 198 1399: 1052 1500: 1053 999999: 531440
Do the Untwist _course
20170212Description Cryptography deals with methods of secret communication that transform a message (the plaintext) into a disguised form (the ciphertext) so that no one seeing the ciphertext will be able to figure out the plaintext except the intended recipient. Transforming the plaintext to the ciphertext is encryption; transforming the ciphertext to the plaintext is decryption. Twisting is a simple encryption method that requires that the sender and recipient both agree on a secret key k, which is a positive integer. The twisting method uses four arrays: plaintext and ciphertext are arrays of characters, and plaincode and ciphercode are arrays of integers. All arrays are of length n, where n is the length of the message to be encrypted. Arrays are origin zero, so the elements are numbered from 0 to n  1. For this problem all messages will contain only lowercase letters, the period, and the underscore (representing a space). The message to be encrypted is stored in plaintext. Given a key k, the encryption method works as follows. First convert the letters in plaintext to integer codes in plaincode according to the following rule: '_' = 0, 'a' = 1, 'b' = 2, ..., 'z' = 26, and '.' = 27. Next, convert each code in plaincode to an encrypted code in ciphercode according to the following formula: for all i from 0 to n  1, ciphercode[i] = (plaincode[ki mod n]  i) mod 28. (Here x mod y is the positive remainder when x is divided by y. For example, 3 mod 7 = 3, 22 mod 8 = 6, and 1 mod 28 = 27. You can use the C '%' operator or Pascal 'mod' operator to compute this as long as you add y if the result is negative.) Finally, convert the codes in ciphercode back to letters in ciphertext according to the rule listed above. The final twisted message is in ciphertext. Twisting the message cat using the key 5 yields the following: Array 0 1 2 plaintext 'c' 'a' 't' plaincode 3 1 20 ciphercode 3 19 27 ciphertext 'c' 's' '.' Your task is to write a program that can untwist messages, i.e., convert the ciphertext back to the original plaintext given the key k. For example, given the key 5 and ciphertext 'cs.', your program must output the plaintext 'cat'. Input The input contains one or more test cases, followed by a line containing only the number 0 that signals the end of the file. Each test case is on a line by itself and consists of the key k, a space, and then a twisted message containing at least one and at most 70 characters. The key k will be a positive integer not greater than 300. Output For each test case, output the untwisted message on a line by itself. Note: you can assume that untwisting a message always yields a unique result. (For those of you with some knowledge of basic number theory or abstract algebra, this will be the case provided that the greatest common divisor of the key k and length n is 1, which it will be for all test cases.) Sample Input 5 cs. 101 thqqxw.lui.qswer 3 b_ylxmhzjsys.virpbkr 0 Sample Output cat this_is_a_secret beware._dogs_barking
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Linux系统编程：入门篇视频教程
20181016Linux系统编程视频课程为《Linux系统编程》入门篇，主要针对零基础的Linux开发学员科普Linux系统编程的概念以及需要掌握的各种技能，掌握Linux命令编写、Linux学习路线并熟悉嵌入式设备编程的方法。为后续的Linux系统编程深入学习打下良好的基础。
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20150301C++语言基础视频培训课程：本课与主讲者在大学开出的程序设计课程直接对接，准确把握知识点，注重教学视频与实践体系的结合，帮助初学者有效学习。本教程详细介绍C++语言中的封装、数据隐藏、继承、多态的实现等入门知识；主要包括类的声明、对象定义、构造函数和析构函数、运算符重载、继承和派生、多态性实现等。 课程需要有C语言程序设计的基础（可以利用本人开出的《C语言与程序设计》系列课学习）。学习者能够通过实践的方式，学会利用C++语言解决问题，具备进一步学习利用C++开发应用程序的基础。
java后台+微信小程序 实现完整的点餐系统
20190727后台技术选型： JDK8 MySQL Springboot Springdatajpa Lombok Freemarker Bootstrap Websocket 小程序端技术选型 微信小程序 小程序端 扫码点餐 菜品分类显示 模拟支付 评论系统 购物车
程序员的算法通关课：知己知彼（第一季）
20191228【超实用课程内容】 程序员对于算法一直又爱又恨！特别是在求职面试时，算法类问题绝对是不可逃避的提问点！本门课程作为算法面试系列的第一季，会从“知己知彼”的角度，聊聊关于算法面试的那些事～ 【哪些人适合学习这门课程？】 求职中的开发者，对于面试算法阶段缺少经验 想了解实际工作中算法相关知识 在职程序员，算法基础薄弱，急需充电 【超人气讲师】 孙秀洋  服务器端工程师 硕士毕业于哈工大计算机科学与技术专业，ACM亚洲区赛铜奖获得者，先后在腾讯和百度从事一线技术研发，对算法和后端技术有深刻见解。 【课程如何观看？】 PC端：https://edu.csdn.net/course/detail/27272 移动端：CSDN 学院APP（注意不是CSDN APP哦） 本课程为录播课，课程无限观看时长，但是大家可以抓紧时间学习后一起讨论哦~
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HTML5+CSS3 精美登陆界面源码
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Mysql数据库基础入门视频教程
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