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Wuhan University of Science and Technology acm:1007: Chinese Edition C language Programming tutorial (second edition) Exercise 7.10

Problem DescriptionEnter n (nInputExample of multi-group test sample. The first row of each group has an integer n indicating that there are n strings. Next there are n rows, one string per line.OutputOutput a sequential string, with each line outputting a string.Sample Input3abaaabcabSample OutputAababacab1#include 2#include string.h>3 voidSortChar* str[],intsize)4 {5 inti,j;6 Char*tmp;7 for(i=0; i1; i++)8 {9 for(j=i+1; j)Ten { One if(strcmp (str[i]

Wuhan University of Science and Technology acm:1001: Chinese Edition C language Programming tutorial (second Edition) after class exercise 3.12

Problem DescriptionEnter n, and the output corresponds to a hollow positive hexagon with a side length of N.For easy viewing, the sample midpoint '. ' Represents a space and prints a space instead of a small dot when plotting a drawing.InputEdge length N. (NOutputSide length is a positive hexagon of nSample Input5Sample Output.....*****....*.....*...*.......*.. *.........*.*...........*.. *.........*...*.......*....*.....*.....*****1#include 2#include 3 voidPrtCharCintcount)4 { 5 while(cou

Wuhan University of Science and Technology acm:1006: Chinese Edition C language Programming tutorial (second edition) Exercise 7.15

Problem DescriptionEnter n strings (nInputMultiple sets of test data. The first row of each set of test data contains an integer n, representing a total of n strings. Next, each line contains a string that consists of printable characters.OutputEach set of test sample output one row. The output finds a string that satisfies the test instructions.Sample Input3djdlkfjsadfjwedlkfjdlkfjl;jf;sfjdsl;al/dljfd2dlkasfjmmlld;femflsad;fiwejdifSample OutputDlkfjdlkfjl;jf;sfjdsl;al/ld;femflsad;fiwejdif1#incl

Wuhan University of Science and Technology acm:1009: Chinese Edition C language Programming tutorial (second Edition) example 4.18

Problem DescriptionVerify Goldbach conjecture that any even number of sufficiently large (>=4) can be represented by a sum of two primes.InputEnter an even n. (2OutputFind A, B make n=a+bWhere A and B are two primes, and aSample Input4100Sample Output2 23 971#include 2 3#include 4 5 using namespacestd;6 7 intPrimeintm)8 9 {Ten One intI,n; A - if(m==1)return 0; - theN= (int) sqrt (Double) m); - - for(i=2; i) - + if(m%i==0)return 0; - +

Wuhan University of Science and Technology acm:1007: Chinese Edition C language Programming tutorial (second Edition) example 4.13

+ if((m%i==0) (n%i==0)) - $ { $ -printf"%d%d\n", i,m*n/i); - the Break; - Wuyi } the - } Wu - } About $ } - -}Other code:1#include 2#include 3 intgcdintAintb)4 {5 if(a%b==0)6 returnb;7 Else8 returnGCD (b,a%b);9 }Ten intLcmintAintb) One { A returna

1046:c Language Programming Tutorial (third Edition) after class exercise 10.4

The title description has n integers, so that the previous number in the order to move backward m position, the last m number into the front number of m, see figure. Write a function: To achieve the above function, in the main function input n number and output adjusted n number. Enter the number of input data n n integer moved position m output moving N number sample input101 2 3 4 5 6 7 8 9 102Sample output1#include 2 #defineN 1003 4 //Move Once5MoveintA[],intN)6 {7 intI, temp = a[n-1];8

1014 C Language Programming Tutorial (third Edition) after class exercise 6.4

The title describes the value of sn=1!+2!+3!+4!+5!+...+n!, where n is a number. Input n output and sample input5Sample output1531#include"stdio.h"2 3 intMainintargcChar Const*argv[])4 {5 6 inti, N;7 Long Long intSn =0, factor =1;8scanf"%d", n);9 Ten for(i =1; I ) One { ASn + =factor; -Factor = factor * (i +1); - } the -printf"%lld\n", Sn); - return 0; -}1014 C Language Programming Tut

1016:c Language Programming Tutorial (third Edition) after class exercise 6.6

The title describes the print out of all "daffodils", the so-called "Narcissus number" refers to a three-digit number, the number of its members of the cubic and equal to that of itself. For example: 153 is a narcissus number, because 153=1^3+5^3+3^3. Output:Input no output All narcissus number, from small start. One per line1#include 2 3 intMainintargcChar Const*argv[])4 {5 inti;6 for(i = -; I +; i++)7 {8 intGE = i%Ten;9 intBai = I/ -;Ten intShi = (I/Ten) %

1012:c Language Programming Tutorial (third Edition) after class exercise 6.2

Title Description Enter a line of characters, respectively, the number of letters, numbers, spaces and other characters. Input line character output statistic sample inputaklsjflj123 sadf918u324 asdf91u32oasdf/. '; 123Sample output23 16 2 41#include"stdio.h"2 3 intMainintargcChar Const*argv[])4 {5 Chars[Bayi];6 intI, Char_count =0, Num_count =0, Space_count =0, Other_count =0;7 //scanf ("%s", s);8 gets (s);9 Ten for(i =0; S[i]! =' /'; i++) One { A if(S[i] >='A' S[i]

1034:c Language Programming Tutorial (third Edition) after class exercise 8.8

The title description writes a function, entering a four-digit number, which requires the output of these four numeric characters, but spaces between each of the two digits. If you enter 1990, you should output "1 9 9 0". Enter a four-digit output to increase the space output sample input1990Sample output1#include 2 3 voidFuncintN)4 {5 intGE = nTen;6 intQian = n/ +;7 intBai = (N-qian * +) / -; 8 intShi = (N-qian * +-Bai * -) /Ten; 9 Tenprintf"%d %d%d%d\n", Qian, Bai, Shi, ge); On

SQL Server-T-code Basic Tutorial--t-sql query and programming background

key, and the primary key is not allowed to be null.ALTER TABLE dbo. Employees ADD constrant pk_employees PRIMARY KEY (empid);Create a PRIMARY KEY constraintUnique constraint: Enforces the uniqueness of the row, allowing the concept of alternate keys for the relational model to be implemented in its own database. Multiple unique constraints are allowed within the same table, no requirement to allow null for the field, but SQL Server rejects duplicate null tokens.ALTER TABLE dbo. Employee

Wuhan University of Science and Technology acm:1003: Chinese Edition C language Programming tutorial (second Edition) example 6.6. Adaptation

Problem DescriptionWowapps most like to learn English, English class never skip class, but English has not been learned, because the class has been sleeping. Why do you sleep, because he thinks the English word is too long. Now Wowapps has a long, long, long word and wants to compress it. Can you help him?Compression: Replaces consecutive repeating letters with one, see sample for details.InputEnter an n first to indicate the number of words Wowapps give (1Next n lines, each behavior of a string

Wuhan University of Science and Technology acm:1002: Chinese Edition C language Programming tutorial (second Edition) example 6.6

Problem DescriptionTomorrow will be the English exam, Wowapps is Midnight Oil recite the words. Wowapps found the word difficult to recite, the back one forget. After careful study, Wowapps found that the reason why the word is difficult to recite is because of the appearance of a character, destroying the beauty of the whole word, resulting in poor memory. Wowapps decided to destroy these discordant characters on behalf of the Moon!! (Applause!!) But the exam has entered the countdown, and Wowa

Wuhan University of Science and Technology acm:1004: Chinese Edition C language Programming tutorial (second Edition) after class exercise 3.7

Problem DescriptionEnter an unsigned short integer k[hex.] and P[oct.], the high byte of K as the low byte of the result, and the high byte of p as the result of the high byte to form a new integer.InputK[hex.] and P[oct.]OutputThe operation gets the new integer n.Sample Input0XD9 01117Sample Output200HINTOutput is 16 binary1#include 2 intMain ()3 {4 5 intk,p;6 while(SCANF ("%x%o", k,p)! =EOF)7 {8 intNewint = (p0xff00)| (k0xff00) >>8;9printf"%x\n", newint);Ten }

Java Programming Tutorial-1th Chapter Introduction

intended to allow Java commands to find the class in the Java library that is needed when executing Java programs5. What kinds of Java programs can be divided into? What's the difference?Java programs can be divided into Java applications and Java applets6. Check the Javaza online Help documentation to list the differences between SYSTEM.OUT.PRINTLN and System.out.print.PRINTLN is wrapped, print is not wrapped7. Try to write a Java program and output the following information in the console win

Wuhan University of Science and Technology acm:1009: Chinese Edition C language Programming tutorial (second edition) Exercise 5.12

Problem Descriptionthis day the teacher as a routine to small Hao out a topic: The teacher to small Hao a string, let small Hao will the string in reverse order output. InputThe first line includes a T, which indicates how many sets of test data are:Next T line, each line includes a string (length not exceeding 65535).OutputFor each test data output it's reverse string.Sample Input1oaHoaiX EvoL ISample OutputI Love Xiaohao1#include 2#include string.h>3 intMain ()4 {5 intL,i,n;6 Chara[655

1028:c Language Programming Tutorial (third Edition) after class exercise 8.1 using functions to find the number of conventions

The title description writes two functions, each for two integers greatest common divisor and least common multiple, calls these two functions with the main function, and outputs the result two integers by the keyboard input. Input two-digit output greatest common divisor least common multiple sample input6 15Sample output3 30Hint Sourceint main(){int a,m,n,cun,bei;scanf("%d %d",m,n);a=m*n;if(mwhile(n!=0){cun=m%n;m=n;n=cun;}bei=a/m;printf("%d %d",m,bei);return 0;}1028:c Language

Wuhan University of Science and Technology acm:1008: Chinese Edition C language Programming tutorial (second edition) Exercise 6.14

Problem DescriptionEnter an octal string, convert it to an equivalent decimal string, and output it in the PRINGF%s format.InputFirst, a positive integer t is entered, indicating that there is a T set of test data (1Next T line, one string per line, representing an octal integer (this integer is no more than 20 bits).OutputFor each test data, output the corresponding decimal string. One line per string.Sample Input11732Sample Output986HINTNote: The given octal number may be large, with a long lo

Wuhan University of Science and Technology Acm:1005:c Language Programming Tutorial (third Edition) after class exercise 6.6

Problem DescriptionTo print out all the "daffodils", the so-called "Narcissus number" refers to a three-digit number, whose numbers are cubic and equal to that of itself. For example: 153 is a narcissus number, because 153=1^3+5^3+3^3. Output:153?????????InputNoOutputThe number of daffodils, the beginning of childhood. One per lineSample InputSample Output1#include 2 3#include 4 5 6 7 intMain ()8 9 {Ten One inti; A - for(i= -;i +; i++) - the if(I==pow (i

Linux Video Tutorial Command Learning introductory development system training operation and maintenance network programming

Video materials are checked one by one, clear high quality, and contains a variety of documents, software installation packages and source code! Perpetual FREE Updates!Technical teams are permanently free to answer technical questions: Hadoop, Redis, Memcached, MongoDB, Spark, Storm, cloud computing, R language, machine learning, Nginx, Linux, MySQL, Java EE,. NET, PHP, Save your time!Get video materials and technical support addresses-------------------------------------------------------------

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