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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

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

1032:c Language Programming Tutorial (third Edition) after class exercise 8.6

Title Description Write a function to concatenate two strings into a string sample input after two lines of string output link123abcSample output123abc1#include 2#include string.h>3 4 voidStr_cat (CharS1[],Chars2[])5 {6 intL1 =strlen (S1);7 intL2 =strlen (S2);8 inti;9 Ten for(i = L1; i ) One { AS1[i] = s2[i-L1]; - } - theS1[i] =' /'; - } - - intMainintargcChar Const*argv[]) + { - Chars1[Bayi], s2[Bayi]; + Ascanf"%s", S1); atscanf"%s", S2); - - str_cat (S1, S2

1013:c Language Programming Tutorial (third Edition) after class exercise 6.3

The title describes the value of SN=A+AA+AAA+...+AA...AAA (with n a), where a is a number. For example: 2+22+222+2222+22222 (n=5), n is entered by the keyboard. Input a output and sample input5Sample output246901#include"stdio.h"2 3 intMainintargcChar Const*argv[])4 {5 intN, Sn =0, num =2;6scanf"%d", n);7 8 while(N >0)9 {TenSn + =num; Onenum = num *Ten+2; AN--; - } - theprintf"%d\n", Sn); - return 0; -}1013:c Language Programming

C Language Programming Tutorial (third Edition) after class exercise 6.1

The title description inputs two positive integers m and N, seeking their greatest common divisor and least common multiple. Input two integer output greatest common divisor, least common multiple sample input5 7Sample output1 351#include 2 3 intMax_common_diviser (intN1,intn2)4 {5 inttemp;6 7 if(N1 n2)8 {9temp = N1; N1 = n2; N2 =temp;Ten } One A - while(n1% N2! =0) - { the intt =N2; -N2 = n1%N2; -N1 =T; - } + -

C Language Programming Case Tutorial (2nd edition) code note (v)-Software development Basics

software development clearly and intuitively, clarify the specific tasks that need to be completed in each stage, and play a guiding and normative role in the development process. Software Development Methodology Programming style documentation; Have a clear name for the identifier; appropriate procedural notes; The procedural writing style of Liang-ha; The form of indentation; A clear statement structure; The writing of expressions shou

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