c programming language 4th edition pdf

The title describes the sum of the diagonal elements of a 3x3 matrix. Input matrix output main diagonal sub-diagonal element and sample input1 2 31 1 13 2 1Sample output3 71#include 2 3 intMainintargcChar Const*argv[])4 {5 inta[3][3], I, J;6 intDiagonal1 =0, Diagonal2 =0;7 8 //input9 for(i =0; I 3; i++)Ten { One for(j =0; J 3; J + +) A { -scanf"%d", a[i][j]); - } the } - -Diagonal1 = a[0][0] + a[1][1] +a[2][2]; -Diagonal2 = a[2][0] + a[1][1] +a[

The title description writes a function that enables a given two-dimensional array (3x3) to transpose, that is, the row and column interchange. Enter a 3x3 matrix output to transpose matrices sample input1 2 34 5 67 8 9Sample output1#include 2 3 4 voidConvertinta[][3],intRowintCol//two-dimensional array when making formal parameters, the second must be written5 {6 intI, J, temp;7 for(i =0; i )8 {9 for(j = i; J )Ten { One //Exchange A if(I! =j

Understand:method is called with ().To be more abstract, the method is a variable name, and "()" is a simple representation of the calling method (also for the compiler to recognize), and the calling method is the result of the operation.Eg:a = 1; A is a variable name, and "=" is a simple expression of the assignment (but we have been in touch with the language for too long and the subconscious has assumed that = is the assignment.) But in fact, "=" i

1026:c Language Programming Tutorial (third Edition) After class exercise 7.4 The topic describes an array of 9 elements that have already been sorted, and entering a number now requires that it be inserted into the array in the same order as it was originally ordered. Enter the first line, the original sequence. The second line, the number that needs to be inser

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 s

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

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

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

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

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; - } + -

later found that it is not at all, and that the error Dup_val_on_index, his book is also wrong.Here is an anonymous blockDate.sqlDECLARE l_date date;BEGIN l_date: = SYSDATE-5; Dbms_output. Put_Line (l_date); end;/I still don't know how to call that process, although I tried it before, but I didn't remember it.I'm going to try to write a call on my own, I write a process for printing time, and then call it in an anonymous block.Get_date.sqlCREATE OR REPLACE PROCEDURE get_date (num_in in number)