C programming:- Arrays,Strings,Structures,Unions - Part 3

C program to calculate determinant of a 3X3 matrix

#include<stdio.h>
int main(){
  int a[3][3],i,j;
  int determinant=0;

  printf("Enter the 9 elements of matrix: ");
  for(i=0;i<3;i++)
      for(j=0;j<3;j++)
           scanf("%d",&a[i][j]);

  printf("\nThe First matrix is\n");
  for(i=0;i<3;i++){
      printf("\n");
      for(j=0;j<3;j++)
           printf("%d\t",a[i][j]);
  }

  for(i=0;i<3;i++)
      determinant = determinant + (a[0][i]*(a[1][(i+1)%3]*a[2][(i+2)%3] - a[1][(i+2)%3]*a[2][(i+1)%3]));

  printf("\nDeterminant of matrix is: %d",determinant);

   return 0;
}



C code for Determinant of 2X2 matrix:

#include<stdio.h>
int main(){
  int a[2][2],i,j;
  long determinant;

  printf("Enter the 4 elements of matrix: ");
  for(i=0;i<2;i++)
      for(j=0;j<2;j++)
           scanf("%d",&a[i][j]);

  printf("\nThe matrix is\n");
  for(i=0;i<2;i++){
      printf("\n");
      for(j=0;j<2;j++)
           printf("%d\t",a[i][j]);
  }

  determinant = a[0][0]*a[1][1] - a[1][0]*a[0][1];

  printf("\nDeterminant of 2X2 matrix: %ld",determinant);
 
   return 0;
}
/*
          WAP to read a nXm matrix and find:
a.       The average of each row.
b.      Average of each column
c.       Average of all mn entries.
*/
#include<stdio.h>
main()
{
          int a[10][10];
          int n,m,i,j,k,sum,sumr,sumc;
          printf("enter order of a matrix:");
          scanf("%d%d",&m,&n);
          printf("enter elements:\n");
          sum=0;
          for(i=0;i<m;i++)
          {
                   for(j=0;j<n;j++)
                   {
                  
                             scanf("%d",&a[i][j]);     
                             sum=sum+a[i][j];
                   }
          }
          printf("total sum of al mn entries:%d avg=%d\n",sum,sum/(m*n));
          for(i=0;i<m;i++)
          {sumr=0;
                   for(j=0;j<n;j++)
                   {
                             sumr=sumr+a[i][j];
         
                            
                   }
                   printf("row %d sum=%d row avg= %d\n",i+1,sumr,sumr/n);
          }
          for(i=0;i<m;i++)
          {sumc=0;
                   for(j=0;j<n;j++)
                   {
                             sumc=sumc+a[j][i];
         
                            
                   }
                   printf("col %d sum=%d col avg=%d\n",i+1,sumc,sumc/m);
          }
}
/*OUTPUT
enter order of a matrix:3 3
enter elements:
1 2 3
4 5 6
7 8 9
total sum of al mn entries:45 avg=5
row 1 sum=6 row avg= 2
row 2 sum=15 row avg= 5
row 3 sum=24 row avg= 8
col 1 sum=12 col avg=4
col 2 sum=15 col avg=5
col 3 sum=18 col avg=6

*/
/* sort letters in a given string */

#include<stdio.h>
#include<string.h>
main()
{
          int i,j,n;
          char ch[30],tmp;
          printf("Enter string : ");
          gets(ch);
          n=strlen(ch);
         
          for(i=0;i<n-1;i++)
          {
                    for(j=0;j<n-1;j++)
                   {
                             if( ch[j] < ch[j+1])
                             {
                              tmp = ch[j];
                              ch[j] = ch[j+1];
                             ch[j+1] = tmp;
                             }
                   }
          }

          printf("\n\nAfter Sorting in descending order\n");
 
      printf("\n%s",ch);

     
}
/*OUTPUT
Enter string : hello


After Sorting in descending order

ollhe
*/

Define a structure called STUDENT that will describe the following information:Std_name,class,reg_no,sub_marks,and total. Using STUDENT, declare an array stu_list with 60 elements. WAP to read the information about all the 60 students and to display the information.
#include<stdio.h>
struct student
{
          char stdname[20];
          char class[10];
          int regno;
          int submarks[6];
          int total;
}stu_list[60];
main()
{
          int i,j,n;
          printf("enter limit of students:");
          scanf("%d",&n);
          printf("enter students details\n");
          for(i=0;i<n;i++)
          {
                   printf("enter name ,  class  ,regno ");
                   scanf("%s%s%d",stu_list[i].stdname,stu_list[i].class,&stu_list[i].regno);
                   printf("enter 6 subjects marks:");
                   stu_list[i].total=0;
                   for(j=0;j<6;j++)
                   {
                             scanf("%d",&stu_list[i].submarks[j]);
                             stu_list[i].total=stu_list[i].total+stu_list[i].submarks[j];
                   }
          }
          printf("\nstudent Record\n");
          for(i=0;i<n;i++)
          {
                   printf("student %d details \n",i+1);
                  
                   printf("name=%s\tclass= %s\tregno=%d\n",stu_list[i].stdname,stu_list[i].class,stu_list[i].regno);
                   printf("6 subjects marks:");
                   for(j=0;j<6;j++)
                   {
                             printf("\nsubject %d= %d\n  ",j+1,stu_list[i].submarks[j]);
                   }
                   printf("total marks:%d",stu_list[i].total);
          }
}

/*
enter limit of students:1
enter students details
enter name ,  class  ,regno
aa s3 111
enter 6 subjects marks:
23 24 25 26 27 28

student Record
student 1 details
name=aa     class= s3     regno=111
6 subjects marks:
subject 1= 23
 
subject 2= 24
 
subject 3= 25
 
subject 4= 26
 
subject 5= 27
 
subject 6= 28
  total marks:153student@ubuntu:~$
*/
/*
7.       Define a structure that describes the set of books in a library. For each book, the members are name of author,publisher,rate and branch of information. WAP to print i) a list of books supplied by a publisher ii) a list of books in a particular branch
*/
#include<stdio.h>
#include<stdlib.h>
struct library
{
          char aname[20];//author name
          char pub[10];//publisher name
          char bname[30];//book name
          char bookbranch[20];
          int rate;
         
}book[30];
main()
{
          int i,c,n,l=0;
          char pname[20],bnam[20];
          printf("enter no of books:");
          scanf("%d",&n);
          printf("enter books details\n");
          for(i=0;i<n;i++)
          {
                   printf("enter author name,publisher,bookname,bookbranch,rate\n");
                   scanf("%s%s%s%s%d",book[i].aname,book[i].pub,book[i].bname,book[i].bookbranch,&book[i].rate);
                  
          }
          printf("\nBook Record\n");
          for(i=0;i<n;i++)
          {
                   printf("book %d details \n",i+1);
                   printf("author\tpublisher\tbookname\tbookbranch\trate\n");
                  
                   printf("%s\t%s\t%s\t%s\t%d\n",book[i].aname,book[i].pub,book[i].bname,book[i].bookbranch,book[i].rate);
                  
          }
         
         
          do
          {
          printf("MENU\n 1) search by publisher \n 2) search by branch\n 3)exit\n enter choice:");
          scanf("%d",&c);
          switch(c)
          {
                   case 1: //a list of books supplied by a publisher
                             printf("enter publisher name:");
                             scanf("%s",pname);
                             for(i=0;i<n;i++)
                             {
                                               
                                      if(strcmp(book[i].pub,pname)==0)
                                      {
                                                printf("author\tpublisher\tbookname\tbookbranch\trate\n");                              
                                                printf("%s\t%s\t%s\t%s\t%d\n",book[i].aname,book[i].pub,book[i].bname,book[i].bookbranch,book[i].rate);
                                      }
                             }
                             break;
                   case 2: // a list of books in a particular branch
                             printf("enter branch name:");
                             scanf("%s",bnam);
                             for(i=0;i<n;i++)
                             {
                                     
                                      if(strcmp(book[i].bookbranch,bnam)==0)
                                      {
                                                printf("author\tpublisher\tbookname\tbookbranch\trate\n");                                        
                                                printf("%s\t%s\t%s\t%s\t%d\n",book[i].aname,book[i].pub,book[i].bname,book[i].bookbranch,book[i].rate);
                                      }
                             }
                             break;
                   case 3: exit(0);
                   default:printf("invalid choice");
          }
          printf("press 1 to continue:");
          scanf("%d",&l);
          }while(l==1);
         
}

/*enter no of books:3
enter books details
enter author name,publisher,bookname,bookbranch,rate
^C
student@ubuntu:~$ ./a.out
enter no of books:3
enter books details
enter author name,publisher,bookname,bookbranch,rate
balaguruswamy    pearson       pgmmininc  computer    200
enter author name,publisher,bookname,bookbranch,rate
byron          pear   pscp  computer    300
enter author name,publisher,bookname,bookbranch,rate
yashwanth  pearson       cp      computer    340

Book Record
book 1 details
author         publisher    bookname   bookbranch rate
balaguruswamy    pearson       pgmmininc  computer    200
book 2 details
author         publisher    bookname   bookbranch rate
byron          pear   pscp  computer    300
book 3 details
author         publisher    bookname   bookbranch rate
yashwanth  pearson       cp      computer    340
MENU
 1) search by publisher
 2) search by branch
 3)exit
 enter choice:1
enter publisher name:pearson
author         publisher    bookname   bookbranch rate
balaguruswamy    pearson       pgmmininc  computer    200
author         publisher    bookname   bookbranch rate
yashwanth  pearson       cp      computer    340
press 1 to continue:1
MENU
 1) search by publisher
 2) search by branch
 3)exit
 enter choice:2
enter branch name:computer
author         publisher    bookname   bookbranch rate
balaguruswamy    pearson       pgmmininc  computer    200
author         publisher    bookname   bookbranch rate
byron          pear   pscp  computer    300
author         publisher    bookname   bookbranch rate
yashwanth  pearson       cp      computer    340

*/
Given two one-dimensional arrays A & B. Read them and sort in ascending order. Then merge them in to a single sorted array C that contains every item from arrays A and B in ascending order.
#include<stdio.h>
void sort(int p[30],int q);
void disp(int p[30],int q);
void merge(int p[30],int r[30],int s1,int s2);
int a[30],b[30],c[60];

main()
{
         
          int i,j,n1,n2,m;
          printf("enter array 1 size:");
          scanf("%d",&n1);
          printf("enter array 2 size:");
          scanf("%d",&n2);
          printf("enter array 1 elements:");
          for(i=0;i<n1;i++)
          scanf("%d",&a[i]);
          printf("enter array 2 elements:");
          for(i=0;i<n2;i++)
          scanf("%d",&b[i]);
          sort(a,n1);
          printf("array 1 elements after sorting\n");
          disp(a,n1);
          sort(b,n2);
          printf("\narray 2 elements after sorting\n");
          disp(b,n2);
          merge(a,b,n1,n2);
          printf("\narray 3 elements after merging\n");
          disp(c,n1+n2);
          sort(c,n1+n2);
          printf("\narray 3 elements after sorting\n");
          disp(c,n1+n2);
}
void sort(int p[30],int q)
{
          int i,j,t;
          for(i=0;i<q-1;i++)
          {
                   for(j=0;j<q-1;j++)
                   {
                             if(p[j]>p[j+1])
                             {
                                      t=p[j];
                                      p[j]=p[j+1];
                                      p[j+1]=t;
                             }
                   }
          }
}
void disp(int p[30],int q)
{
          int i;
          for(i=0;i<q;i++)
          printf("%d ",p[i]);
}
void merge(int p[30],int r[30],int s1,int s2)
{
          int i,j,s;
          s=s1+s2;
          j=0;
          for(i=0;i<s;i++)
          {
                   if(i<s1)
                   c[i]=p[i];
                   else
                   {c[i]=r[j];j++;}
                  
          }
}

/* OUTPUT
student@ubuntu:~$ ./a.out
enter array 1 size:3
enter array 2 size:3
enter array 1 elements:1 55 33
enter array 2 elements:3 77 44
array 1 elements after sorting
1 33 55
array 2 elements after sorting
3 44 77
array 3 elements after merging
1 33 55 3 44 77
array 3 elements after sorting
1 3 33 44 55 77
*/
Read the employee number,name and salary of 500 employees in a firm. Prepare the payroll in the ascending order of salaries and print it. .(MG University NOV 2010)
/*Read the employee number,name and salary of 500 employees in a firm. Prepare the payroll in the ascending order of salaries and print it. .(NOV 2010)*/

#include<stdio.h>
#include<string.h>
struct employee
{
          int empno;
          char name[30];
          int sal;
}emp[500],t;
main()
{
          int n,i,j;
          printf("enter emp limit:");
          scanf("%d",&n);
          printf("enter %d employees empno,emp name and salary:",n);
          for(i=0;i<n;i++)
          {
                   scanf("%d%s%d",&emp[i].empno,emp[i].name,&emp[i].sal);
          }

          printf("employee database based on salary order:\n");
          for(i=0;i<n-1;i++)
          {
                   for(j=0;j<n-1;j++)
                   {
                             if(emp[j].sal>emp[j+1].sal)
                             {
                                      t=emp[j];
                                      emp[j]=emp[j+1];
                                       emp[j+1]=t;
                             }
                   }
          }
         
          printf(" empno\tempname\tempsal\n");
          for(i=0;i<n;i++)
          {
                   printf("%d\t%s\t%d\n",emp[i].empno,emp[i].name,emp[i].sal);
          }
          printf("employee database based on name order:\n");
          for(i=0;i<n-1;i++)
          {
                   for(j=0;j<n-1;j++)
                   {
                             if(strcmp(emp[j].name,emp[j+1].name)>0)
                             {
                                      t=emp[j];
                                      emp[j]=emp[j+1];
                                      emp[j+1]=t;
                             }
                   }
          }
         
          printf(" empno\tempname\tempsal\n");
          for(i=0;i<n;i++)
          {
                   printf("%d\t%s\t%d\n",emp[i].empno,emp[i].name,emp[i].sal);
          }
}
/*
enter emp limit:5
enter 5 employees empno,emp name and salary:
111 arun 4500
123 anu  3000
231 athira 200
34  arjun 2300
45  basil 23000
employee database based on salary order:
 empno       empname    empsal
231    athira 200
34      arjun 2300
123    anu    3000
111    arun  4500
45      basil  23000
employee database based on name order:
 empno       empname    empsal
123    anu    3000
34      arjun 2300
111    arun  4500
231    athira 200
45      basil  23000

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