To solve multiplication of numbers in few seconds

Example 1
Calculate 98 * 93
Solution
Select a closest base ( a power of 10). In this case we can select 100 as a base. Write these numbers with the difference from the base. That is 98-100 = -2 and 93-100 is -7. So write it as
98   :   -2
93   :   -7
Left side of the answer will be the diagonal sum including their signs. That is 98-7=91. So 91 will be the left side of the product. (You can take the other diagonal sum also, that is 93-2=91. Always these diagonal sums will be same).
To find the right hand side, just multiply the differences including their signs. That is -2 * -7 = 14
So we got that left side of the product is 91 and right side is 14. So answer is 9114

---

Example 2
Calculate 96 * 112
Solution
Use the same method. Let's select 100 as the base (Closed to both of the numbers).Write these numbers with the difference from the base. That is 96-100 = -4 and 112-100 is 12. So write it as
96    :   -4
112  :   12
Take the diagonal sum to get the left side of the product. That is 96+12 = 108 (Or 112-4 = 108).
For the right side, find the product of the differences. That is -4*12 = -48. Since it is -ve, we need to make it as +ve. For this borrow 1 from our left hand side 108. This borrowed 1 becomes 100 and our right hand side becomes 100+(-48) = 52. This is our right hand side
Since we borrowed one from left hand side, left hand side is 107
So answer is 10752

---

Example 3
Calculate 103 * 115
Solution
Use the same method. Let's select 100 as the base (Closed to both of the numbers).Write these numbers with the difference from the base. That is 103-100 = 3 and 115-100 is 15. So write it as
103  :   3
115  :   15
Take the diagonal sum to get the left side of the product. That is 103+15 = 118 (Or 115+3= 118).
For the right side , find the product of the differences. That is 3*15 = 45.
So answer is 11845

---

Example 4
Calculate 122 * 89
Solution
Use the same method. Let's select 100 as the base (Closed to both of the numbers).Write these numbers with the difference from the base. That is 122-100 = 22 and 89-100 is -11. So write it as
122  :  22
89    :   -11
Take the diagonal sum to get the left side of the product. That is 122+(-11) = 111 (Or 89+22= 111).
For the right side , find the product of the differences. That is 22*(-11) = -242. Since it is -ve, we need to make it as +ve. For this borrow 3 from our left hand side 111. This borrowed 3 becomes 300 and our right hand side becomes 300+(-242) = 58. This is our right hand side
Since we borrowed 3 from left hand side, left hand side is 108
So answer is 10858

---

Example 5
Calculate 1024 * 989
Use the same method. Let's select 1000 as the base (Closed to both of the numbers).Write these numbers with the difference from the base. That is 1024-1000 = 24 and 989-1000 is -11. So write it as
1024  :  24
989    :  -11
Take the diagonal sum to get the left side of the product. That is 1024+(-11) = 1013 (Or 989+24= 1013).
For the right side, find the product of the differences. That is 24*(-11) = -264. Since it is -ve, we need to make it as +ve. For this borrow 1 from our left hand side 1013. This borrowed 1 becomes 1000 and our right hand side becomes 1000+(-264) = 736.
Since we borrowed 1 from left hand side, left hand side is 1012 now.
So our answer is 1012736

---

Example 6
Calculate 997 * 986
Use the same method. Let's select 1000 as the base (Closed to both of the numbers).Write these numbers with the difference from the base. That is 997-1000 = -3 and 986-1000 is -14. So write it as
997  : -3
986  :  -14
Take the diagonal sum to get the left side of the product. That is 997+(-14) = 983 (Or 986-3= 983).
For the right side, find the product of the differences. That is (-3)*(-14) = 42. But here 42 has only 2 digits whereas our base 1000 has three zeros. So write 42 as 042.
So our answer is 98304

---

---

Quotes and thoughts for peace of mind and to think positive.

C program for parenthesized expression using stack

/*PARENTHESIZED EXPRESSION USING STACK */

#include<stdio.h>
#include<conio.h>
#define MAX 10
int top=-1;
int stack[MAX];
void push(char);
char pop();
void main()
{
   char exp[MAX],ch;
   int i,flag=1;
   clrscr();
   printf("\n Enter Infix Expression:-\n");
   gets(exp);
   i=0;
   while(exp[i]!='\0')
   {
     if(exp[i]=='('||exp[i]=='{'||exp[i]=='[')
     {
push(exp[i]);
     }
     if(exp[i]==')'||exp[i]=='}'||exp[i]==']')
     {
       ch=pop();
       if(exp[i]==')'&&(ch=='{'||ch=='['))
       {
flag=0;
       }
       if(exp[i]=='}'&&(ch=='('||ch=='['))
       {
flag=0;
       }
       if(exp[i]==']'&&(ch=='('||ch=='{'))
       {
flag=0;
       }
     }
     i++;
   }
   if(top>=0)
   {
     flag=0;
   }
   if(flag==1)
   {
     printf("\n Valid Expression");
   }
   else
   {
     printf("\n Invalid Expression");
   }
   getch();
}
void push(char c)
{
  if(top==MAX-1)
  {
   printf("\n stack is full");
  }
  else
  {
    top++;
    stack[top]=c;
  }
}
char pop()
{
  char c;
  if(top==-1)
  {
   printf("\n Stack is empty");
  }
  else
  {
   c=stack[top];
   top--;
   return c;
  }
}

C program in stack to concatenate two strings using push and pop functions

#include<stdio.h>
#include<conio.h>
#include<string.h>
#define max 30
void push(char ch);
char pop();
char stack[max];
int top=-1;
void main()
{
char a[20],b[20],k[40],ch;
int i,l1,l2;
clrscr();
printf("\n Enter First String:-\n");
scanf("%s",a);
printf("\n Enter Second String:-\n");
scanf("%s",b);
l1=strlen(a);
l2=strlen(b);
for(i=0;i<l2;i++)
{
 push(b[i]); //push second string into stack
}
push(' ');
for(i=0;i<l1;i++)
{
 push(a[i]); //push first string into stack
}
printf("\n After concatenation of two string:-\n");
i=0;
while(top!=-1)
{
 ch=pop();
 if(ch==' ')
{
k[i]='\0';
printf("%s",strrev(k));
i=0;
 }
else
 {
  k[i]=ch;
  i++;
 }
}
k[i]='\0';
printf("%s",strrev(k));
getch();
}
void push(char ch)
{
if(top==max-1)
{
 printf("\n Stack is full");
}
else
{
 top++;
 stack[top]=ch;
}
}

char pop()
{
char t;
if(top==-1)
{
 printf("\n Stack is Empty");
}
else
{
 t=stack[top];
 top--;
 return(t);
}
}

C program for binary search

#include<stdio.h>
#include<conio.h>
void main()
{
 int a[20],n,i,k,mid,s,e,f=0;
 clrscr();
 printf("\n Enter how many number u want:-");
 scanf("%d",&n);
 printf("\n Please enter %d sorted numbers:-",n);
 for(i=0;i<n;i++)
 {
  scanf("%d",&a[i]);
 }
 printf("\n numbers are:-");
 for(i=0;i<n;i++)
 {
  printf("\t%d",a[i]);
 }
 printf("\n\n Enter search key number:-\n");
 scanf("%d",&k);
 s=0;
 e=n-1;
 while(s<=e)
 {
  mid=(s+e)/2;
  if(k==a[mid])
  {
   printf("\n %d number is found on %d position.",k,mid+1);
   f=1;
   break;
  }
  else
  {
   if(k>a[mid])
   {
    s=mid+1;
   }
   else
   {
    e=mid-1;
   }
  }
 }
 if(f==0)
 {
  printf("\n number is not found");
 }
 getch();
}



/* OUTPUT
 Enter how many number u want:-5                                              
                                                                               
 Please enter 5 sorted numbers:-10 11 12 13 14                                
                                                                               
 numbers are:-  10      11      12      13      14                            
                                                                               
 Enter search key number:-                                                    
 12                                                                            
                                                                               
 12 number is found on 3 position.   */

C program for linear searching

#include<stdio.h>
#include<conio.h>
void main()
{
 int a[10],n,i,k,f=0;
 clrscr();
 printf("\n Enter how many number u want:-");
 scanf("%d",&n);
 printf("\n Enter %d numbers:-",n);
 for(i=0;i<n;i++)
 {
  scanf(" %d",&a[i]);
 }
 printf("\n numbers are:-\n");
 for(i=0;i<n;i++)
 {
  printf("\t%d",a[i]);
 }
 printf("\n Enter search key number:-");
 scanf("%d",&k);
 for(i=0;i<n;i++)
 {
  if(k==a[i])
  {
   printf("\n %d number is found on %d position.",k,i+1);
   f=1;
   break;
  }
 }
 if(f==0)
 {
  printf("\n Number is not found");
 }
 getch();
}
/*
 Enter how many number u want:-5                                              
                                                                               
 Enter 5 numbers:-4 6 8 9 14                                                  
                                                                               
 numbers are:-                                                                
        4       6       8       9       14                                    
 Enter search key number:-9                                                    
                                                                               
 9 number is found on 4 position.      */

C program in data structure to make circular linked list

#include<stdio.h>
#include<conio.h>
#include<process.h>
void create();
void display();
void addbeg();
void addlast();
struct node
{
 int info;
 struct node *link;
}*start=NULL;
void main()
{
 int n;
 clrscr();
 while(1)
 {
  printf("\n **menu**\n");
  printf("\n 1. to create list \n 2. to display list");
  printf("\n 3. to add node at beginning of linked list");
  printf("\n 4. to add node at the last of linked list");
  printf("\n 5. exit");
  printf("\n Enter your choice:-");
  scanf("%d",&n);
  switch(n)
  {
   case 1 : create();
   break;
   case 2 : display();
   break;
   case 3 : addbeg();
   break;
   case 4 : addlast();
   break;
   case 5 : exit(0);

  }
 }
 getch();
}

void create()
{
 struct node *temp,*q;
 int x;
 temp=(struct node *)malloc(sizeof(struct node));
 if(temp==NULL)
 {
  printf("\n Insufficient memory");
 }
 else
 {
  printf("\n Enter number:-");
  scanf("%d",&x);
  temp->info=x;
  temp->link=NULL;
  if(start==NULL)
  {
   start=temp;
  }
  else
  {
   q=start;
   while(q->link!=start)
   {
    q=q->link;
   }
   q->link=temp;
  }
  temp->link=start;
 }
}

void display()
{
  struct node *q;
  printf("\n Circular linked list is:-\n\n");
  q=start;
  while(q->link!=start)
  {
   printf("   %d",q->info);
   q=q->link;
  }
  printf("   %d",q->info);
}

void addbeg()
{
 int x;
 struct node *temp,*q;
 temp=(struct node*)malloc(sizeof(struct node));
 if(temp==NULL)
 {
  printf("\n Insufficient memory");
 }
 else
 {
  printf("\n Enter number:-");
  scanf("%d",&x);
  temp->info=x;
  temp->link=NULL;

  q=start;
  while(q->link!=start)
  {
   q=q->link;
  }
  temp->link=start;
  start=temp;
  q->link=start;
 }
}
void addlast()
{
 struct node *temp,*q;
 int x;
 temp=(struct node*)malloc(sizeof(struct node));
 if(temp==NULL)
 {
  printf("\n Insufficient memory");
 }
 else
 {
  printf("\n Enter number:-");
  scanf("%d",&x);
  temp->info=x;
  temp->link=NULL;

  q=start;
  while(q->link!=start)
  {
   q=q->link;
  }
  q->link=temp;
  temp->link=start;
 }
}

C program in data structure to find the intersection between two linked list

#include<stdio.h>
#include<conio.h>
#include<process.h>
#include<stdlib.h>
struct node
{
 int info;
 struct node *link;
}*start1=NULL,*start2=NULL;
void main()
{
 struct node *temp,*q,*p;
 int n,i,x,f=0;
 clrscr();
 printf("\n First linked list");
 printf("\n Enter How many number:-");
 scanf("%d",&n);
 for(i=0;i<n;i++)
 {
  temp=(struct node*)malloc(sizeof(struct node));
  if(temp==NULL)
  {
   printf("\n Insufficient memory");
   getch();
   exit(0);
  }
  printf("\n Enter number:-");
  scanf("%d",&x);
  temp->info=x;
  temp->link=NULL;
  if(start1==NULL)
  {
   start1=temp;
  }
  else
  {
   q=start1;
   while(q->link!=NULL)
   {
     q=q->link;
   }
   q->link=temp;
  }
 }
 printf("\n first list\n");
 q=start1;
 while(q!=NULL)
 {
  printf("\t%d",q->info);
  q=q->link;
 }
 start2=NULL;
 printf("\n Second linked list");
 printf("\n Enter How many number:-");
 scanf("%d",&n);
 for(i=0;i<n;i++)
 {
  temp=(struct node*)malloc(sizeof(struct node));
  if(temp==NULL)
  {
   printf("\n Insufficient memory");
   getch();
   exit(0);
  }
  printf("\n Enter number:-");
  scanf("%d",&x);
  temp->info=x;
  temp->link=NULL;
  if(start2==NULL)
  {
   start2=temp;
  }
  else
  {
   q=start2;
   while(q->link!=NULL)
   {
     q=q->link;
   }
   q->link=temp;
  }
 }
 printf("\n second list\n");
 q=start2;
 while(q!=NULL)
 {
  printf("\t%d",q->info);
  q=q->link;
 }
 printf("\n intersection of two list is:-\n");
 q=start1;
 while(q!=NULL)
 {
  p=start2;
  while(p!=NULL)
  {
   if(q->info==p->info)
   {
    printf("\t%d",q->info);
   }
   p=p->link;
  }
  q=q->link;
 }
 getch();
}

/*   OUTPUT
 First linked list                                                            
 Enter How many number:-5                                                      
                                                                               
 Enter number:-1                                                              
                                                                               
 Enter number:-2                                                              
                                                                               
 Enter number:-3                                                              
                                                                               
 Enter number:-4                                                              
                                                                               
 Enter number:-5                                                              
                                                                               
 first list                                                                    
        1       2       3       4       5                                      
 Second linked list                                                            
 Enter How many number:-4

 Enter number:-1

 Enter number:-2

 Enter number:-5

 Enter number:-6

 second list
1       2       5       6
 intersection of two list is:-
1       2       5            
*/

C program in data structure to reverse the singly linked list

/* Reverse the singly linked list */

#include<stdio.h>
#include<conio.h>
#include<process.h>
struct node
{
 int info;
 struct node *link;
}*start=NULL;
void main()
{
 struct node *temp,*q,*p,*t;
 int n,i,x;
 clrscr();
 printf("\n Enter How many number:-");
 scanf("%d",&n);
 for(i=0;i<n;i++)
 {
  temp=(struct node*)malloc(sizeof(struct node));
  if(temp==NULL)
  {
   printf("\n Insufficient memory");
   getch();
   exit(0);
  }
  printf("\n Enter number:-");
  scanf("%d",&x);
  temp->info=x;
  temp->link=NULL;
  if(start==NULL)
  {
   start=temp;
  }
  else
  {
   q=start;
   while(q->link!=NULL)
   {
     q=q->link;
   }
   q->link=temp;
  }
 }
 t=NULL;
 for(i=0;i<n;i++)
 {
  q=start;
  while(q->link!=t)
  {
   q=q->link;
  }
  t=q;
  printf("\n %d",q->info);
 }
 getch();
}


/* OUTPUT
 Enter How many number:-5                                                      
                                                                               
 Enter number:-1                                                              
                                                                               
 Enter number:-2                                                              

 Enter number:-3                                                              
                                                                               
 Enter number:-4                                                              
                                                                               
 Enter number:-5                                                              
                                                                               
 5                                                                            
 4                                                                            
 3                                                                            
 2
 1              

C program in data structure for concatenate two linked list

#include<stdio.h>
#include<conio.h>
#include<process.h>
#include<string.h>
struct node
{
 int info;
 struct node *link;
}*start1=NULL,*start2;
void main()
{
 struct node *temp,*q;
 int n,i,x;
 clrscr();
 printf("\n First linked list");
 printf("\n Enter How many number:-");
 scanf("%d",&n);
 for(i=0;i<n;i++)
 {
  temp=(struct node*)malloc(sizeof(struct node));
  if(temp==NULL)
  {
   printf("\n Insufficient memory");
   getch();
   exit(0);
  }
  printf("\n Enter number:-");
  scanf("%d",&x);
  temp->info=x;
  temp->link=NULL;
  if(start1==NULL)
  {
   start1=temp;
  }
  else
  {
   q=start1;
   while(q->link!=NULL)
   {
     q=q->link;
   }
   q->link=temp;
  }
 }
 printf("\n first list\n");
 q=start1;
 while(q!=NULL)
 {
  printf("\t%d",q->info);
  q=q->link;
 }
 start2=NULL;
 printf("\n Second linked list");
 printf("\n Enter How many number:-");
 scanf("%d",&n);
 for(i=0;i<n;i++)
 {
  temp=(struct node*)malloc(sizeof(struct node));
  if(temp==NULL)
  {
   printf("\n Insufficient memory");
   getch();
   exit(0);
  }
  printf("\n Enter number:-");
  scanf("%d",&x);
  temp->info=x;
  temp->link=NULL;
  if(start2==NULL)
  {
   start2=temp;
  }
  else
  {
   q=start2;
   while(q->link!=NULL)
   {
     q=q->link;
   }
   q->link=temp;
  }
 }
 printf("\n second list\n");
 q=start2;
 while(q!=NULL)
 {
  printf("\t%d",q->info);
  q=q->link;
 }

 q=start1;
 while(q->link!=NULL)
 {
      //concatenate two linked list
  q=q->link;
 }
 q->link=start2;
 printf("\n After concatenate linked list\n");
 q=start1;
 while(q!=NULL)
 {
  printf("\t%d",q->info);
  q=q->link;
 }
 getch();
}

RDBMS example for project with different quiries

SQL> create table project
  2  (pno number(5) primary key,
  3  pname varchar2(20),
  4  status varchar2(5)
  5  check (status in('c','p','i')));

Table created.
SQL> insert into project values (&pno,'&pname','&status');
SQL> select * from project;

       PNO PNAME                STATU
---------- -------------------- -----
        11 voice_recognition    i
        12 speech_recognition   c
        13 video_recognition    i
        14 steganography        p
        15 cryptography         i

SQL> create table dept
  2  (deptno number(5) primary key,
  3  deptname varchar2(20),
  4  HOD varchar2(20),
  5  location varchar2(20));

Table created.
SQL> insert into dept values (&deptno,'&deptname','&HOD','&location');
SQL> select * from dept;

    DEPTNO DEPTNAME             HOD                  LOCATION
---------- -------------------- -------------------- --------------------
         1 mathematics          vijay                pune
         2 computer             ranjit               pune
         3 chemical             rahul                nasik
         4 civil                ajay                 nagpur
         5 IT                   santosh              nanded

SQL> create table dep_pro
  2  (deptno number(5) references dept,
  3  pno number(5) references project,
  4  primary key (deptno,pno));

Table created.
SQL> insert into dep_pro values (&deptno,&pno);
SQL> select * from dep_pro;

    DEPTNO        PNO
---------- ----------
         1         11
         2         12
         3         13
         4         14
         5         15





Queries:-

1 Find HOD of computer Department located in ‘Pune’.
SQL> select HOD from dept
  2  where deptname='computer' and location='pune';

HOD
--------------------
Ranjit

2 List all projects of mathematics department which are Incomplete.
  SQL>select project.*
from dept,project
where dept.deptno=project.deptno and statu='i'
and deptname='mathmatics';
 
  PNO      PNAME                     STATU
---------- --------------------             -----
  11        voice_recognition             i



3 Display the Project details of Computer Department.
SQL>select project.*
from dept,project
where dept.deptno=project.deptno and deptname='Computer';
       PNO PNAME              
---------- --------------------
        12 speech_recognition  

4 List department wise project along with status.

SQL> select deptname,pname,status from project,dept,dep_pro
  2  where project.pno=dep_pro.pno and
  3  dep_pro.deptno=dept.deptno
  4  group by deptname,pname,status;

DEPTNAME             PNAME                STATU
-------------------- --------------------         -----
IT                   cryptography                    i
chemical             video_recognition       i
civil                steganography                 p
computer             speech_recognition    c
mathematics          voice_recognition    c

C program in data structure in queue for character DMA Queue

C program in data structure in queue. For character DMA Queue

#include<stdio.h>
#include<conio.h>
#include<process.h>
void insert();
void deletes();
void display();
struct node
{
 char info;
 struct node *link;
}*front=NULL,*rear=NULL;
void main()
{
 int n;
 clrscr();
 while(1)
 {
  printf("\n **menu**\n");
  printf("\n 1. insert \n 2. deletes");
  printf("\n 3. display \n 4. exit");
  printf("\n Enter your choice:- ");
  scanf("%d",&n);
  switch(n)
  {
   case 1 : insert();
   break;
   case 2 : deletes();
   break;
   case 3 : display();
   break;
   case 4 : exit(0);
  }
 }
 getch();
}

void insert()
{
 struct node *temp,*q;
 char x;
 temp=(struct node *)malloc(sizeof(struct node));
 if(temp==NULL)
 {
  printf("\n Insufficient memory");
 }
 else
 {
  printf("\n Enter character:-");
  scanf("%c",&x);
  temp->info=x;
  temp->link=NULL;
  if(rear==NULL&&front==NULL)
  {
   rear=temp;
   front=temp;
  }
  else
  {
   q=rear;
   while(q->link!=NULL)
   {
     q=q->link;
   }
   q->link=temp;
  }
 }
}

void deletes()
{
 struct node *q;
 if(front==NULL&&rear==NULL)
 {
  printf("\n Queue is empty");
 }
 else
 {
   front=rear;
   rear=front->link;
   printf("\n deleted node is %c",front->info);
   free(front);
 }
}

void display()
{
 struct node *q;
 if(front==NULL&&rear==NULL)
 {
  printf("\n Queue is empty");
 }
 else
 {
  front=rear;
  while(front!=NULL)
  {
   printf("\n %c",front->info);
   front=front->link;
  }
 }
}

C program in data structure for polynomial subtraction

C program in data structure for polynomial subtraction 

#include<stdio.h>
#include<conio.h>
void main()
{
 int a[10],b[10],c[10];
 int n,k,i,p,coeff;
 clrscr();
 for(i=0;i<10;i++)
 {
  a[i]=0;
  b[i]=0;
  c[i]=0;
 }
 printf("\n first polynomial");
 printf("\n enter number of terms=");
 scanf("%d",&n);
 for(i=0;i<n;i++)
 {
  printf("\n enter power & coefficient=");
  scanf("%d%d",&p,&coeff);
  a[p]=coeff;
 }
 printf("\n second polynomial");
 printf("\n enter number of terms=");
 scanf("%d",&k);
 for(i=0;i<k;i++)
 {
  printf("\n enter power & coefficient=");
  scanf("%d%d",&p,&coeff);
  b[p]=coeff;
 }
 for(i=0;i<10;i++)
 {
  c[i]=a[i]-b[i];
 }
 printf("\n polynomial subtraction is=");
 for(i=9;i>=0;i--)
 {
  if(c[i]!=0)
  {
   if(i!=0)
   {
    printf("%dx^%d+",c[i],i);
   }
   else
   {
    printf("%dx^%d",c[i],i);
   }
  }
 }
 getch();
}

C program for multiplication of two polynomial in data structure

#include<stdio.h>
#include<conio.h>
void main()
{
 int a[10],b[10],c[10];
 int i,j,n,k,power,coeff;
 clrscr();
 for(i=0;i<10;i++)
 {
  a[i]=0;
  b[i]=0;
  c[i]=0;
 }
 printf("\n first polynomel");
 printf("\n enter number of terms=");
 scanf("%d",&n);
 for(i=0;i<n;i++)
 {
  printf("\n enter coefficient=");
  scanf("%d",&coeff);
  printf("\n enter power");
  scanf("%d",&power);
  a[power]=coeff;
 }
 printf("\n second polynomel");
 printf("\n enter number of terms=");
 scanf("%d",&k);
 for(i=0;i<k;i++)
 {
  printf("\n enter coefficient=");
  scanf("%d",&coeff);
  printf("\n enter power");
  scanf("%d",&power);
  b[power]=coeff;
 }
 for(i=9;i>=0;i--)
 {
  if(a[i]!=0)
  {
   for(j=9;j>=0;j--)
   {
    if(b[j]!=0)
    {
     c[i+j]=c[i+j]+a[i]*b[j];
    }
   }
  }
 }
 printf("\n our results is=");
 for(i=9;i>=0;i--)
 {
  if(c[i]!=0)
  {
   if(i!=0)
   {
    printf("%dx^%d+",c[i],i);
   }
   else
   {
    printf("%dx^%d",c[i],i);
   }
  }
 }
 getch();
}

C program in data structure for addition of two polynomial

#include<stdio.h>
#include<conio.h>
void show(int p[]);
void main()
{
 int a[10],b[10],c[10];
 int n,k,i,p,coeff;
 clrscr();
 for(i=0;i<10;i++)
 {
  a[i]=0;
  b[i]=0;
  c[i]=0;
 }
 printf("\n First polynomial");
 printf("\n Enter number of terms=");
 scanf("%d",&n);
 for(i=0;i<n;i++)
 {
  printf("\n Enter power & coefficient=");
  scanf("%d%d",&p,&coeff);
  a[p]=coeff;
 }
 printf("\n Second polynomial");
 printf("\n Enter number of terms=");
 scanf("%d",&k);
 for(i=0;i<k;i++)
 {
  printf("\n Enter power & coefficient=");
  scanf("%d%d",&p,&coeff);
  b[p]=coeff;
 }
 printf("\n First polynomial is:-\n\n");
 show(a);
 printf("\n Second polynomial is:-\n\n");
 show(b);
 for(i=0;i<10;i++)
 {
  c[i]=a[i]+b[i];
 }
 printf("\n Addition of two polynomials is=\n\n");
 show(c);
 getch();
}
void show(int p[])
{
  int i;
  for(i=9;i>0;i--)
  {
   if(p[i]!=0)
   {
    printf("%dx^%d+",p[i],i);
   }
  }
  printf("%d",p[i]);
}

C program in data structure for evaluation of polynomial

#include<stdio.h>
#include<conio.h>
#include<math.h>
int eval(int b[],int n,int x);
void main()
{
 int a[10],i,e,n,x;
 clrscr();
 for(i=0;i>10;i++)
 {
  a[i]=0;
 }
 printf("\n Enter number of term:-\n");
 scanf("%d",&n);
 printf("\n Enter co-efficient:-\n");
 for(i=n;i>=0;i--)
 {
  printf("\n Enter co-efficient for A[%d]:-",i);
  scanf("%d",&a[i]);
 }
 printf("\n Polynomial Expression is:-\n");
 for(i=n;i>0;i--)
 {
  if(a[i]!=0)
  {
   printf("%dx^%d+",a[i],i);
  }
 }
 printf("%d",a[i]);
 printf("\n Enter value for x:-\n");
 scanf("%d",&x);
 e=eval(a,n,x);
 printf("\n Evaluation of Polynomial is:-\t%d",e);
 getch();
}
int eval(int b[],int n,int x)
{
 int i,s=0;
 for(i=n;i>=0;i--)
 {
  s=s+b[i]*pow(x,i);
 }
 return(s);
}

C program in data structure of polynomial for implementation of malloc and realloc function

 IMPLEMENTATION OF MALLOC AND REALLOC FUNCTION

#include<stdio.h>
#include<conio.h>
void main()
{
  int s,i,newsize,*p;
  clrscr();
  printf("\n Enter size of an array:-");
  scanf("%d",&s);
  p=(int *)malloc(s*sizeof(int));
  printf("\n Enter the number:-\n");
  for(i=0;i<s;i++)
  {
    scanf("%d",&p[i]);
  }
  printf("\n Entered numbers are:- \n");
  for(i=0;i<s;i++)
  {
    printf(" %d",p[i]);
  }
  printf("\nEnter newsize of an array:-");
  scanf("%d",&newsize);
  p=(int *)realloc(p,newsize);
  printf("\n Enter the numbers for newsize:-\n");
  for(i=s;i<newsize;i++)
  {
    scanf("%d",&p[i]);
  }
  printf("\n All elements of an array is:-\n");
  for(i=0;i<newsize;i++)
  {
    printf(" %d",p[i]);
  }
  getch();
 }

C program in data structure for storage representation of 2-D array

#include<stdio.h>
#include<conio.h>
void main()
{
 int a[10][10],i,j,r,c;
 clrscr();
 printf("\n Enter row and column number:-");
 scanf("%d%d",&r,&c);
 printf("\n Enter %d*%d matrix:-",r,c);
 for(i=0;i<r;i++)
 {
  for(j=0;j<c;j++)
  {
   scanf("%d",&a[i][j]);
  }
 }
 printf("\n storage representation of 2-D array:-");
 for(i=0;i<r;i++)
 {
  for(j=0;j<c;j++)
  {
   printf("\n[%d][%d]= %d",i,j,a[i][j]);
  }
  printf("\n");
 }
 getch();
}



/* OUTPUT
 Enter row and column number:-3 3                                              
                                                                               
 Enter 3*3 matrix:-1 2 3                                                      
                   4 5 6                                                      
                   7 8 9                                                      
                                                                               
 storage representation of 2-D array:-                                        
[0][0]= 1                                                                      
[0][1]= 2                                                                      
[0][2]= 3                                                                      
                                                                               
[1][0]= 4                                                                      
[1][1]= 5                                                                      
[1][2]= 6                                                                      
                                                                               
[2][0]= 7                                                                      
[2][1]= 8                                                                      
[2][2]= 9                                                                      
     */

C program for preorder inorder postorder in data structure

#include<stdio.h>
#include<conio.h>
#include<process.h>
#include<stdlib.h>
struct node
{
 int info;
 struct node *llink;
 struct node *rlink;
}*root=NULL;
void insert();
void preorder(struct node *q);
void inorder(struct node *q);
void postorder(struct node *q);
int stack[10];
int top=-1;
void main()
{
 int n;
 clrscr();
 while(1)
 {
  printf("\n **menu**");
  printf("\n 1.insert \n 2.preorder");
  printf("\n 3.inorder \n 4.postorder");
  printf("\n 5.exit");
  printf("\n enter your choice:-");
  scanf("%d",&n);
  switch(n)
  {
   case 1 : insert();
   break;
   case 2 : preorder(root);
   break;
   case 3 : inorder(root);
   break;
   case 4 : postorder(root);
   break;
   case 5 : exit(0);
  }
 }
 getch();
}
void insert()
{
 struct node *temp,*q;
 int x,i,n;
 printf("\n\n enter how many nodes you want:-");
 scanf(" %d",&n);
 for(i=0;i<n;i++)
 {

  temp=(struct node*)malloc(sizeof(struct node));
  if(temp==NULL)
  {
    printf("\n insufficient memory");
  }
  else
  {
   printf("\n\n enter number:-");
   scanf("%d",&x);
   temp->info=x;
   temp->llink=NULL;
   temp->rlink=NULL;
   if(root==NULL)
   {
    root=temp;
   }
   else
   {
    q=root;
    while(1)
    {
     if(temp->info==q->info)
     {
       printf("\n\n element already exist");
       break;
     }
     if(temp->info>q->info)
     {
      if(q->rlink==NULL)
      {
       q->rlink=temp;
       break;
      }
      q=q->rlink;
     }
     if(temp->info<q->info)
     {
      if(q->llink==NULL)
      {
       q->llink=temp;
       break;
      }
      q=q->llink;
     }
    }
   }
  }
 }
}
void preorder(struct node *q)
{
 top++;
 stack[top]=q;
 while(top!=-1)
 {
  q=stack[top];
  top--;
  if(q!=NULL)
  {
   printf("  %d->",q->info);
   top++;
   stack[top]=q->rlink;
   top++;
   stack[top]=q->llink;
  }
 }
}
void inorder(struct node *q)
{

  while(top!=-1||q!=NULL)
  {
   if(q!=NULL)
   {
    top++;
    stack[top]=q;
    q=q->llink;
   }
   else
   {
    q=stack[top];
    top--;
    printf(" %d->",q->info);
    q=q->rlink;
   }
  }
}


void postorder(struct node *q)
{
 int f[5];
 int top_p;
 stack[++top]=NULL;
 do
 {
  while(q!=NULL)
  {
   stack[++top]=q;
   f[top]=1;
   if(q->rlink!=NULL)
   {
    stack[++top]=q->rlink;
    f[top]=-1;
   }
   q=q->llink;
  }
  top_p=top;
  q=stack[top--];
  while(f[top_p]==1)
  {
   printf(" %d->",q->info);
   top_p=top;
   q=stack[top--];
  }
 }
  while(q!=NULL);

}

C program in data structure for tree

#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
struct node
{
 int info;
 struct node *llink,*rlink;
}*root=NULL;
void create();
void node(struct node *q);
void degree(struct node *q);
void leaf(struct node *q);
void interior(struct node *q);
void child(struct node *q);
int no,l,in,ch,p;
void main()
{
 int n;
 clrscr();
 while(1)
 {
  printf("\n **menu**");
  printf("\n 1.create a tree \n 2.count num of nodes");
  printf("\n 3.degree of tree \n 4.leaf nodes \n 5.interior nodes");
  printf("\n 6.childrens and parent \n 7.exit");
  printf("\n enter your choice:-\t");
  scanf("%d",&n);
  switch(n)
  {
   case 1 : create();
   break;
   case 2 : node(root);
   printf("\n\n total nodes are:-  %d",no);
   break;
   case 3 : degree(root);
   break;
   case 4 : leaf(root);
   printf("\n\n leaf nodes is:-  %d",l);
   break;
   case 5 : interior(root);
   printf("\n\n interior nodes are:-  %d",in);
   break;
   case 6 : child(root);
   printf("\n\n childrean is:- %d and parents is:-  %d",ch,p);
   break;
   case 7 : exit(0);
  }
 }
}
void create()
{
 struct node *temp,*q;
 int x,n,i;
 printf("\nEnter how many nodes you want:-");
 scanf("  %d",&n);
 for(i=0;i<n;i++)
 {
  temp=(struct node *)malloc(sizeof(struct node));
  if(temp==NULL)
  {
   printf("\n insufficient memory");
  }
  else
  {
   printf("\n enter number:- ");
   scanf(" %d",&x);
   temp->info=x;
   temp->llink=NULL;
   temp->rlink=NULL;
   if(root==NULL)
   {
    root=temp;
   }
   else
   {
    q=root;
    while(1)
    {
     if(temp->info==q->info)
     {
      printf("\n number is aleardy exist");
      break;
     }
     if(temp->info>q->info)
     {
      if(q->rlink==NULL)
      {
       q->rlink=temp;
       break;
      }
      q=q->rlink;
     }
     if(temp->info<q->info)
     {
      if(q->llink==NULL)
      {
       q->llink=temp;
       break;
      }
      q=q->llink;
     }
    }
   }
  }
 }
}
void node(struct node *q)
{
 if(q!=NULL)
 {
  node(q->llink);
  node(q->rlink);
  no=no+1;
 }
}
void degree(struct node *q)
{
 if(q!=NULL)
 {
  if(q->llink!=NULL&&q->rlink!=NULL)
  {
   printf("\n\n degree of %d node is:= 2",q->info);
  }
  if(q->llink!=NULL&&q->rlink==NULL)
  {
   printf("\n\n degree of %d node is:= 1",q->info);
  }
  if(q->llink==NULL&&q->rlink!=NULL)
  {
   printf("\n\n degree of %d node is:= 1",q->info);
  }
  if(q->llink==NULL&&q->rlink==NULL)
  {
   printf("\n\n degree of %d node is:= 0",q->info);
  }
  degree(q->llink);
  degree(q->rlink);
 }
}
void leaf(struct node *q)
{
 if(q!=NULL)
 {
  leaf(q->llink);
  leaf(q->rlink);
  if(q->llink==NULL||q->rlink==NULL)
  {
   l=l+1;
  }
 }
}
void interior(struct node *q)
{
 if(q!=NULL)
 {
  interior(q->llink);
  interior(q->rlink);
  if(q->llink!=NULL||q->rlink!=NULL)
  {
   in=in+1;
  }
 }
}
void child(struct node *q)
{
 if(q!=NULL)
 {
  child(q->llink);
  child(q->rlink);
  if(q->llink!=NULL||q->rlink!=NULL)
  {
   p=p+1;
  }
  if(q->llink==NULL||q->rlink==NULL)
  {
   ch=ch+1;
  }
 }
}

C program in data structure for mirror of a tree

#include<stdio.h>
#include<conio.h>
#include<process.h>
#include<stdlib.h>
struct node
{
 int info;
 struct node *llink;
 struct node *rlink;
}*root=NULL;
void insert();
void preorder(struct node *q);
void mirror(struct node *q);
void main()
{
 int n;
 clrscr();
 while(1)
 {
  printf("\n **menu**");
  printf("\n 1.insert \n 2.preorder");
  printf("\n 3.mirror \n 4.exit");
  printf("\n enter your choice:-");
  scanf("%d",&n);
  switch(n)
  {
   case 1 : insert();
     break;
   case 2 : printf("\n\n our tree in preorder \n\n");
     preorder(root);
     break;
   case 3 :
     mirror(root);
     printf("\n\n mirror image is:-\n\n");
     preorder(root);
     break;
   case 4 : exit(0);
  }
 }
 getch();
}




void insert()
{
 struct node *temp,*q;
 int x,i,n;
 printf("\n\n enter how many node you want:=");
 scanf("%d",&n);
 for(i=0;i<n;i++)
 {

  temp=(struct node*)malloc(sizeof(struct node));
  if(temp==NULL)
  {
   printf("\n insufficient memory");
  }
  else
  {
   printf("\n\n enter number:-");
   scanf("%d",&x);
   temp->info=x;
   temp->llink=NULL;
   temp->rlink=NULL;
   if(root==NULL)
   {
    root=temp;
   }
   else
   {
    q=root;
    while(1)
    {
     if(temp->info==q->info)
     {
      printf("\n\n element already exist");
      break;
     }
     if(temp->info>q->info)
     {
      if(q->rlink==NULL)
      {
       q->rlink=temp;
       break;
      }
      q=q->rlink;
     }
     if(temp->info<q->info)
     {
      if(q->llink==NULL)
      {
       q->llink=temp;
       break;
      }
      q=q->llink;
     }//end of if
    }//end of while
   }//end of else2
  }//end of else1
 }//end of for loop
}
void preorder(struct node *q)
{
 if(q!=NULL)
 {
  printf(" %d->",q->info);
  preorder(q->llink);
  preorder(q->rlink);
 }
}
void mirror(struct node *q)
{
 struct node *temp1;
 if(q!=NULL)
 {
  mirror(q->llink);
  mirror(q->rlink);
  temp1=q->llink;
  q->llink=q->rlink;
  q->rlink=temp1;
 }
}

C program in stack data structure for infix and postfix expression

#define max 50
#include<stdio.h>
#include<conio.h>
#include<process.h>
void push(char ch);
char pop();
char stack[max];
int top=-1;
int prec(char chr);
void main()
{
 char a[60],b[60],ch;
 int i,j,p,r,x,y,re;
 clrscr();
 printf("\n Enter Infix Expression:-\n");
 scanf("%s",&a);
 i=0;
 j=0;
 for(i=0;a[i]!='\0';i++)
 {
  switch(a[i])
  {
   case '^':
   case '$':
   case '*':
   case '/':
   case '+':
   case '-': p=prec(a[i]);
    while(top!=-1&&p<=prec(stack[top]))
    {
      b[j]=pop();
      j++;
    }
    push(a[i]);  //push operator into stack
    break;
   case '(': push(a[i]);
    break;

   case ')':
    do
    {
     ch=pop();
     if(ch!='(')
     {
      b[j]=ch;
      j++;
     }
    }while(ch!='(');
     break;
   default : b[j]=a[i];  //output exp
    j++;
  }
 }
 while(top!=-1)
 {
  b[j]=pop();  //last pop the all operators.
  j++;
 }
 b[j]='\0';
 printf("\n Postfix Expression Is:-\n %s",b);
 printf("\n Evaluation of the Postfix expression:-\n");
 i=0;
 while(b[i]!='\0')
 {
  if(b[i]>='0'&&b[i]<='9')
  {
   push(b[i]-48); //convert char into nuumber
  }
  else
  {
   y=pop(); //pop the two poerands and prefrom operation
   x=pop();
   switch(b[i])
   {
      case '+' : r=x+y;
break;
      case '-' : r=x-y;
break;
      case '*' : r=x*y;
break;
      case '/' : r=x/y;
break;
      case '$' :
      case '^' : r=pow(x,y);
break;
   }
   push(r); //push the result into stack
  }
  i++;
 }
 re=pop(); // at the last pop the result and display.
 printf("\n %d",re);
 getch();
}

int prec(char chr)
{
 switch(chr)   // check the priority of the operators.
 {
  case '$' :
  case '^' : return(5);
  case '*' :
  case '/' : return(4);
  case '+' :
  case '-' : return(3);
  case '(' : return(2);
 }
 return(0);
}
void push(char x)
{
  top++;
  stack[top]=x;
}
char pop()
{
 char y;
 y=stack[top];
 top--;
 return(y);
}

RDBMS example for one to many relationship

Q.).
         Customer (cno,cname,city)
         Account (ano,acc_type,balance)

          Relationships between  customer and account  is one-to-many.
 
         Constraints :-primary key,
                                 Balance should be>100

SQL> create table customer2
  2  ( cno number(5)primary key,
  3    cname varchar2(30),
  4    city varchar2(40)
  5  );
Table created.

SQL> insert into customer2
  2  values('&cno','&cname','&city');
Enter value for cno: 101
Enter value for cname: mahendra
Enter value for city: pali
old   2: values('&cno','&cname','&city')
new   2: values('101','mahendra','pali')
1 row created.
SQL> select * from customer2;

  CNO   CNAME   CITY
--------   -----------  ------------------
  101 mahendra       pali
  102 babulal          pune
 103 amit              mumbai

 CNO   CNAME   CITY
--------   -----------  ------------------
 104        raju                 pune
105          akash          Ajmer

SQL> create table account
  2  ( ano number(5)primary key,
  3    acc_type varchar2(20),
  4    balance number(5) check(balance>100),
  5    cno number(5)references customer2(cno)
  6  );
Table created.

SQL> insert into account
2 values('&ano','&acc_type','&balance','&cno');
SQL> select * from account;

       ANO ACC_TYPE                          BALANCE        CNO
---------- ------------------------------ ---------- ----------
      1201 saving                              13000        101
      1202 current                             20000        102
      1203 saving                              14000        103
      1204 saving                              50000        104
      1205 current                             14000        105
      1206 saving                              25000        104
      1207 current                             19000        105


B). wirte a cursor to add interest of 3% to the balance of all account whose balance is greater than 10000.

  1  declare
  2     cursor c1 is select ano from account
  3     where balance>10000;
  4  begin
  5     for x in c1 loop
  6       update account
  7       set balance=balance+balance*3/100
  8       where ano=x.ano;
  9       commit;
 10     end loop;
 11* end;
SQL> /
PL/SQL procedure successfully completed.

SQL> select * from account;

       ANO ACC_TYPE                          BALANCE        CNO
---------- ------------------------------ ---------- ----------
      1201 saving                              13390        101
      1202 current                             20600        102
      1203 saving                              14420        103
      1204 saving                              51500        104
      1205 current                             14420        105
      1206 saving                              25750        104
      1207 current                             19570        105

7 rows selected.


a). create  or replace a PL/ SQL procedure to find total balance of all customers of pune city.
  1  create or replace procedure dis(x in varchar)
  2  is
  3      b number(20);
  4  begin
  5      select sum(a.balance) into b from customer2 c,
  6      account a where c.cno=a.cno and
  7      c.city=x;
  8        dbms_output.put_line('total balance of all customers of pune city:-'||b);
  9* end;
 10  /

Procedure created.

SQL> declare
  2
  3  begin
  4    dis('pune');
  5  end;
  6  /
total balance of all customers of pune city:-97850

PL/SQL procedure successfully completed.

RDBMS example of many to many relationship

Q.).    book ( bno, bname, pubname, price)
 
         Author ( ano, aname)
 
         Relationships between book and author is many to many.

         Constraints:-primary key,
                                Aname and pubname should NOT NULL.


SQL> select * from book_1;

       BNO BNAME                PUBNAME                             PRICE
---------- -------------------- ------------------------------ ----------
       101 C++                           nirali                                150
       102 M A/C                      vision                                250
       103 RDBMS                    BPB                                   175
       104 data structre              BPB                                   165
       105 software engineering   nirali                                135

SQL> select * from author;

       ANO ANAME
---------- --------------------
      1201  mr.dewasi
      1202  mr.shiravi b
      1203  mr.pankaj
      1204  kanetkar
      1205  babulal

SQL> select * from bookauth;

    BNO        ANO
---------- ----------
       101       1201
       102       1202
       103       1203
       104       1204
       102       1204
       105       1205
       101       1203

7 rows selected.


************************************************************************
a). create or replace a PL/SQL procedure to display details of all books written by ‘kanetkar’.

SQL>
  1  create or replace procedure disa(t in varchar2)
  2  is cursor ca is select b.bno,b.bname,b.pubname,b.price from book_1 b,
  3     author a, bookauth ba where b.bno=ba.bno and a.ano=ba.ano and
  4     a.aname=t;
  5  begin
  6     for x in ca loop
  7        dbms_output.put_line(x.bno||' '||x.bname||' '||x.pubname||' '||x.price);
  8     end loop;
  9* end;
SQL> /

Procedure created.

SQL>
  1  declare
  2  begin
  3     disa('kanetkar');
  4* end;
  5  /
104 data structre  BPB 165
102 M A/C vision 250

PL/SQL procedure successfully completed.


************************************************************************

b). create or replace a trigger that restricts insertion or updation of books having price less than 0.

SQL>
  1  create or replace trigger ta1
  2  before insert or update
  3  on book_1
  4  for each row
  5  declare
  6    p book_1.price%type;
  7  begin
  8    p:=:new.price;
  9    if(p<0) then
 10      raise_application_error(-20089,'book price must be>0');
 11    end if;
 12* end;
SQL> /

Trigger created.

SQL> insert into book_1
  2  values(106,'SE','nirali',-45);
insert into book_1
            *
ERROR at line 1:
ORA-20089: book price must be>0
ORA-06512: at "SCOTT.TA1", line 6
ORA-04088: error during execution of trigger 'SCOTT.TA1'


SQL> update book_1
  2  set price=-45
  3  where bno=101;
update book_1
       *
ERROR at line 1:
ORA-20089: book price must be>0
ORA-06512: at "SCOTT.TA1", line 6
ORA-04088: error during execution of trigger 'SCOTT.TA1'

************************************************************************

RDBMS example of book and dept many to one relationship

Q.) Book( bno, bname, pudname, price)

        Department( dno, dname)

        Relationships between book and department is many to one.
   
        Constraints:-primary key,  Price should bo>0.

SQL> select * from department_1;
    DNO DNAME
---------- ------------------------------
       101 computer
       102 science
       103 arts
       104 MBA
       105 MSC
SQL> select * from books_1;
 BNO     BNAME   PUBNAME    PRICE  DNO
-------   -------------  ---------------   --------- --------
201      C++              nirali               150      101
1202    numerical     BPB                180      102
             method  
1203     RDBMS     vision               140      101


  BNO     BNAME   PUBNAME    PRICE  DNO
-------   -------------  ---------------   --------- --------

1204    data structure       nirali          145       104

1205     social sic         vision            120      105
1206       POM              BPB             130      103

6 rows selected.
************************ ** ** ******************************************

a).create or replace a PL/SQL function to return total expenditure on books of a given department.

SQL>
  1  create or replace function disb(x in varchar2)return number
  2  is
  3    c number(5);
  4  begin
  5    select sum(b.price) into c from department_1 d,books_1 b
  6    where d.dno=b.dno and d.dname=x;
  7    return(c);
  8* end;
SQL> /
Function created.

SQL>
  1  declare
  2     dn department_1.dname%type;
  3     t number(10);
  4  begin
  5     dn:='&dn';
  6     t:=disb(dn);
  7     dbms_output.put_line('total expenditure on books of a'||' '||dn||' '||'is:-'||t);
  8* end;
SQL> /
Enter value for dn: computer
old   5:    dn:='&dn';
new   5:    dn:='computer';
total expenditure on books of a computer is:-290

PL/SQL procedure successfully completed.


************************************************************************
************************************************************************

b). write a cursor to display details of all books brought for a ‘computer’ department.

SQL>
  1  declare
  2    cursor c8 is select b.bno,b.bname,b.pubname,b.price from department_1 d,
  3    books_1 b where d.dno=b.dno and d.dname='computer';
  4  begin
  5    for x in c8
  6    loop
  7      dbms_output.put_line(x.bno||' '||x.bname||' '||x.pubname||' '||x.price);
  8    end loop;
  9* end;
SQL> /
1201 C++ nirali 150
1203 RDBMS vision 140

PL/SQL procedure successfully completed.


************************************************************************

RDBMS examples

Q.3). movie (mvno,mvname,releaseyear)

         Actor (actno,actname)

         Relationships between movie and actor is many to many with descriptive attribute rate of actor for movie.
   
         Constraints:- primary key.


SQL> select * from movie;

 MVNO    MVNAME         RELEASEYEAR
----------  --------------------  -----------
         1     dhoom                       2002
         2     DDLJ                        1999
         3    MBBS                        2000
         4    om kare                     2002
         5    singham                     2011

SQL> select * from actor;

 ACTNO  ACTNAME
----------   ------------------------------
       101    hrithik
       102    amitabh
       103    abhishek
       104    ajay
       105    salaman

SQL> select * from movact;

MVNO  ACTNO    RATE
---------- ---------- ----------
         1        101     500000
         2        102     800000
         3        103     900000
         4        105    1200000
         5        104    1500000
         2        104     700000
         1        101    1200000
7 rows selected.

SQL> set serveroutput on;

a).create or replace a PL/SQL procedure to display details of all movies of actor ‘amitabh’.

SQL>
  1  create or replace procedure showma(t in varchar2)
  2  is cursor c5 is select m.mvno,m.mvname,m.releaseyear
  3     from movie m,actor a,movact ma
  4     where m.mvno=ma.mvno and a.actno=ma.actno and
  5     a.actname=t;
  6  begin
  7     for x in c5 loop
  8       dbms_output.put_line(x.mvno||' '||x.mvname||' '||x.releaseyear);
  9     end loop;
 10* end;
SQL> /

Procedure created.

SQL> declare
  2
  3  begin
  4     showma('amitabh');
  5  end;
  6  /
2 DDLJ 1999

PL/SQL procedure successfully completed.


b).write a cursor to display names of all movies which are released in year 2002.
SQL>
  1  declare
  2    cursor c7 is select mvname from movie
  3    where releaseyear=2002;
  4  begin
  5    for x in c7 loop
  6      dbms_output.put_line(x.mvname);
  7    end loop;
  8* end;
SQL> /
dhoom
om kare

PL/SQL procedure successfully completed.