Easy C/C++ Codes

How to use Stack in C/C++ programs

2014-12-16T17:38:00.000+05:30

This is an example program showing the syntax for stack declaration and manipulation on that stack using push and pop operations defined in the <stack> library. Other auxiliary functions of stack library like size() and top() are also used in the program.

Push operation is used to insert an element in a stack

Pop operation is used to remove an element from a stack

Property of Stack:

The element pushed in to the stack will always reside at the top of the stack.

The element popped up from stack will always be the top element from the stack.

Depiction of Stack push and pop Functions.

Try to run the below code in your C compiler and understand the syntax and usage of the Stack Library

#include<iostream>

#include<stack>

using namespace std;

int main()

{

stack <string> cards;

\\ Declaration of Stack which

\\can contain elements of "String" Type and

\\name of the stack assigned is "cards"

cards.push("kings of hearts"); \\ push an element

cards.push("kings of club");

cards.push("kings of spade");

cards.push("kings of diamond");

\\ size() function gives the

\\ number of elements in the stack

cout<<"there are"<<cards.size()<<"cards in the deck";

cards.push("ace of Spade");

cards.push("ace of hearts");

\\ top() function of stack library returns

\\ the element residing at the top position of the stack

string topcard=cards.top();

cout<<"\ntop card is\t"<<topcard;

\\pop() function of stack library will
\\simply delete the top element from the top
\\and the next element below it will be updated
\\ as new top of the stack

cards.pop();

cout<<"\ntop card is\t"<<cards.top();

}

Call by value Call by reference C++ Programming

2014-01-12T00:32:00.001+05:30

Struct in C

2011-09-14T22:47:00.003+05:30

A struct in C programming language is a structured (record) type that aggregates a fixed set of labelled objects, possibly of different types, into a single object.
A struct declaration consists of a list of fields, each of which can have any type. The total storage required for a struct object is the sum of the storage requirements of all the fields, plus any internal padding.

/* Objective: To prepere a menu which offer following facilities:
        -Register a student
        -delete record
        -search record and print its average
        -display list of student as per percentage obtained     */

        // Source Code //

#include<stdio.h>            //including header files//
#include<string.h>
#include<conio.h>
struct info
      //declaration of structure//
{
char name[20];
char branch[20];
int year;
int marks[5];
float percent;
char eno[20];
}student[50];

//Function Declaration//

void enter(int);
void del(int);
void search(int);
void display(int);
int main()
{
int i,n;
char choice;

printf("enter the no of students u want to register\t");
scanf("%d",&n);         // taking no of records to enter//
enter(n);               // calling enter function//
do
{
printf("enter\n 1:Delete \n 2:search \n 3:Display");
scanf("%d",&i);         //taking user's choice//
switch(i)
{
case 1:
del(n);                  // calling del function//
break;
case 2:
search(n);              // calling search function//
break;
case 3:
display(n);            // calling display function//
break;

default:
printf("invalid choice");
break;
}
printf("do u wanna continue press y or n");
scanf("%c",&choice);
}
while(choice=='y' || choice =='Y');
getch();
}
void enter(int n)      // function definition//
{
int i,j;

for(i=0;i<n;i++)
{
printf("Enter name,branch,year,enrolement no. of student");
scanf("%s %s %d %s",student[i].name,student[i]

.branch,&student[i].year,student[i].eno);
        //taking student info//
printf("enter marks of 5 subject");
for(j=0;j<5;j++)
scanf("%d",&student[i].marks[j]); // taking student's marks//
}
}
void del(int n)    // function definition//
{
char sname[20];
int i,j,flag;
printf("enter the name of student whose records u wanna delete:");
scanf("%s",sname);
for(i=0;i<n;i++)
if(strcmp(student[i].name,sname)==0)      //checking//
flag=i;
for(j=flag;j<n-1;j++)
{
strcpy(student[j].name,student[j+1].name); //swapping//
student[j].year=student[j+1].year;
strcpy(student[j].branch,student[j+1].branch);
for(i=0;i<5;i++)
student[j].marks[i]=student[j+1].marks[i];
strcpy(student[j].eno,student[j+1].eno);
}
printf("The records after deleting info are:");
display(n-1);               //function call for display function//
}
void search(int n)          //function definition//
{
int i, avg,j, flag;
char eno[20];
printf("enter eno of student whose marks are to be calculated");
scanf("%s",eno);
for(i=0;i<n;i++)

                if(strcmp(student[i].eno,eno)==0)
                flag=i;

                for(j=0;j<n;j++)
                                {
                                avg+=student[flag].marks[j];
                                }
printf("\nThe avg is %d",avg);
}

void display(int n)           // function definition//
{
int i,j,sum[50],pos,a[50],k;
float small;
struct info s1;
for(i=0;i<n;i++)
{
                 sum[i]=0;
                for(j=0;j<5;j++)

                sum[i]+=student[i].marks[j];
                student[i].percent=(float)sum[i]/5;


}
printf("The information of students are:");

for(i=0;i<n;i++)
{
    // sorting info in increasing order of percentage//
    for(j=0;j<n-i-1;j++)
    {
        if(student[j].percent>student[j+1].percent)
        {
        strcpy(s1.name,student[i].name);
        strcpy(s1.branch,student[i].branch);
        strcpy(s1.eno,student[i].eno);
        s1.year=student[i].year;
        for(k=0;k<5;k++)
        s1.marks[k]=student[i].marks[k];
        s1.percent=student[i].percent;
        strcpy(student[i].name,student[i+1].name);
        strcpy(student[i].branch,student[i+1].branch);
        strcpy(student[i].eno,student[i+1].eno);
        student[i].year=student[i+1].year;
        for(k=0;k<5;k++)
        student[i].marks[k]=student[i+1].marks[k];
        student[i].percent=student[i+1].percent;
        strcpy(student[i+1].name,s1.name) ;
        strcpy(student[i+1].branch,s1.branch);
        strcpy(student[i+1].eno,s1.eno) ;
        student[i+1].year=s1.year;
        for(k=0;k<5;k++)
        student[i+1].marks[k]=s1.marks[k];
        student[i+1].percent=s1.percent;
        }
    }

    }
for(i=0;i<n;i++)
{
    printf("\nName: %s \n branch: %s \n enrolement no: %s",student[i].name,student[i].branch,student[i].eno) ;
    printf("\nThe marks of 5 subjects are:");
    for(j=0;j<5;j++)
    printf("\n%d",student[i].marks[j]);
    printf("\npercentage %f",student[i].percent);
}
}

Array Data Type & Functions

2011-09-14T22:39:00.002+05:30

An array is a systematic arrangement of objects, usually in rows and columns.

Generally, a collection of data items that can be selected by indices computed at run-time, including:

Array data structure, an arrangement of items at equally spaced addresses in computer memory
Array data type, used in a programming language to specify a variable that can be indexed
Associative array, an abstract data structure model that generalizes arrays to arbitrary indices

Objective: Given an array of N integers, write a program to perform following operations on the array based user's choice:
          -Insertion of elements at perticular location
          -Deletion of specific element
          -searching of perticular element
          -merging of list of elements
          -array traversing

          // Source Code   //

#include<stdio.h>     //including header files//
#include<conio.h>     //including header files//
void insertion(int[],int);         //function declaration//
void deletion(int[],int);          //function declaration//
void duplidel(int[],int);          //function declaration//
void search(int[],int);            //function declaration//
void merge(int[],int[],int);       //function declaration//

int main()
{
int a[20],b[20],n,e,i;

printf("enter the no of elements you wanna enter:");
scanf("%d",&n);                    //enter no of elements of array//
printf("enter array elements in assending order:\n");
for(i=0;i<n;i++)
scanf("%d",&a[i]);                 //entering elements of array//
printf("select\n 1: insertion \n 2: deletion \n 3: deletion of duplicate \n 4:search \n 5:merge");
scanf("%d",&e);
switch (e)              //testing user's choice //
{
case 1:
insertion( a,n);        //calling insertion function//
break;
case 2:
deletion( a, n);       //calling deletion function//
break;
case 3:
duplidel( a, n);      //calling duplicate delete function//
break;

case 4:

search ( a, n);     //calling search function//
break;
case 5:
{
printf("enter the second array:\n");
for(i=0;i<n;i++)
scanf("%d",&b[i]);
merge( a,b,n);
}
break;
default:
printf("invalid selection");
break;
}
getch();
}

void insertion (int a[20],int n) //function definition//
{ int e,pos,t,i;                  //variable declaration//
printf("enter the element to be inserted ");
scanf("%d",&e);
printf("enter the position at which the above element is to be inserted");
scanf("%d",&pos);
for(i=pos-1;i<n+1;i++)
{
t=a[i];
a[i]=e;
e=t;
}

printf("the array after insertion is:");
for(i=0;i<n+1;i++)
printf("%d\t",a[i]);
getch();
}
void deletion( int a[20], int n) //function definition//
{ int i,e,t,pos;
printf("enter the element u wanna delete:");
scanf("%d",&e);
for(i=0;i<n;i++)
{
if( a[i]==e)
{
pos=i;
break;
}
}
for(i=pos;i<n;i++)
a[i]=a[i+1];
for(i=0;i<n-1;i++)
printf("%d\t",a[i]);
getch();
}
void search(int a[],int n)        //function definition//
{
int i, flag=0,no;
printf("enter the no u wanna search:");
scanf("%d",&no);
for(i=0;i<n;i++)
{
if (a[i]==no);
        {
            printf("%d found at %d position",no,i+1);
            flag=1;
            break;
        }
}
if (flag==0)
printf("%d not found",no);
getch();
}
void merge(int a[],int b[],int n)     //function definition//
{ int c[20],i,j;
for(i=0;i<n;i++)
c[i]=a[i];
for(j=0;j<n;j++)
c[j+i]=b[j];
printf("The array after merging is:");
for(i=0;i<2*n;i++)

printf("%d\t",c[i]);
}
void duplidel(int a[], int n)
{
int i, j, k,temp,m=0,pos;
for(i=0;i<n-m;i++)
{
    temp = a[i];
    for(j=i+1;j<n-m;j++)
    {
        if(a[j]==temp)
        {
            m++;
            for(k=j;k<n-m;k++)
            a[k]=a[k+1];
        }
    }

}
printf("The array after deletion of duplicate element is:");
for(i=0;i<n-m;i++)
printf("%d\t",a[i]);
}

Divide And Conquer Approach

2011-09-04T13:58:00.000+05:30

Divide-and-conquer is a top-down technique for designing algorithms that consists of dividing the problem into smaller sub-problems, we achieve the desired result by keep dividing the original problem in to many sub problems and solve those small (sub-problems)
recursively as its easy to calculate some small problem then to complete big one and finally combine all those sub problems to get the result of main problem.

divide-and-conquer paradigm consists of following major phases:

Breaking the problem into several sub-problems that are similar to the original problem but smaller in size,
Solve the sub-problem recursively (successively and independently), and then
Combine these solutions to sub-problems to create a solution to the original problem.

Quick sort is the best example of the divide and conquer течникуе

Merge Sort Explanation

2011-09-03T23:45:00.003+05:30

Merge sort is a divide and conquer algorithm. It is an O(n log n) comparison-based sorting algorithm. Most implementations produce a stable sort, meaning that the implementation preserves the input order of equal elements in the sorted output.

Conceptually, a merge sort works as follows

If the list is of length 0 or 1, then it is already sorted. Otherwise:
Divide the unsorted list into two sublists of about half the size.
Sort each sublist recursively by re-applying the merge sort.
Merge the two sublists back into one sorted list.

A step by step procedure, Approach, Theoretical Evaluation , Results and Analysis of MERGE SORT algorithm is give in this short pdf file Click Here To Download This File Now

Here Is the Code Of the Merge Sort Implementation

Click HereTo go to that code

if you need any further information or help regarding Merge Sort Algorithm Then Plzz let me Know by Commenting below, I'll Feel great to help u

FLOYD WARSHALL ALGORITHM

2011-06-12T12:45:00.000+05:30

#include<stdio.h>
#include<stdlib.h>
#define MAX 10

void floyd(int);
int w[MAX][MAX], d[MAX][MAX][MAX];

void main()
{
int i, j,k,v;
clrscr();
printf("enter the no. of vertices\n");
scanf("%d",&v);
printf("enter the weights \n");

for(i=1;i<=v;i++)
{
   for(j=1;j<=v;j++)
    scanf("%d",&w[i][j]);
   }

floyd(v);
getch();
}//end of main

void floyd(int v)
{
int k, i,j;

k=0;
for(i=1;i<=v;i++)
{
for(j=1;j<=v;j++)
d[k][i][j]=w[i][j];
}

      for(k=1;k<=v;k++)
      {
       for(i=1;i<=v;i++)
    {
    for(j=1;j<=v;j++)
       d[k][i][j]=min(d[k-1][i][j], d[k-1][i][k]+ d[k-1][k][j]);
    }
    }

//displayin matrix

for(k=0;k<=v;k++)
{
printf(" k=%d \n",k);
   for(i=1;i<=v;i++)
   {
    printf("\n");
     for(j=1;j<=v;j++)
     printf("\t %d",d[k][i][j]);
    }
    printf("\n \n ");
}
}

OUTPUT:

Enter the no. of vertices in the graph : 5

Enter the weight matrix :

0 3 8 1000 -4

1000 0 1000 1 7

1000 4 0 1000 1000

2 1000 -5 0 1000

1000 1000 1000 6 0

k=0

0 3 8 1000 -4

1000 0 1000 1 7

1000 4 0 1000 1000

2 1000 -5 0 1000

1000 1000 1000 6 0

k=1

0 3 8 1000 -4

1000 0 1000 1 7

1000 4 0 1000 996

2 5 -5 0 -2

1000 1000 1000 6 0

k=2

0 3 8 4 -4

1000 0 1000 1 7

1000 4 0 5 11

2 5 -5 0 -2

1000 1000 1000 6 0

k=3

0 3 8 4 -4

1000 0 1000 1 7

1000 4 0 5 11

2 -1 -5 0 -2

1000 1000 1000 6 0

k=4

0 3 -1 4 -4

3 0 -4 1 -1

7 4 0 5 3

2 -1 -5 0 -2

8 5 1 6 0

k=5

0 1 -3 2 -4

3 0 -4 1 -1

7 4 0 5 3

2 -1 -5 0 -2

8 5 1 6 0

SUTHERLAND HODGEMAN POLYGON CLIPPING

2011-05-19T17:49:00.000+05:30

The algorithm begins with an input list of all vertices in the subject polygon. Next, one side of the clip polygon is extended infinitely in both directions, and the path of the subject polygon is traversed. Vertices from the input list are inserted into an output list if they lie on the visible side of the extended clip polygon line, and new vertices are added to the output list where the subject polygon path crosses the extended clip polygon line.

#include <stdio.h>
#include <graphics.h>
#include <conio.h>
#include <math.h>
#include <process.h>
#define TRUE 1
#define FALSE 0
typedef unsigned int outcode;
outcode CompOutCode(float x,float y);
enum { TOP = 0x1,
BOTTOM = 0x2,
RIGHT = 0x4,
LEFT = 0x8
};
float xmin,xmax,ymin,ymax;
void clip(float x0,float y0,float x1,float y1)
{
outcode outcode0,outcode1,outcodeOut;
int accept = FALSE,done = FALSE;
outcode0 = CompOutCode(x0,y0);
outcode1 = CompOutCode(x1,y1);
do
{
if(!(outcode0|outcode1))
{
accept = TRUE;
done = TRUE;
}
else
if(outcode0 & outcode1)
done = TRUE;
else
{
float x,y;

outcodeOut = outcode0?outcode0:outcode1;
if(outcodeOut & TOP)
{
x = x0+(x1-x0)*(ymax-y0)/(y1-y0);
y = ymax;
}
else
if(outcodeOut & BOTTOM)
{
x = x0+(x1-x0)*(ymin-y0)/(y1-y0);
y = ymin;
}
else
if(outcodeOut & RIGHT)
{
y = y0+(y1-y0)*(xmax-x0)/(x1-x0);
x = xmax;
}
else
{
y = y0+(y1-y0)*(xmin-x0)/(x1-x0);
x = xmin;
}
if(outcodeOut==outcode0)
{
x0 = x;
y0 = y;
outcode0 = CompOutCode(x0,y0);
}
else
{
x1 = x;
y1 = y;
outcode1 = CompOutCode(x1,y1);
}
}
}while(done==FALSE);
if(accept)
line(x0,y0,x1,y1);
outtextxy(150,20,"POLYGON AFTER CLIPPING");

rectangle(xmin,ymin,xmax,ymax);
}
outcode CompOutCode(float x,float y)
{
outcode code = 0;
if(y>ymax)
code|=TOP;
else
if(y<ymin)
code|=BOTTOM;
if(x>xmax)
code|=RIGHT;
else
if(x<xmin)
code|=LEFT;
return code;
}
void main( )
{
float x1,y1,x2,y2;
/* request auto detection */
int gdriver = DETECT, gmode, n,poly[14],i;
clrscr( );
printf("Enter the no of sides of polygon:");
scanf("%d",&n);
printf("\nEnter the coordinates of polygon\n");
for(i=0;i<2*n;i++)
{
scanf("%d",&poly[i]);
}
poly[2*n]=poly[0];
poly[2*n+1]=poly[1];
printf("Enter the rectangular coordinates of clipping window\n");
scanf("%f%f%f%f",&xmin,&ymin,&xmax,&ymax);
/* initialize graphics and local variables */
initgraph(&gdriver, &gmode, "c:\\tc\\bgi");

outtextxy(150,20,"POLYGON BEFORE CLIPPING");
drawpoly(n+1,poly);
rectangle(xmin,ymin,xmax,ymax);
getch( );
cleardevice( );
for(i=0;i<n;i++)
clip(poly[2*i],poly[(2*i)+1],poly[(2*i)+2],poly[(2*i)+3]);
getch( );
restorecrtmode( );
}

/*
OUTPUT:
Enter the no of sides of polygon:5
Enter the coordinates of polygon
50
50
200
100
350
350
80
200
40
80
Enter the rectangular coordinates of clipping window
150
150
300
300*/

Video lecture on Clipping Lines and Polygons...
Click Here to Watch lecture Now..

COHEN SUTHERLAND LINE CLIPPING ALGORITHM

2011-05-19T17:39:00.000+05:30

#include <stdio.h>
#include <graphics.h>
#include <conio.h>
#include <math.h>
#define TRUE 1
#define FALSE 0
typedef unsigned int outcode;
outcode CompOutCode(float x,float y);
enum { TOP = 0x1,
BOTTOM = 0x2,
RIGHT = 0x4,
LEFT = 0x8
};
float xmin,xmax,ymin,ymax;
void clip(float x0,float y0,float x1,float y1)
{
outcode outcode0,outcode1,outcodeOut;
int accept = FALSE,done = FALSE;
outcode0 = CompOutCode(x0,y0);
outcode1 = CompOutCode(x1,y1);
do
{
if(!(outcode0|outcode1))
{
accept = TRUE;
done = TRUE;
}
else
if(outcode0 & outcode1)
done = TRUE;
else
{
float x,y;
outcodeOut = outcode0?outcode0:outcode1;
if(outcodeOut & TOP)
{

x = x0+(x1-x0)*(ymax-y0)/(y1-y0);
y = ymax;
}
else
if(outcodeOut & BOTTOM)
{
x = x0+(x1-x0)*(ymin-y0)/(y1-y0);
y = ymin;
}
else
if(outcodeOut & RIGHT)
{
y = y0+(y1-y0)*(xmax-x0)/(x1-x0);
x = xmax;
}
else
{
y = y0+(y1-y0)*(xmin-x0)/(x1-x0);
x = xmin;
}
if(outcodeOut==outcode0)
{
x0 = x;
y0 = y;
outcode0 = CompOutCode(x0,y0);
}
else
{
x1 = x;
y1 = y;
outcode1 = CompOutCode(x1,y1);
}
}
}while(done==FALSE);
if(accept)
line(x0,y0,x1,y1);
outtextxy(200,20,"LINE AFTER CLIPPING");
rectangle(xmin,ymin,xmax,ymax);
}

outcode CompOutCode(float x,float y)
{
outcode code = 0;
if(y>ymax)
code|=TOP;
else
if(y<ymin)
code|=BOTTOM;
if(x>xmax)
code|=RIGHT;
else
if(x<xmin)
code|=LEFT;
return code;
}
void main( )
{
float x1,y1,x2,y2;
int gdriver = DETECT, gmode ;
printf("\nEnter the endpoints of line\n");
scanf("%f%f%f%f",&x1,&y1,&x2,&y2);
printf("Enter the rectangular coordinates of clipping window\n");
scanf("%f%f%f%f",&xmin,&ymin,&xmax,&ymax);
/* initialize graphics and local variables */
initgraph(&gdriver, &gmode, "c:\\tc\\bgi");
outtextxy(200,20,"LINE BEFORE CLIPPING");
line(x1,y1,x2,y2);
rectangle(xmin,ymin,xmax,ymax);
getch( );
cleardevice( );
clip(x1,y1,x2,y2);
getch( );
restorecrtmode( );
}

/* OUTPUT

Enter the endpoints of line

100

470

340

Enter the rectangular coordinates of clipping window

180

150

300

280

SHEARING OF LINE

2011-05-18T21:18:00.000+05:30

#include <graphics.h>
#include <stdio.h>
#include <conio.h>
void main( )
{
/* request auto detection */
int gdriver = DETECT, gmode;
int x1,y1,x2,y2,a,b;
/* initialize graphics and local variables */
initgraph(&gdriver, &gmode, "c:\\tc\\bgi");
printf("Enter the value of line coordinates:");
scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
printf("Enter the value of x-shearing factor:");
scanf("%d",&a);
printf("Enter the value of y-shearing factor:");
scanf("%d",&b);
cleardevice( );
outtextxy(200,20,"LINE BEFORE SHEARING");
line(x1,y1,x2,y2);
x1 = x1+a*y1;
y1 = b*x1+y1;
x2 = x2+a*y2;
y2 = b*x2+y2;
getch( );
cleardevice( );
outtextxy(200,20,"LINE AFTER SHEARING");
line(x1,y1,x2,y2);
getch( );
closegraph( );
restorecrtmode( );
}

/* OUTPUT

Enter the value of line coordinates:

100

Enter the value of x-shearing factor:1

Enter the value of y-shearing:1

MIDPOINT ALGORITH FOR DRAWING ELLIPSE

2011-05-18T20:52:00.000+05:30

#include <graphics.h>
#include <stdio.h>
#include <conio.h>
#include <math.h>
void symmetry(int xc,int yc,int x,int y);
void main( )
{

/* request auto detection */

int gdriver = DETECT, gmode ;
int xc,yc,a,b,x,y,fx,fy,p;
int aa = a*a,bb = b*b,aa2 = aa*2,bb2 = bb*2;
clrscr( );
printf("Enter the coordinates of center of ellipse:");
scanf("%d%d",&xc,&yc);
printf("Enter the x-radius and y-radius of ellipse:");
scanf("%d%d",&a,&b);

/* initialize graphics mode */

initgraph(&gdriver, &gmode, "c:\\tc\\bgi");
x = 0; /*starting point*/
y = b;
symmetry(xc,yc,x,y);
fx = 0; /*initial partial derivatives*/
fy = aa2*y;
p = bb-aa*b+(0.25*aa); /*compute and round off p1*/
while(fx < fy) /* |slope|<1 */
{
x++;
fx = fx+bb2;
if(p<0)
p = p+fx+bb;
else

{
y--;
fy = fy-aa2;
p = p+fx+bb-fy;
}
symmetry(xc,yc,x,y);
}
p = bb*(x+0.5)*(x+0.5)+aa*(y-1)*(y-1)-aa*bb;
while(y>0)
{
y--;
fy = fy-aa2;
if(p>=0)
p = p-fy+aa;
else
{
x++;
fx = fx+bb2;
p=p+fx-fy+aa;
}
symmetry(xc,yc,x,y);
}
outtextxy(120,20,"ILLUSTRATION OF MIDPOINT ELLIPSE ALGORITHM");
outtextxy(xc-25,yc,"(xc,yc)");
getch( );
closegraph( );
restorecrtmode( );
}
void symmetry(int xc,int yc,int x,int y)
{
putpixel(xc+x,yc+y,WHITE);
putpixel(xc-x,yc+y,WHITE);
putpixel(xc+x,yc-y,WHITE);
putpixel(xc-x,yc-y,WHITE);
}
/ * OUTPUT

Enter the coordinates of center of ellipse:320

240

Enter the x-radius and y-radius of ellipse:100 80

LONGEST COMMON SUBSEQUENCE (L.C.S.)

2011-05-17T22:54:00.000+05:30

LONGEST COMMON SUB-SEQUENCE algorithm is mainly used to find the common string/characters in the given string and the string given by user , its main application in found in DNA testing where a particular DNA contains a long series of character which is to b matched with another DNA sample of long string character..
#include<stdio.h>
#include<conio.h>
#include<string.h>
#include<graphics.h>

int printlcs(int,int);

struct lcs
{
int dir;
int num;
};
struct lcs b[20][20];
char str1[20],str2[20];
void main()
{ int gd=DETECT,gm;
   int l1,l2,i,j,row,col;
   initgraph(&gd,&gm,"..\\bgi");
   printf("Enter the first string");
   gets(str1);

   printf("Enter the second string");
   gets(str2);
   l1=strlen(str1);
   l2=strlen(str2);
   for(i=0;i<=l1;i++)
   b[i][0].num=0;
   for(i=0;i<=l2;i++)
   b[0][i].num=0;
   row=1;
   col=1;
   for(i=0;i<l1;i++)
   { col=1;
     for(j=0;j<l2;j++)
     {
    if(str1[i]==str2[j])
    { b[row][col].num=b[row-1][col-1].num+1;
    b[row][col].dir=92;
    col++;
    }
    else
    {
       if(b[row][col-1].num>b[row-1][col].num)
       { b[row][col].num=b[row][col-1].num;
         b[row][col].dir=27;
         col++;
       }
       else
       { b[row][col].num=b[row-1][col].num;
         b[row][col].dir=24;
         col++;
        }
    }

      }
      row++;
   }

    for(j=0;j<=(l1+1);j++)
    line(30,100+(j*33),500,100+(j*33));
    for(i=0;i<=(l2+1);i++)
    line(30+(i*62),100,30+(i*62),400);
    printf("\nThe Table for finding Longest common subsequence is\n\n");
    printf("\t");
    for(i=0;i<l2;i++)
    printf(" \t%c",str2[i]);
    printf("\n\n");
    for(i=0;i<=l1;i++)
    {
     for(j=0;j<=l2;j++)
     {
     if(i==0 || j==0)
     {
     printf("\t");
     printf("%d",b[i][j].num);
     }
     else
     printf("   (%c %d)",b[i][j].dir,b[i][j].num);
     }
    printf("\n\n %c",str1[i]);
    }
   printf("\n\n\n\nThe Longest common subsequence is- ");
   printlcs(l1,l2);

   getch();
}
int printlcs(int l1,int l2)
{ if(l1==0|| l2==0)
     return(0);
   if(b[l1][l2].dir==92)
    {
     printlcs(l1-1,l2-1);
     printf("%c",str1[l1-1]);
    }
    else
    {
   if(b[l1][l2].dir==24)
     printlcs(l1-1,l2);
   else
     printlcs(l1,l2-1);
   }
}

DIJKSTRA'S ALGORITHM

2011-05-17T22:47:00.000+05:30

#include<stdio.h>
#include<conio.h>
#define MAXNODES 50
#define MAX1 150
#define INFINITY 5000

using namespace std;

int weight[MAXNODES][MAXNODES],i,j,distance[MAXNODES],visit[MAXNODES];
int precede[MAXNODES],final=0;
int path[MAX1];
int smalldist,newdist,k,s,d,current,n,distcurr;

void Display_Result()
{
i=d;
path[final]=d;
final++;
while(precede[i]!=s)
{
j=precede[i];
i=j;
path[final]=i;
final++;
}
path[final]=s;
printf("\nThe shortest path followed is :\n\n");
for(i=final;i>0;i--)
printf("\t\t(%d -> %d) with cost = %d\n\n",path[i],path[i-1],weight[path[i]][path[i-1]]);
printf("\nFor total cost = %d",distance[d]);
}

main()
{
printf("\nEnter the number of nodes(Less than 50)in the matrix : ");
scanf("%d",&n);
printf("\nEnter the cost matrix :\n\n");
for(i=0;i<n;i++)
for(j=0;j<n;j++)
    scanf("%d",&weight[i][j]);
printf("\nEnter the source node (0 to %d) : ",n-1);
scanf("%d",&s);
printf("\nEnter the destination node (0 to %d) : ",n-1);
scanf("%d",&d);
for(i=0;i<n;i++)
{
distance[i]=INFINITY;
precede[i]=INFINITY;
}
distance[s]=0;
current=s;
visit[current]=1;
while(current!=d)
{
distcurr=distance[current];
smalldist=INFINITY;
for(i=0;i<n;i++)
    if(visit[i]==0)
    {
      newdist=distcurr+weight[current][i];
      if(newdist<distance[i])
      {
    distance[i]=newdist;
    precede[i]=current;
      }
      if(distance[i]<smalldist)
      {
    smalldist=distance[i];
    k=i;
      }
    }
current=k;
visit[current]=1;
}
Display_Result();
getch();
}

/*----------------OUTPUT----------------*/
/*
Enter the number of nodes(Less than 50)in the matrix:6

Enter the cost matrix:

5000    6       3       5000    5000    5000
6       5000    2       5       5000    5000
3       2       5000    3       4       5000
5000    5       3       5000    2       3
5000    5000    4       2       5000    5
5000    5000    5000    3       5       5000

Enter the source code:0

Enter the destination code:5

The shortest path followed is:

        (0 -> 2)with cost=3

        (2 -> 3)with cost=3

        (3 -> 5)with cost=3

For total cost=9
*/

PRIM'S ALGORITHM

2011-05-05T21:01:00.000+05:30

#include<iostream.h>
#include<stdio.h>
#include<conio.h>
int weight[20][20],visited[20],d[20],p[20];
int v,e;

void creategraph()
{
int i,j,a,b,w;
cout<<"\nEnter number of vertices";
cin>>v;
cout<<"\nEnter number of edges";
cin>>e;
for(i=1;i<=v;i++)
for(j=1;j<=v;j++)
    weight[i][j]=0;
for(i=1;i<=v;i++)
{
p[i]=visited[i]=0;
d[i]=32767;
}
for(i=1;i<=e;i++)
{
cout<<"\nEnter edge a,b and weight w:";
cin>>a>>b>>w;
weight[a][b]=weight[b][a]=w;
}
}

void prim()
{
int current,totalvisited,mincost,i;
current=1;d[current]=0;
totalvisited=1;
visited[current]=1;
while(totalvisited!=v)
{
for(i=1;i<=v;i++)
{
    if(weight[current][i]!=0)
      if(visited[i]==0)
    if(d[i]>weight[current][i])
    {
    d[i]=weight[current][i];
    p[i]=current;
    }
}
mincost=32767;
for(i=1;i<=v;i++)
{
    if(visited[i]==0)
      if(d[i]<mincost)
      {
    mincost=d[i];
    current=i;
      }
}
visited[current]=1;
totalvisited++;
}
mincost=0;
for(i=1;i<=v;i++)
mincost+=d[i];
cout<<"\nMinimum cost="<<mincost;
cout<<"\nMinimum span tree is";
for(i=1;i<=v;i++)
cout<<"\nVertex "<<i<<" is connected to"<<p[i];
}

main()
{
clrscr();
creategraph();
prim();
getch();
}

/*-------------------OUTPUT---------------------*/

/*
Enter number of vertices6

Enter number of edges9

Enter edge a,b and weight w:1 2 6

Enter edge a,b and weight w:1 3 3

Enter edge a,b and weight w:2 3 2

Enter edge a,b and weight w:2 4 5

Enter edge a,b and weight w:3 5 4

Enter edge a,b and weight w:3 4 3

Enter edge a,b and weight w:4 5 2

Enter edge a,b and weight w:4 6 3

Enter edge a,b and weight w:5 6 5

Minimum cost=13
Minimum span tree is
Vertex 1 is connected to 0
Vertex 2 is connected to 3
Vertex 3 is connected to 1
Vertex 4 is connected to 3
Vertex 5 is connected to 4
Vertex 6 is connected to 4
*/

Step By Step Choosing of nodes Graphically shown Here
Click Here to watch the steps now..

ACTIVITY SELECTION PROBLEM

2011-05-04T20:16:00.000+05:30

An activity selection is a problem of scheduling a resource among several competing activity. The activity selection problem is also defines as : " Given a set of n activities with start time si, and fi as finish time of an ith activity. the problem is to find the maximum size set of mutually compatible activities.

#include<iostream.h>
#include<stdio.h>
#include<conio.h>
struct activity
{ int st;
int end;
} act[20];
void activity(int,int);
int ans[10];
static count=1;
void main()
{
int i,j,n;
clrscr();
cout<<"Enter the number of activities";
cin>>n;
printf<<Enter the start point and end points of the activities\n";
printf("\t START \t END");
for(i=0;i<n;i++)
scanf("%d %d",&act[i].st,&act[i].end);
activity(0,n);
   cout<<"\n The set of avtivities selected are";
   ans[0]=0;
   count--;
   for(j=0;j<=count;j++)
   printf(" A%d",ans[j]+1);
   getch();
}
void activity(int i,int j)
{
   int m;
   m=i+1;
   while(m<=j && act[m].st<act[i].end)
   m++;
   if(m<j)
   {
         ans[count++]=m;
       activity(m,j);
   }
}

MERGE SORT

2011-05-02T22:45:00.000+05:30

#include<iostream.h>
#include<conio.h>
void mergesort(int[],int,int);
void merge(int[],int,int,int);
void main()
{
int a[10],p,q,r,i,n;
clrscr();
cout<<"Enter the number of elements";
cin>>n;
p=0;
r=n-1;
cout<<"Enter the array";
for(i=0;i<n;i++)
{
     cin>>a[i];
}
mergesort(a,p,r);
cout<<"The sorted array is:";
for(i=0;i<n;i++)
{
     cout<<"\n"<<a[i];
}
getch();
}

void mergesort(int a[],int p,int r)
{
     if( p < r)
    {
       int q=(p+r)/2;
       mergesort(a,p,q);
         mergesort(a,q+1,r) ;
         merge(a,p,q,r);
     }
}
void merge(int a[],int p,int q,int r)
{
int c[10];
int i=p;
int j=q+1;
int k=p;
while((i<=q)&&(j<=r))
{
if(a[i]<a[j])
{
     c[k]=a[i];
      i=i+1;
     k=k+1;
}
else
{
     c[k]=a[j];
      j=j+1;
      k=k+1;
}
}
while(i<=q)
{
     c[k] =a[i];
     i=i+1;
     k=k+1;
}
while(j<=r)
{
     c[k]=a[j];
     j=j+1;
     k=k+1;
}
int l=p;
while(l<=r)
{
a[l]=c[l];
l=l+1;
}
}

PATTERN PRINTING

2011-03-23T17:44:00.003+05:30

THIS PROGRAM WILL GIVE THE FOLLOWING OUTPUT DEPENDING ON THE NUMBER N GIVEN BY USER.

1

12

123

1234

12345

#include<iostream.h>
#include<conio.h>
void main()
{
clrscr();
int n;
cout<<"enter a number";
cin>>n;
for(int i=1;i<=n;i++)
{
    for(int j=1;j<=i;j++)
    {
    cout<<j;
    }
    cout<<"\n";
}
getch();
}

IF YOU WANT TO PRINT THE NUMBERS IN FOLLOWING MANNER ,THE PROGRAM IS GIVEN BELOW :

1

22

333

4444

55555

#include<iostream.h>
#include<conio.h>
void main()
{
clrscr();
int n;
cout<<"enter a number";
cin>>n;
for(int i=1;i<=n;i++)
{
    for(int j=1;j<=i;j++)
    {
    cout<<i;
    }
    cout<<"\n";
}
getch();
}

Just copy and paste the above programs in the notepad file ans save it with .cpp extension and open it in ur compiler, and the work is done , enjoy :)

RECTANGLE DRAWING ALGORITHM

2011-03-23T11:58:00.000+05:30

#include<iostream.h>
#include<conio.h>
#include<math.h>
#include<process.h>
#include<graphics.h>
void main()
{
int gdriver=DETECT,gmode,errorcode;
initgraph(&gdriver,&gmode,"c:\\tc\\bgi ");
errorcode=graphresult();
if(errorcode!=grOk)
{
cout<<"graphics error"<<grapherrormsg(errorcode)<<endl;
cout<<"press any key to start";
getch();
exit(1);
}
int x1,y1,x2,y2;
cout<<"///////////////// RECTANGLE/////////////////";
cout<<"\n enter x - cordinate of upper right side of rectangle";
cin>>x1;
cout<<"\n enter y - cordinate of upper right side of rectangle";
cin>>y1;
cout<<"\n enter x - cordinate of lower left side of rectangle";
cin>>x2;
cout<<"\n enter y - cordinate of lower left side of rectangle";
cin>>y2;
clrscr();
cout<<" DESIRED RECTANGLE";
line(x1,y1,x1,y2);
line(x2,y2,x1,y2);
line(x1,y1,x2,y1);
line(x2,y2,x2,y1);
getch();
closegraph();
}

DRAWING TRIANGLE & ITS TRANSLATION

2011-03-23T11:35:00.000+05:30

In this program we will first draw a triangle according to the coordinates given by user and then applying the translation matrix or translation factor we will draw that triangle with translated coordinates

#include<iostream.h>
#include<conio.h>
#include<math.h>
#include<process.h>
#include<graphics.h>
void main()
{
int gdriver=DETECT,gmode,errorcode;
initgraph(&gdriver,&gmode,"c:\\tc\\bgi ");
errorcode=graphresult();
if(errorcode!=grOk)
{
cout<<"graphics error"<<grapherrormsg(errorcode)<<endl;
cout<<"press any key to start";
getch();
exit(1);
}
int i,j,k,x1,x2,x3,y1,y2,y3,x,y;
int c[3][3],a[3][3];
cout<<"ENTER COORDINATES OF TRIANGLE";
cout<<"\n 1st coordinate";
cout<<"\n x - coordinate ";
cin>>a[0][0];
cout<<"\n y - coordinate";
cin>>a[1][0];
a[2][0]=1;
cout<<"\n 2nd coordinate";
cout<<"\n x - coordinate";
cin>>a[0][1];
cout<<"\n y - coordinate";
cin>>a[1][1];
a[2][1]=1;
cout<<"\n 3rd coordinate";
cout<<"\n x - coordinate";
cin>>a[0][2];
cout<<"\n y - coordinate";
cin>>a[1][2];
a[2][2]=1;
cout<<"\n\n\n enter the translation coordinate";
cout<<" enter x - coordinate";
cin>>x;
cout<<"enter y - coordinate";
cin>>y;
int t[3][3]={{1,0,x},{0,1,y},{0,0,1}};
for(i=0;i<3;i++)
{
    for(j=0;j<3;j++)
    {
        c[i][j]=0;
        for(k=0;k<3;k++)
        {
            c[i][j]+=t[i][k]*a[k][j];
        }
    }
}
clrscr();
line(a[0][0],a[1][0],a[0][1],a[1][1]);
line(a[0][1],a[1][1],a[0][2],a[1][2]);
line(a[0][0],a[1][0],a[0][2],a[1][2]);
cout<<" BEFORE TRANSFORMATION";
cout<<"\n press any key to get translated triangle";
getch();
clrscr();
line(c[0][0],c[1][0],c[0][1],c[1][1]);
line(c[0][1],c[1][1],c[0][2],c[1][2]);
line(c[0][0],c[1][0],c[0][2],c[1][2]);
cout<<" AFTER TRANSLATION";
getch();
closegraph();
}

FIBONACCI SERIES

2011-03-21T20:11:00.000+05:30

#include<iostream.h>
#include<conio.h>
void main()
{
clrscr();
int n, first , second , third;
cout<<"/////////////////fibonacci series////////////////////";
cout<<"\n enter the number of terms in series you want";
cin>>n;
first=1;
second=1;
cout<<first<<"\t"<<second<<"\t";
for(int i=3;i<=n;i++)
{
          third=first+second;
          cout<<third<<"\t";
          first=second;
          second=third;
}
getch();
}

BINARY SEARCH

2011-02-21T12:32:00.000+05:30

#include<iostream.h>
#include<conio.h>
void main()
{
clrscr();
int a[50],x,n,beg,end,mid,flag;
cout<<"////////////////BINARY SEARCH//////////////////";
cout<<"\n enter the size of array";
cin>>n;
cout<<" enter the sorted array";
for(int i=1;i<=n;i++)
cin>>a[i];
cout<<" enter the element to search";
cin>>x;
beg=1;
end=n;
mid=(beg+end)/2;
while(beg<=end)
{
mid=(beg+end)/2;
if(a[mid]==x)
{
flag=1;break;

}
else if(x<a[mid])
{
end=mid-1;
}
else
{
beg=mid+1;
}
}
if(flag==1)
cout<<"element found at position "<<mid;
else
cout<<" element is not in above array";
getch();
}

LINEAR SEARCH

2011-02-20T22:42:00.000+05:30

#include<iostream.h>
#include<conio.h>
void main()
{
clrscr();
int a[50],n,x,flag;
cout<<"/////////////////// LINEAR SEARCH//////////////////";
cout<<"\n enter the size of array (max 50)";
cin>>n;
cout<<"\n enter array elements";
for(int i=1;i<=n;i++)
cin>>a[i];
cout<<"\n enter an element you want to search";
cin>>x;
for(int j=1;j<=n;j++)
{
if(a[j]==x)
{
flag=1;break;
}
else
flag=0;

}
if(flag==1)
cout<<"\n element found at position "<<j;
else cout<<"element not found";
getch();
}

BRESENHAM'S CIRCLE ALGORITHM

2011-02-17T20:34:00.001+05:30

This algorithm will dram an arc from 90 degree angle to 45 degree angle.... after that to construct the whole circle we have to take its mirror image in all 8 quardrants..... the other algorithm for drawing a circle is MID POINT ALGORITHM....

#include<iostream.h>
#include<conio.h>
#include<graphics.h>
#include<process.h>
#include<math.h>
void main()
{
int gdriver=DETECT,gmode,errorcode;
initgraph(&gdriver,&gmode,"c:\\tc\\bgi ");
errorcode=graphresult();
if(errorcode!=grOk)
{
cout<<"graphics error"<<grapherrormsg(errorcode)<<endl; cout<<"press any key to start"; getch(); exit(1); } clrscr(); int r,x,d,y; cout<<"enter the radius of the circle"; cin>>r;
x=0; y=r; d=3-(2*r);
while(x<=y)
{
putpixel(x,y,RED);
if(d<0)
{
d=d+4*x+6;
}
else
{

d=d+4*(x-y)+10;
y--;
}
x++;
}
getch();
closegraph();
}

BRESENHAM'S LINE ALGORITHM

2011-02-17T20:26:00.000+05:30

#include<iostream.h>
#include<conio.h>
#include<graphics.h>
#include<process.h>
#include<math.h>
void main()
{
int gdriver=DETECT,gmode,errorcode;
initgraph(&gdriver,&gmode,"c:\\tc\\bgi ");
errorcode=graphresult();
if(errorcode!=grOk)
{
cout<<"graphics error"<<grapherrormsg(errorcode)<<endl;
cout<<"press any key to start";
getch();
exit(1);
}
clrscr();
int x1,y1,x2,y2,dx,dy,d,loc1,loc2;
cout<<"enter x and y (starting point)";
cin>>x1>>y1;
cout<<"enter x and y (end point)";
cin>>x2>>y2;
dy=y2-y1;
dx=x2-x1;
d=2*dy-dx;
loc1=2*dy;
loc2=2*(dy-dx);
while(x1<=x2)
{
if(d<0)
{
putpixel(x1,y1,BLUE);
d=d+loc1;
}
else
{
putpixel(x1,y1,RED);
d=d+loc2;
y1++;

}
x1++;
}
getch();
closegraph();
}

SIMPLE LINE DRAWING ALGORITHM

2011-02-17T20:16:00.000+05:30

#include<iostream.h>
#include<conio.h>
#include<graphics.h>
#include<process.h>
#include<math.h>
void main()
{
int gdriver=DETECT,gmode,errorcode;
initgraph(&gdriver,&gmode,"c:\\tc\\bgi ");
errorcode=graphresult();
if(errorcode!=grOk)
{
cout<<"graphics error"<<grapherrormsg(errorcode)<<endl;
cout<<"press any key to start";
getch();
exit(1);
}
clrscr();
int x1,y1,x2,y2,x,y; float m,c;
cout<<"enter valve of x and y(starting point)";
cin>>x1>>y1;
cout<<"enter value of x and y (end point)";
cin>>x2>>y2;
x=x1; y=y1;
m=float (y2-y1)/(x2-x1);
c=y1-m*x1;
cout<<"m="<<m;
cout<<"c="<<c;
if(m==0)
{
while(x<=x2)
{putpixel(x,y,RED);
x++;
}
}
if(abs(m)<=1)
{
while(x<=x2)
{
putpixel(x,y,RED);
x++;
y=m*x+c;
}
}
else
{
while (y<=y2)
{
putpixel(x,y,CYAN);
y++;
x=(y-c)/m;
}
}
getch();
closegraph();
}