C Programming

#include<stdio.h>

#include<conio.h>

#include<string.h>

void main()

{ char str1[20];

char str2[20];

int x,y,n,r,counter1,counter2,points=0;

clrscr();

printf ("Please enter a name: \n\n");

gets(str1);

printf ("\nEnter another name: \n\n");

gets(str2);

counter1=strlen(str1);

counter2=strlen(str2);

for (x=0; x<counter1; x++)

{ for (y=0; y<counter2; y++)

{ if (str1[x]==str2[y])

points++;

else;}

if(points!=0)

points++;

}

for (n=1; n<counter1; n++)

{ if (str1[0]==str1[n])

points++;

else;}

printf ("\n%d \n",points);

r=points%6;

switch®

{

case 1: printf ("\nFriends"); break;

case 2: printf ("\nLovers"); break;

case 3: printf ("\nAffection"); break;

case 4: printf ("\nMarriage"); break;

case 5: printf ("\nEnemies"); break;

case 0: printf ("\nSister"); break;

}

getch();

}

In C++, enum signifies a slightly stronger type than in C.

For example, in C, one could write:

enum Direction { UP, DOWN };

Direction d = 1;

In C++, this would be illegal, only UP or DOWN can be assigned to a variable of type Direction, though there is still implicit casting to integer, so one could still write:

int i = UP;

Another difference has to do with the way variable are declared in general in C++. In C, once the enum was declared as above, declaring variables of type Direction would have to be done through the enum keyword, like this:

enum Direction d = UP;

In C++, the name "Direction" becomes a type in itself, so you can write:

Direction d = UP;

He considered the "goto" statement to be harmful.

constants are values that does not chnage through out the program exceution..

FIFO, means "First In, First Out". An example of such a data structure is a queue.

#include #include
#include
void main()
{
char a[50],r[50];
int i,j;
printf("\n Enter the string:");
gets(a);
for(i=strlen(a)-1,j=0;i>=0;i--,j++)
r[j]=a[i];
r[j]='\0';
printf("\n Reverse string is %s",r);
if(strcmpi(a,r)==0)
printf("\n the original string is a pallindrome");
else
printf("\n the original string is not pallindrome");
getch();
}








Note:-If you want it case sensitive then use 'strcmp' instead of 'strcmpi'

There is no builtin function 'counta' in C.

class Abc

{

public static void main(String[] args)

{

System.out.println("Hello World!");

}

}

echo enter n value

read n

sumodd=0

sumeven=0

i=0

while [ $i -ne $n ]

do

echo "Enter Number"

read num

if [ `expr $num % 2` -ne 0 ]

then

sumodd=`expr $sumodd + $num`

sumeven=`expr $sumeven+$num`

fi

i=`expr $i + 1`

done

echo Sum of odd numbers = $sumodd

echo Sum of even numbers = $sumeven

design a flowchart that will input three numbers and get their sum. If the sum is greater than 20, then print "sum>20",else print the sum.

i cant sorry

by this program you can find the factorial:

#include<iostream>

using namespace std;

main()

{

int n,x,f=1;

cin>> n;

x=0;

while(x<n)

{

x++;

f= f*x;

}

cout<<"factorial is"<<f<<"\n";

system("pause");

return 0;

}

For a full binary tree of height 3 there are 4 leaf nodes. E.g., 1 root, 2 children and 4 grandchildren.

#include<iostream>

using namespace std;

#include<conio.h>

void main()

{

char m[100];

int units[100],i,j,n;

float x,tb;

cout<<"enter the total number of bills to be calculate:\n";

cin>>n;

for(i=1;i<=n;i++)

{

cout<<"\nenter the name \n:==>";

cin>>m;

for(i=0;i<n;i++)

{

cout<<"\nenter the total units consumed \n:==>";

cin>>units[i];

}

}

if(units[i]<=100)

{

x=units[i]*0.6;

tb=50+x;

if(tb>300)

{

tb=0.15*tb+tb;

}

cout<<"\ncalculated bill:==>";

cout<<m<<tb;

}

else if(units[i]>100 && units[i]<=300)

{

tb=100*0.6+(units[i]-100)*0.8+50;

if(tb>300)

{

tb=0.15*tb+tb;

}

cout<<"\ncalculated bill:==>";

cout<<m<<tb;

}

else if(units[i]>300)

{

tb=100*0.6+200*0.8+(units[i]-300)*0.9+50;

if(tb>300)

{

tb=0.15*tb+tb;

}

cout<<"\ncalculated bill:==>";

cout<<m<<tb;

}

getch();

}

Strong number is a number for which sum of factorials of the digits is equal to the given number.

Eg: let us consider 145 where sum of factorials of the digits= 1!+4!+5!=1+24+120=145 i,e., the given number.

c program for strong number is:

#include<stdio.h>

#include<conio.h>

void main()

{

int sof=0,n,dn,ctr,prod,rem;

printf("Enter the number\n");

scanf("%d",&n);

dn=n;

while(n!=0)

{

prod=1,ctr=1;

rem=n%10;

while(ctr<=rem)

{

prod=prod*ctr;

ctr=ctr+1;

}

sof=sof+prod;

n=n/10;

}

if(sof==dn)

{

printf("The number entered is strong number");

}

else

{

printf("The number entered is not a strong number");

}

}

By

Nagarjuna

sri indu ECE,

IBP.

What is the difference between data communication and telecommunication ?

The native Assembly language of the given platform.

For example it would be stupid to write anything in Motorola 68000 Assembly for Intel x86 platform: it wouldn't work.

if one dimension means it have only rows but in two dimension means it have both row and coloumns

#include<stdio.h> main() { int a,b,c,d; // The four integers to be asked printf("Give the first integer: "); //asks for the first integer scanf("%d",&a); // puts the user input in the address of the integer "a" printf("Give the second integer: "); //same explanations scanf("%d",&b); printf("Give the third integer: "); scanf("%d",&c); printf("Give the fourth integer: "); scanf("%d",&d); printf("1. The sum of the four integers is: %d",a+b+c+d); //prints the sum of the four integers given by the user, notice the "a+b+c+d" at the end) printf("2. The sum of the first two numbers minus the sum of the last: %d",a+b-c-d); //prints the second condition by putting the correct operations return 0; //ends the program } I never tested this program though, but i think it would work.

Unfortunately, there is no power function in C/C++, so you'll have to make one yourself. I made one a few days ago, here it is:

int power (int one, int two)

{

int temp = one, counter = 1;

while (counter

{

temp *= one;

counter++;

}

return temp;

}

/* PROGRAM TO CALCULATE THE VALUE OF X RAISED TO GIVEN POWER*/

#include <stdio.h>

#include <math.h>

#include <conio.h>

main()

{

int x,power;

float result;

clrscr();

printf("Enter the value of x:");

scanf("%d",&x);

printf("Enter the value of power:");

scanf("%d",&power);

result=pow(x,power);

printf("The value of %d^%d is %f",x,power,result);

getch();

return;

}

Game Maker 7 (you can find it on "YoYoGames" )

A Sample java method that can do this sum of all numbers between 1 to 50

public int sumNumbers(){

int retVal = 0;

for(int i = 0; i <=50; i++){

retVal = retVal + i;

}

return retVal;

}

Enter the number, then press the button marked with +/-