Roots of Quadratic Equation
Algorithm:
Step 1: start
Step 2: Read the values of a, b, c
and go to step3
Step 3: Is a=0 then go to step 4
otherwise go to step5
Step 4: Print the given equation
is linear equation and go to step 11
Step 5: Compute d=(b*b)-(4*a*c))
and go to step 6
Step 6: Is d<0 then go to
step7 otherwise go to step8
Step 7: Print the roots are
imaginary and go to step 11
Step 8: Is d=o then go to step 9
otherwise go to step10
Step 9: Compute r 
and print the roots are real
and equal lent to ‘r’ and go to step 11
and print the roots are real
and equal lent to ‘r’ and go to step 11
Step 10: Is d >0 Then compute 
and r2 
, print r1, r2 and go to
step11
and r2 , print r1, r2 and go to
step11
Step 11: Stop
/* WAP in C To Find Roots of
given Quadratic Equations*/
#include<stdio.h>
#include<math.h>
float d,r1,r2,r;
main( )
{
float a,b,c;
clrscr();
printf("\nENTER a, b, c
VALUES OF QUDRATIC EQUATION\n");
scanf("%f%f%f",&a,&b,&c);
if( a==0)
{
printf("\nTHE GIVEN EQUATION IS
LINEAR");
}
else
{
d=(b*b)-(4*a*c);
if(d>0)
{
printf("\n ROOTS ARE UNEQUAL");
r1=((-b)+(sqrt(d)))/(2*a);
r2=((-b)-(sqrt(d)))/(2*a);
printf("\n ROOT1= %f\n ROOT2=
%f",r1,r2);
}
else if(d<0)
{
printf("\n ROOTS ARE IMAGINERY AND
COMPLEX");
}
else if(d==0)
{
printf("\n ROOTS ARE EQUAL");
r=((-b)+(sqrt(d)))/(2*a);
printf("\n ROOT= %f",r);
}
}
getch( );
}
Output:
ENTER a, b, c VALUES OF
QUADRETIC EQUATION
0
1
2
The given equation is linear
ENTER a, b, c VALUES OF
QUADRETIC EQUATION
3
6
3
roots are equal
ROOT= -6.000000
ENTER a, b, c VALUES OF
QUADRETIC EQUATION
1
2
3
roots are IMAGINERY AND COMPLEX
ENTER
a, b, c VALUES OF QUADRETIC EQUATION
1
4
3
roots are UNEQUAL
ROOT1= -1.000000
ROOT2= -3.000000
Calculate Area,
Circumference, of Circle and Volume of Sphere & Hemisphere using Functions
and Switch statement.
Algorithm:
Step1: Start
Step2: Read ch
Step3: Is ch=1 then go to step5 otherwise go to step 6
Step4: call aoc( )
Step5: Is ch=2 then go to step7 otherwise go to step 8
Step6: call coc( )
Step7: Is ch=3 then go to step 9 otherwise go to step 10
Step8: call vos( )
Step9: Is ch=4 then go to step11 otherwise go to step 12
Step10: call vohs( )
Step11: print the entered choice not matched
Step12: stop
Algorithm for
functions designing:
(1)aoc( ):
Step1: Start
Step2 : Read r
Step3: Compute a = 3.141*(r*r)
Step4: Print a and go to step 5
Step5: Stop
(2)coc( ):
Step1: Start
Step2: Read r
Step3: Compute c=2*3.141*r
Step4: Print ‘c’ and go to step5
Step5: Stop
(3) vos( ):
Step1: Start
Step2: Read r
Step3: Compute v1 =(4/3)*(3.141)*(r*r*r)
Step4: Print v1 and go to step 5
Step5: Stop
(4) vohs():-
Step1: Start
Step2: Read r
Step3: Compute v2=(2/3)*3.141*(r*r*r)
Step4: Print v2 and go to step5
Step5: Stop
/* WAP in C to Calculate Area,
Circumference, of Circle and Volume of Sphere & Hemisphere using Functions
and Switch statement */
#include<stdio.h>
main( )
{
int choice;
clrscr();
printf("\tMAIN
MENU\n");
printf("\t----
----\n");
printf("\t1.AREA OF
CIRCLE\n");
printf("\t2.CIRCUMFRENCE
OF CIRCLE\n");
printf("\t3.VOLUME OF
SPHERE\n");
printf("\t4.VOLUME OF
HEMI SPHERE\n");
printf("\n\tENTER YOUR
CHOICE:\n\t");
scanf("%d",&choice);
switch(choice)
{
case 1 :
aoc( );
break;
case 2 :
coc( );
break;
case 3 :
vos( );
break;
case 4 :
vosh( );
break;
default:
printf("\n\t ENTER VALID
CHOICE");
break;
}
getch();
}
aoc( )
{
int r,a;
printf("\n\tENTER THE VALUE OF
RADIUS:");
scanf("%d",&r);
a=(3.141*r*r);
printf("\n\tTHE AREA OF CIRCLE
IS:%d",a);
}
coc( )
{
int r,c;
printf("\n\tENTER THE
VALUE OF R:");
scanf("%d",&r);
c=(2*3.141*r);
printf("\n\tTHE
CIRCUMFRENCE OF CIRCLE IS: %d",c);
}
vos( )
{
int r,v1;
printf("\n\tENTER THE
VALUE OF R:");
scanf("%d",&r);
v1= ((4*3.141*r*r*r)/3);
printf("\n\tTHE VOLUME
OF SPHERE IS:%d",v1);
}
vosh( )
{
int r;
float v2;
printf("\n\tENTER THE
VALUE OF R:");
scanf("%d",&r);
v2= ((2*3.141*r*r*r)/3);
printf("\n\tTHE VOLUME
OF HEMISPHERE IS:%f",v2);
}
Output:
MAIN MENU
-------- ---------
1.AREA OF CIRCLE
2.CIRCUMFRENCE OF CIRCLE
3.VOLUME OF SPHERE
4.VOLUME OF HEMI SPHERE
ENTER YOUR CHOICE
1
ENTER THE VALUE OF R:3
THE AREA OF CIRCLE IS: 28
MAIN MENU
-------- ---------
1.AREA OF CIRCLE
2.CIRCUMFRENCE OF CIRCLE
3.VOLUME OF SPHERE
4.VOLUME OF HEMI SPHERE
ENTER YOUR CHOICE
2
ENTER THE VALUE OF R:2
THE CIRCUMFRENCE OF CIRCLE IS: 12
MAIN MENU
-------- ---------
1.AREA OF CIRCLE
2.CIRCUMFRENCE OF CIRCLE
3.VOLUME OF SPHERE
4.VOLUME OF HEMI SPHERE
ENTER YOUR CHOICE
3
ENTER THE VALUE OF R:2
THE VOLUME OF SPHERE IS: 25
MAIN MENU
-------- ---------
1.AREA OF CIRCLE
2.CIRCUMFRENCE OF CIRCLE
3.VOLUME OF SPHERE
4.VOLUME OF HEMI SPHERE
ENTER YOUR CHOICE
4
ENTER THE VALUE OF R:4
THE VOLUME OF HEMISPHERE IS:
134.016006
MAIN MENU
-------- ---------
1.AREA OF CIRCLE
2.CIRCUMFRENCE OF CIRCLE
3.VOLUME OF SPHERE
4.VOLUME OF HEMI SPHERE
ENTER YOUR CHOICE
6
ENTER THE VALID CHOICE
FIBBIONACCI SERIES
Algorithm:
Step 1: Start
Step 2: Read n
Step 3: Set x=1,y=o,z=o
Step 4: While n≠o then go to step5 otherwise go to step 10
Step 5: Write z and go to step16
Step 6: Compute z=x+y , Set x=y and y=z and go to step 7
Step 7: Compute n=n-1 and go to step8
Step 8: Go to step4
Step 9: Repeat step 5 to step 7 until n reaches to 0 and go
to step 10
Step 10: Stop
/* WAP in C to find Fibonacci up to given range */
#include<stdio.h>
main( )
{
int n,z=0,x,y;
clrscr( );
printf("ENTER
NUMBER\n");
scanf("%d",&n);
printf("FIBANOCII SERIES\n");
printf("********* *******\n");
x=1;
y=0;
z=0;
while(n!=0)
{
printf("
%d",z);
z=x+y;
x=y;
y=z;
n=n-1;
}
getch( );
}
Out put:
Enter number
9
FIBANOCII SERIES
********* ********
0 1 1
2 3 5 8 13 21
Enter number
5
FIBANOCII SERIES
********* ********
0 1 1
2 3
Armstrong Number
Algorithm:
Step 1: Start
Step 2: Read n
Step 3: Set rim=0, num, a=0, num=n
Step 4: While n≠0 then go to step5 otherwise go to step8
Step 5: Compute r=n%10, n = n/10, a= a+(rim * rim * rim)
Step 6: Is num= =a then print the given the number is Armstrong
number and go to step8 otherwise go to step7
Step 7: Print the given number is not Armstrong
Step 8: Stop
/* WAP in C to find given
number is Armstrong or not */
#include<stdio.h>
main( )
{
int n,rim=0,num,a=0;
clrscr( );
printf("ENTER
NUMBER\n");
scanf("%d",&n);
num=n;
while(n!=0)
{
rim=n%10;
a= a+(rim * rim * rim);
n=n/10;
}
num= =a ? printf("\n%d IS
ARMSTRONG NUMBER",num): printf("\n%d IS NOT ARMSTRONG
NUMBER",num);
getch( );
}
Output:
ENTER NUMBER
371
371 IS ARMSTRONG NUMBER
ENTER NUMBER
541
541 IS NOT ARMSTRONG NUMBER
Prime Number in given range.
Algorithm:
Step 1: Start
Step 2: Read n
Step 3: Repeat step 4 until i reaches to n
Step 4: Set dc=0, Repeat step 5 until j reaches to i
Step 5: Is i % j=0 then
Set dc=dc+1, go to step 5
Step 6: Is dc=2 then print i and go to step 3
Step 7: Stop
/* WAP in C to find Prime numbers up to given number */
#include<stdio.h>
main( )
{
int n,i,j,dc=0;
clrscr();
printf("ENTER NUMBER\n");
scanf("%d",&n);
printf("\nPRIME NUMBERS\t");
printf("\n*****
*******\t");
for
(i=1;i<=n;i++)
{
dc=0;
for(j=1;j<=i;j++)
{
if(i%j==0)
{
dc++;
}
}
if(dc==2)
{
printf("\n\t%d",i);
}
}
getch( );
}
OUTPUT:
Enter number
10
PRIME NUMBER
****** ********
2
3
5
7
Enter number
15
PRIME NUMBER
****** ********
2
3
5
7
11
13
Perfect Number
Algorithm:
Step1: Start
Step2: Read n, i
Step3: Set p=0 and
Step4: Repeat step5 and steps6 until i<=n-1
Step5: Is n%i=o then
print the given number is perfect number and go to step 10 otherwise go to step 7
Step6: Compute p=p+i
Step7: Go to step 4
Step8: Is p==n then print the given number is perfect number
and go to step 10. Otherwise go to step 9
Step9: Print the given number is not a perfect number and go
to step 10
Step10: Stop
/*WAP in c to find given number is perfect or not*/
#include<stdio.h>
main( )
{
int
n,i,p=0;
clrscr(
);
printf("ENTER
YOUR NUM\n");
scanf("%d",&n);
for(i=1;i<n;i++)
{
if(n%i==0)
{
p=p+i;
}
}
if(p==n)
{
printf("\n%d IS PERFECT NUMBER", n);
}
else
{
printf("\n%d IS NOT PERFECT NUMBER",n);
}
getch( );
}
Output:
ENTER YOUR NUM
6
6 IS PERFECT NUMBER
ENTER YOUR NUM
9
9 IS NOT PERFECT NUMBER
Factorial using Recursion
Algorithm:
Step 1: Start
Step 2: Read n
Step 3: fact=factorial (n)
Step 4: Print the factorial value of given number and fact.
Step 5: Stop
Algorithm for
function designing:
Step 1: Is n=1 then go to step5. Otherwise go to step6
Step 2: Return the value “l”
Step 3: Compute fact=n*factorial (n-l)
Step 4: Return fact value
/*WAP in c to find Factorial of given number using Recursion*/
#include<stdio.h>
main( )
{
int n,f=0;
clrscr( );
printf("ENTER NUMBER\n");
printf("****** *********\n");
scanf("%d",&n);
f=fact(n);
printf("\n\nFACTORIAL OF %d IS : %d",n,f);
getch( );
}
int fact(int x)
{
if(x<1)
{
return(1);
}
else
{
return(x*fact(x-1));
}
}
OUT PUT:
ENTER NUMBER:
****** *********
5
FACTORIAL OF 5 IS : 120
ENTER NUMBER:
****** *********
7
FACTORIAL OF IS : 5040
Biggest & Smallest
elements in the array
Algorithm:
Step 1: Start
Step 2: Read length of array (number of elements placed in
the array) and store this value in ‘n’
Step 3: Repeat step 4 ‘n’ times until ‘I’ reaches ‘n’ from
‘o’
Step 4: Read a[i]
Step 5: Set big=a[0]
small=a[0]
Step 6: Repeat step7 and step8 ‘n-1’ times where i=1 to
‘n-1’
Step 7: Is a[i]>big then place big=a[i]
Step 8: Is a[i]<small then place small=a[i]
Step 9: Print biggest and smallest elements placed in the
array
Step 10: Stop
/* WAP in C to find BIGGEST AND SMALLEST NUMBER from given list of
number */
#include<stdio.h>
main( )
{
int a[20],s,i,j,big,small;
clrscr();
printf("ENTER SIZE OF
ARRAY\n");
scanf("%d",&s);
printf("ENTER NUMBERS IN
ARRAY\n");
for(i=1;i<=s;i++)
{
scanf("%d",&a[i]);
}
big = small = a[1];
for(i=1;i<=s;i++)
{
if(big < a[i])
{
big=a[i];
}
if(small > a[i])
{
small=a[i];
}
}
printf("\t\nTHE BIGGEST
NUMBER IS: %d",big);
printf("\t\nTHE SMALLEST
NUMBER IS:%d",small);
getch();
}
OUTPUT:
ENTER THE SIZE OF ARRAY
5
ENTER NUMBERS IN ARRAY
15
65
98
63
65
THE BIGGEST NUMBER IS: 98
THE SMALLEST NUMBER IS: 15
ENTER THE SIZE OF ARRAY
8
ENTER NUMBERS IN ARRAY
5
65
98
63
65
102
980
15
THE BIGGEST NUMBER IS: 980
THE SMALLEST NUMBER IS: 5
Sorting Elements of array
Algorithm:
Step 1: Start
Step 2: Read length of array and place this value in n
Step 3: Read step 4 “n” times until ‘i’ reaches to n from
‘o’
Step 4: Print a[i]
Step 5: Repeat step6 for ‘n’ times until ‘i’ reaches to ‘n’
from ‘a’
Step 6: Print a[i]
Step 7: Repeat step8 for ‘n-1’ times until ‘i’ reaches to
‘n-1’ from ‘1’
Step 8: Repeat step9 for ‘n’ times until ‘j’ reaches to ‘n’
from ‘i+1’
Step 9: Is a[i]>a[j] then go to step 10 otherwise go to
step 7
Step 10: Set t=a[i]
a[i]=a[j]
a[j]= t
Step 11: Repeat step 10 for n times until ‘I’ reaches to ‘n’
from ‘o’
Step 12: Print a [i ] and go to step 13
Step 13: Stop
/* WAP in C to Sorting Elements of array*/
#include<stdio.h>
main( )
{
int a[20],i,j,temp=0,n;
clrscr( );
printf("ENTER SIZE OF
ARRAY\n");
scanf("%d",&n);
printf("ENTER NUMBERS IN
ARRAY\n");
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
printf("\nBEFORE SORTING
NUMBERS\n");
printf("\n****** *******
*******\n");
for(i=1;i<=n;i++)
{
printf("%d\n",a[i]);
}
for(i=1;i<=n;i++)
{
for(j=i+1;j<=n;j++)
{
if(a[i] < a[j])
{
temp=a[i];
a[i]=a[j];
a[j]=temp;
}
}
}
printf("AFTER SORTING NUMBERS\n");
printf("***** ******* *******\n");
for(i=1;i<=n;i++)
{
printf("%d\n",a[i]);
}
getch();
}
OUTPUT:
ENTER SIZE OF ARRAY
5
ENTER NUMBERS IN ARRAY
12
65
36
54
98
BEFORE SORTING NUMBERS
******* ********* **********
12
65
36
54
98
AFTER SORTING NUMBERS
******* ******** *********
12
36
54
65
98
Find given Number from an array
Algorithm:
Step 1: Start
Step 2: Read length of array and place this value in s
Step 3: go to step 4 “n” times until ‘i’ reaches to n from
‘0’
Step 4: read a[i]
Step 5 : Read the number want to search n
Step 6 : go to step 7 “n” times until ‘i’ reaches to n from
‘0’
Step7 : Is a[i]=n than
Set
found=1 otherwise fount=0
Step 8: Is found=1 than
Display number is found otherwise not found
Step 9: stop
/* WAP in C to Find given number from an array*/
#include<stdio.h>
main( )
{
int a[20],i,s,n,found=0;
clrscr( );
printf("ENTER SIZE OF
ARRAY\n");
scanf("%d",&s);
printf("ENTER NUMBERS IN
ARRAY\n");
for(i=1;i<=s;i++)
{
scanf("%d",&a[i]);
}
printf("\nENTER NUMBER
FOR SEARCHING\n");
scanf("%d",&n);
for(i=1;i<=s;i++)
{
if(n==a[i])
found=1;
}
if(found==1)
{
printf("ELEMENT IS FOUND IN
ARRAY\n");
}
if(found==0)
{
printf("ELEMENT IS NOT FOUND IN
ARRAY\n");
}
getch( );
}
OUTPUT:
ENTER SIZE OF ARRAY
5
ENTER NUMBERS IN ARRAY
12
23
65
98
96
ENTER NUMBER FOR SEARCHING
98
ELEMENT IS FOUND IN ARRAY
ENTER SIZE OF ARRAY
5
ENTER NUMBERS IN ARRAY
12
23
65
98
96
ENTER NUMBER FOR SEARCHING
44
ELEMENT IS NOT FOUND IN ARRAY
Sum of Diagonal elements of Square Matrix
Algorithm:
Step 1: start
Step 2: Read number of rows and columns of matrix and store
the values in r,c
Step 3: set sum=0
Step 4: Repeat step6 ‘n’ times until ‘I’ reaches to ‘n’ from
‘o’
Step 5: Repeat step6
‘m’ times until j reaches to m from ‘o’
Step 6: Read a[i] [j]
Step 7: r=c then go to step8 otherwise go to step 12
Step 8: Repeat step10 until ‘i’ reaches to ‘r from ‘o’
Step 9: repeat step10 until ‘j’ reaches to c from ‘o’
Step 10: Is i=j then
Compute
sum=sum+a[i][j] otherwise go to step8
Step 11: Display sum and go to step 13
Step 12: Print the given matrix is not square matrix and go
to step 13
Step 13: stop
/* WAP in C to find Sum of Diagonal of Square Matrix */
#include<stdio.h>
main()
{
int a[5][5], r,c,i,j,sum=0;
clrscr( );
printf("ENTER SIZE OF
MATRIX:\n");
scanf("%d%d"
,&r,&c);
if((r==c)
{
printf("ENTER ELEMNTS
OF MATRIX\n");
for(i=0;i<r;i++)
{
for(j=0;j<c;j++)
{
scanf("%d",&a[i][j]);
}
}
printf("SUM OF
DIGONAL OF MATRIX IS:");
for(i=0;i<r;i++)
{
for(j=0;j<c;j++)
{
if( i == j) sum= c[i][j] +a[i][j];
}
}
printf(" %d",sum);
}
else
{
printf("MATRIX IS NOT SQUARE
MATRIX");
}
getch();
}
OUTPUT:
ENTER SIZE OF MATRIX:
2
2
ENTER ELEMNTS OF MATRIX
1
2
3
4
SUM OF DIGONAL OF MATRIX IS: 6
ENTER SIZE OF MATRIX:
2
3
MATRIX IS NOT SQUARE MATRIX
Matrix Multiplication
Algorithm:
Step 1: Start
Step 2: Read number of rows and columns of first matrix and
store the values in r1,c1
Step 3: Repeat step4 and step5 until i reaches to ‘r1’ from
‘0’
Step 4: Repeat step5 until j reaches to ‘c1’ from ‘0’
Step 5: Read a[i][j]
Step 6: Read number of rows and columns for second matrix
and store them in r2, c2
Step 7: Repeat step8 and step9 until i reaches to ‘r2’ from
‘0’
Step 8: Repeat step9 until j reaches to ‘c2’ from ‘0’
Step 9: Read b[i][j]
Step 10: Is c2=r1
then go to step11 otherwise go to step 19
Step 11: Repeat step12 to step15 until i reaches to ‘r1’
from ‘0’
Step 12: Repeat step13 to step 15 until j reaches to ‘c2’
from ‘0’
Step13: Set c[i][j]=o
Step14: Repeat step15 until k reaches to ‘c1’ from ‘0’
Step15: Compute c[i][j]=c[i][j]+a[i][k]*b[k][i]
Step16: Repeat step17 and step 18 until i reaches to ‘r1’ from ‘0’
Step17: Repeat step18 until j reaches to ‘c2’ from ‘0’
Step18: Print c[i][j] and go to step20
Step19: Matrix multiplication is not possible and goes to
step 20
Step 20: Stop
/* WAP in C to Find Multiplication of two Matrix */
#include<stdio.h>
main( )
{
int
a[5][5],b[5][5],c[5][5],r1,c1,r2,c2,k,i,j;
clrscr( );
printf("ENTER SIZE OF
FIRST(A) MATRIX:\n");
scanf("%d%d"
,&r1,&c1);
printf("ENTER ELEMNTS OF
FIRST(A) MATRIX\n");
for(i=0;i<r1;i++)
{
for(j=0;j<c1;j++)
{
scanf("%d",&a[i][j]);
}
}
printf("ENTER SIZE OF
SECOND(B) MATRIX:\n");
scanf("%d%d"
,&r2,&c2);
printf("ENTER ELEMENTS OF
SECOND(B) MATRIX\n");
for(i=0;i<r2;i++)
{
for(j=0;j<c2;j++)
{
scanf("%d",&b[i][j]);
}
}
if(c1==r2)
{
printf("MATRIX
MULTIPLICATION IS POSSIBLE\n");
printf("RESULTANT
MATRIX IS:\n");
for(i=0;i<r1;i++)
{
for(j=0;j<c2;j++)
{
c[i][j]=0;
for(k=0;k<r1;k++)
{
c[i][j] = c[i][j]+(a[i][k]*b[k][j]) ;
}
}
}
for(i=0;i<r1;i++)
{
for(j=0;j<c2;j++)
{
printf(" %d",c[i][j]);
}
printf("\n");
}
}
else
{
printf("MATRIX
MULTIPLICATION IS NOT POSSIBLE");
}
getch( );
}
OUTPUT:
ENTER SIZE OF FIRST(A) MATRIX:
2
2
ENTER ELEMNTS OF FIRST(A) MATRIX
1
2
3
4
ENTER SIZE OF SECOND(B) MATRIX:
2
1
ENTER ELEMENTS OF SECOND(B) MATRIX
7
8
MATRIX MULTIPLICATION IS POSSIBLE
RESULTANT MATRIX IS:
23
53
String Functions
Algorithm:
Step 1: Start
Step 2: Read input strings into s1,s2
Step 3: Compute length of string s1
Step 4: Concate s1
and s2 save on s1, display s1
Step 5: Convert
string s1 into lower case and display s1
Step 6: Convert string s2 into upper case and display s2
Step 7: Reverse the
string s1 and save on s1 and display s1
Step 8: Copy s2 into s1 and display s1
Step 9: Stop
/* Program to Demonstrate String Functions in C Language*/
#include<stdio.h>
#include<string.h>
main( )
{
char s1[100],s2[100];
clrscr( );
printf("ENTER
STRINGS\n");
gets(s1);
gets(s2);
printf("\nSTRING
OPERATIONS USING STRING FUNCTIONS\n\n");
puts("LENGTH OF STRING S1
IS: ");
printf("%d",strlen(s1));
puts("\nCONCATING OF S1
AND S2 IS: ");
strcat(s1,s2);
puts(s1);
puts("\nLOWER CASE OF
STRING S1 IS: ");
strlwr(s1);
puts(s2);
puts("\nUPPER CASE OF
STRING S2 IS: " );
strupr(s2);
puts(s2);
puts("\n\nREVERSE OF
STRING S1 IS: ");
strrev(s1);
puts(s1);
strrev(s1);
puts("\nCOPYING S2 INTO
S1: ");
strcpy(s1,s2);
puts(s1);
getch( );
}
OUTPUT:
ENTER STRINGS
computer
science
STRING OPERATIONS USING STRING FUNCTIONS
LENGTH OF STRING S1 IS:
9
CONCATING OF S1 AND S2 IS:
COMPUTER science
UPPER CASE OF STRING S2 IS:
SCIENCE
LOWER CASE OF STRING S1 IS:
computer
REVERSE OF STRING S1 IS:
retupmoc
COPYING S2 INTO S1:
SCIENCE
String Palindrome
Algorithm:
Step 1: Start
Step 2: Read input string into ‘s1’
Step 3: Set s2=null
Step 4: Reverse s1
and save on s2
Step 5: Is s1=s2
then go to step 6 otherwise go to step 7
Step 6: Print the given string is palindrome and go step 8
Step 7: Print the given string is not a palindrome and go to
step 8
Step 8: stop
/*Write a C Program to Find Given String is Palindrome or not*/
#include<stdio.h>
#include<string.h>
main( )
{
char s1[100],s2[100];
int i;
clrscr();
printf("ENTER STRING
S1\n");
gets(s1);
strcpy(s2,s1);
puts("\n\nREVERSE OF
STRING S1 IS: ");
strrev(s1);
if(strcmp(s1,s2)==0)
{
printf("\n%s IS PALINDROM",s2);
}
else
{
printf("\n%s IS NOT
PALINDROM",s2);
}
getch( );
}
OUTPUT:
ENTER STRINGS
computer
REVERSE OF STRING S1 IS:
retupmoc
computer IS NOT PALINDROM
ENTER STRINGS
amma
REVERSE OF STRING S1 IS:
amma
amma IS PALINDROM
Pointers
Algorithm:
Step 1: Start
Step 2: Read *p, *q, *r
Step 3: Set a=10, b=20, c=0
Step 4: Set a=*p,
b=*q, c=*r
Step 5: Compute *r =
*p +*q
Step 6: Print address of a is p, Print value of a is *p
Step 7: Print address
of b is q, Print value of b is *q
Step 8: Print address of c is r, Print value of c is *r
Step 9: Stop
/*Program to Demonstrate Pointers in C Language*/
#include<stdio.h>
main( )
{
int a=10, b=20, c=0;
int *p,*q,*r;
clrscr( );
p=&a;
q=&b;
r=&c;
*r=*p+*q;
printf("\n\n
a=%d",a);
printf("\n\n address
of a=%u",p);
printf("\n\n
b=%d",b);
printf("\n\n address
of b=%u",q);
printf("\n\n
c=%d",c);
printf("\n\n address of c=%u",r);
printf("\n\n final
result");
printf("\n\n%d",*r);
getch ( );
}
OUTPUT:
a = 30
address of a = 65486
b = 20
address of a = 65488
c = 30
address of a = 65490
final result
30
Design “Student Profile’’ using Structures
Algorithm:
Step 1: Start
Step 2: Define structure with name ‘faculty’
Step 3: Declare the members of structure as follows.
Step 4: (1) roll as integer
(2) name
as character
(3)cls as
character
(4) fee as
float
Step 5: Declare structure variable stu[10]
Step 6: Read ‘n’ for strength of faculty
Step 7: Repeat step8 to step11 until ‘i’ reaches to ‘n’ from
‘0’
Step 8: Read stu[i].roll
Step 9: Read stu[i].name
Step 10: Read stu[i].cls
Step 11: Read stu[i].fee
Step 12: Repeat step13 to step16 until ‘i’ reaches to ‘n’ from
‘0’
Step 13: Print stu[i].roll
Step 14: Print stu[i].name
Step 15: Print stu[i].cls
Step 16: Print stu[i].fee
Step 17:-stop
/*WAP in C to Create Structure for Student Profile*/
#include<stdio.h>
#include<string.h>
main( )
{
int n,i;
struct studet
{
int roll;
char
name[20];
char cls[5];
int fee;
};
struct studet stu[10];
clrscr( );
printf("ENTER NUMBER OF
STUDENTS\n");
scanf("%d",&n);
for(i=1;i<=n;i++)
{
printf("\n ENTER %d
STUDENT DETAIL", i);
puts("\nROLL:");
scanf("%d",&stu[i].roll);
puts("\nNAME:");
scanf("%s",stu[i].name);
puts("\nCLASS:");
scanf("%s",stu[i].cls);
puts("\nFEE:");
scanf("%d",&stu[i].fee);
}
for(i=1;i<=n;i++)
{
printf("\n %d
STUDENT DETAIL",i);
printf("\nROLL:%d",stu[i].roll);
printf("\nNAME:%s",stu[i].name);
printf("\nCLASS:%s",stu[i].cls);
printf("\nFEE:%d",stu[i].fee);
}
getch();
}
Output:
ENTER NUMBER OF STUDENTS
5
ENTER 1 STUDENT DETAIL
ROLL:
01
NAME:
satish
CLASS:
b.sc
FEE:
45000
ENTER 2 STUDENT DETAIL
ROLL:
02
NAME:
lakshman
CLASS:
b.com
FEE:
25000
ENTER 3 STUDENT DETAIL
ROLL:
03
NAME:
d.v.v.satish
CLASS:
bba
FEE:
35000
ENTER 4 STUDENT DETAIL
ROLL:
04
NAME:
sarma
CLASS:
bca
FEE:
25000
ENTER 5 STUDENT DETAIL
ROLL:
05
NAME:
durga rao
CLASS:
b.sc
FEE:
15000
1 STUDENT DETAIL
ROLL: 01
NAME: satish
CLASS: bba
FEE:35000
2 STUDENT DETAIL
ROLL: 02
NAME: lakshman
CLASS: b.com
FEE:25000
3 STUDENT DETAIL
ROLL: 03
NAME: d.v.v.satish
CLASS: bba
FEE:35000
4 STUDENT DETAIL
ROLL: 04
NAME: sarma
CLASS: bca
FEE:25000
5 STUDENT DETAIL
ROLL: 05
NAME: durgarao
CLASS: b.sc
FEE:15000
Operations on Files
Algorithm:
Step 1: Start
Step 2: Open the file in “write mode”
Step 3: Read some text into the file until the file is end
Step 4: Close the file
Step 5: Open the file in read mode
Step 6: Display the information placed on files
Step 7: Close the file
Step 8: Stop
/*WAP in C to Create File of Student Details*/
#include<stdio.h>
#include<string.h>
main( )
{
FILE *fp;
int roll[20],n,i;
char name[20][20];
char cls[5][5];
clrscr();
fp=fopen("d:\\sridhar\\clab\\final\\studet.doc","w+");
printf("ENTER NUMBER OF
STUDENTS\n");
scanf("%d",&n);
for(i=1;i<=n;i++)
{
printf("\n ENTER %d
STUDENT DETAIL", i);
puts("\nROLL:");
scanf("%d",&roll[i]);
fprintf(fp,"\n%d",roll[i]);
puts("\nNAME:");
scanf("%s",name[i]);
fprintf(fp,"\t%s",name[i]);
puts("\nCLASS:");
scanf("%s",cls[i]);
fprintf(fp,"\t%s",cls[i]);
}
printf("\n STUDENT
DETAIL");
printf("\nROLL\tNAME\tCLASS");
if(fp==NULL)
{
printf("\nOOPS NO
DATA\n");
}
else
{
for(i=1;i<=n;i++)
{
fscanf(fp,"%d",&roll[i]);
fscanf(fp,"%s",name[i]);
fscanf(fp,"%s",cls[i]);
printf("\n%d\t%s\t%s",roll[i],name[i],cls[i]);
}
}
fclose(fp);
getch();
}
OUTPUT:
ENTER NUMBER OF STUDENTS
5
ENTER 1 STUDENT DETAIL
ROLL:
1
NAME:
satish
CLASS:
b.sc
ENTER 2 STUDENT DETAIL
ROLL:
2
NAME:
lakshman
CLASS:
b.com
ENTER 3 STUDENT DETAIL
ROLL:
3
NAME:
d.satish
CLASS:
bba
ENTER 4 STUDENT DETAIL
ROLL:
4
NAME:
sarma
CLASS:
bca
ENTER 5 STUDENT DETAIL
ROLL:
5
NAME:
durgarao
CLASS:
b.sc
FILE CREATED AT “d:\\sridhar\\clab\\final\\studet.doc”
1 satish b.sc
2 lakshman b.com
3 d.satish bba
4 sarma bca
5 durgarao b.sc
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