C program-3

1. /* Write a C program to find the biggest of three numbers*/

#include
#include

main()
{
int a, b, c;
clrscr();

printf("Enter the values of a,b and c\n");
scanf ("%d %d %d", &a, &b, &c);

printf ("a = %d\tb = %d\tc = %d\n", a,b,c);

if ( a > b)
{
if ( a > c)
{
printf ("A is the greatest among three\n");
}
else
{
printf ("C is the greatest among three\n");
}
}
else if (b > c)
{
printf ("B is the greatest among three\n");
}
else
printf ("C is the greatest among three\n");


}
/*-----------------------------
Output
Enter the values of a,b and c
23 32 45
a = 23 b = 32 c = 45
C is the greatest among three

RUN2
Enter the values of a,b and c
234
678
195
a = 234 b = 678 c = 195
B is the greatest among three

RUN3
Enter the values of a,b and c
30 20 10
a = 30 b = 20 c = 10
A is the greatest among three
------------------------------*/

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

2. /* write a C program to find and output all the roots of a *
* quadratic equation, for non-zero coefficients. In case *
* of errors your program should report suitable error message*/

#include
#include
#include

main()
{
float A, B, C, root1, root2;
float realp, imagp, disc;
clrscr();

printf("Enter the values of A, B and C\n");
scanf("%f %f %f", &A,&B,&C);

if( A==0 || B==0 || C==0)
{
printf("Error: Roots cannot be determined\n");
exit(1);
}
else
{
disc = B*B - 4.0*A*C;
if(disc < 0) { printf("Imaginary Roots\n"); realp = -B/(2.0*A) ; imagp = sqrt(abs(disc))/(2.0*A); printf("Root1 = %f +i %f\n",realp, imagp); printf("Root2 = %f -i %f\n",realp, imagp); } else if(disc == 0) { printf("Roots are real and equal\n"); root1 = -B/(2.0*A); root2 = root1; printf("Root1 = %f \n",root1); printf("Root2 = %f \n",root2); } else if(disc > 0 )
{
printf("Roots are real and distinct\n");
root1 =(-B+sqrt(disc))/(2.0*A);
root2 =(-B-sqrt(disc))/(2.0*A);
printf("Root1 = %f \n",root1);
printf("Root2 = %f \n",root2);
}
}

} /* End of main() */

/*---------------------------
Output
RUN 1
Enter the values of A, B and C
3 2 1
Imaginary Roots
Root1 = -0.333333 +i 0.471405
Root2 = -0.333333 -i 0.471405

RUN 2
Enter the values of A, B and C
1 2 1
Roots are real and equal
Root1 = -1.000000
Root2 = -1.000000

RUN 3
Enter the values of A, B and C
3 5 2
Roots are real and distinct
Root1 = -0.666667
Root2 = -1.000000
---------------------------------*/


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

3. /* Write a C program to simulate a simple calculator to perform *
* arithmetic operations like addition, subtraction,multiplication *
* and division only on integers. Error message should be repoetrd *
* if any attempt is made to divide by zero */

#include

main()
{
char operator;
float n1, n2;
float sum, diff, prod, quot, result;

clrscr();

printf ("Simulation of Simple Calculator\n\n");

printf("Enter two numbers\n");
scanf ("%f %f", &n1, &n2);

fflush (stdin);

printf("Enter the operator [+,-,*,/]\n");
scanf ("%c", &operator);


switch (operator)
{
case '+': result = n1 + n2;
break;
case '-': result = n1 - n2;
break;
case '*': result = n1 * n2;
break;
case '/': result = n1 / n2;
break;
default : printf ("Error in operation\n");
break;
}
printf ("\n%5.2f %c %5.2f= %5.2f\n", n1,operator, n2, result);

}
/*-----------------------------
Output
Simulation of Simple Calculator

Enter two numbers
3 5
Enter the operator [+,-,*,/]
+

3.00 + 5.00= 8.00


RUN2
Simulation of Simple Calculator

Enter two numbers
12.75
8.45
Enter the operator [+,-,*,/]
-

12.75 - 8.45= 4.30

RUN3
Simulation of Simple Calculator

Enter two numbers
12 12
Enter the operator [+,-,*,/]
*

12.00 * 12.00= 144.00


RUN4
Simulation of Simple Calculator

Enter two numbers
5
9
Enter the operator [+,-,*,/]
/

5.00 / 9.00= 0.56


------------------------------*/

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


4. /* Write a C program to simulate a simple calculator to perform *
* arithmetic operations like addition, subtraction,multiplication *
* and division only on integers. Error message should be repoetrd *
* if any attempt is made to divide by zero */

#include

main()
{
char operator;
float n1, n2;
float sum, diff, prod, quot, result;

clrscr();

printf ("Simulation of Simple Calculator\n\n");

printf("Enter two numbers\n");
scanf ("%f %f", &n1, &n2);

fflush (stdin);

printf("Enter the operator [+,-,*,/]\n");
scanf ("%c", &operator);


switch (operator)
{
case '+': result = n1 + n2;
break;
case '-': result = n1 - n2;
break;
case '*': result = n1 * n2;
break;
case '/': result = n1 / n2;
break;
default : printf ("Error in operation\n");
break;
}
printf ("\n%5.2f %c %5.2f= %5.2f\n", n1,operator, n2, result);

}
/*-----------------------------
Output
Simulation of Simple Calculator

Enter two numbers
3 5
Enter the operator [+,-,*,/]
+

3.00 + 5.00= 8.00


RUN2
Simulation of Simple Calculator

Enter two numbers
12.75
8.45
Enter the operator [+,-,*,/]
-

12.75 - 8.45= 4.30

RUN3
Simulation of Simple Calculator

Enter two numbers
12 12
Enter the operator [+,-,*,/]
*
12.00 * 12.00= 144.00


RUN4
Simulation of Simple Calculator

Enter two numbers
5
9
Enter the operator [+,-,*,/]
/

5.00 / 9.00= 0.56


------------------------------*/

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

5. /*Write a C program to generate and print first N FIBONACCI numbers*/

#include
main()
{
int fib1=0, fib2=1, fib3, N, count=0;

printf("Enter the value of N\n");
scanf("%d", &N);

printf("First %d FIBONACCI numbers are ...\n", N);
printf("%d\n",fib1);
printf("%d\n",fib2);
count = 2; /* fib1 and fib2 are already used */

while( count < N) { fib3 = fib1 + fib2; count ++; printf("%d\n",fib3); fib1 = fib2; fib2 = fib3; } } /* End of main() */ /*-------------------------- Enter the value of N 10 First 5 FIBONACCI numbers are ... 0 1 1 2 3 5 8 13 21 34 -------------------------------*/ ******************************************************************** 6. /* Write a C program to find the GCD and LCM of two integers * * output the results along with the given integers. Use Euclids' algorithm*/ #include

main()
{
int num1, num2, gcd, lcm, remainder, numerator, denominator;
clrscr();

printf("Enter two numbers\n");
scanf("%d %d", &num1,&num2);

if (num1 > num2)
{
numerator = num1;
denominator = num2;
}
else
{
numerator = num2;
denominator = num1;
}
remainder = num1 % num2;
while(remainder !=0)
{
numerator = denominator;
denominator = remainder;
remainder = numerator % denominator;
}
gcd = denominator;
lcm = num1 * num2 / gcd;
printf("GCD of %d and %d = %d \n", num1,num2,gcd);
printf("LCM of %d and %d = %d \n", num1,num2,lcm);
} /* End of main() */
/*------------------------
Output
RUN 1
Enter two numbers
5
15
GCD of 5 and 15 = 5
LCM of 5 and 15 = 15
------------------------------*/

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