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printf("Difference of %d and %d is : %d",a,b,a-b);
break;
case 3 :
printf("Multiplication of %d and %d is : %d",a,b,a*b);
break;
case 4 :
printf("Division of Two Numbers is %d : ",a/b);
break;
case 5:
printf("Remainder of %d and %d is : %d",a,b,a%b);
break;
default :
printf(" Enter Your Correct Choice.");
break;
}
}
Output:
Enter the values of a & b:
5
5
Enter your choice
1
Sum of a and b is : 10
2. Develop a program to compute the roots of a quadratic equation by accepting
the coefficients. Print the appropriate messages
Algorithm
Step-1: [Read values of A, B, C]-Read A, B, C.
Step-2: [Check for Co-efficient]-If A=B=C=0 then roots are non-determinant
Step-3:[Compute the value Disc]- Disc=B*B-4*A*C;
Step-4:[Different roots are obtained depending on the value of disc]-
If Disc is lesser than 0 than goto step 5
If Disc is equals to 0 than goto step 6
If Disc is greater than 0 than goto step 7.](https://image.slidesharecdn.com/labmanualcpl21scheme-2-221016024033-45d0dea8/75/LAB_MANUAL_cpl_21scheme-2-pdf-4-2048.jpg)
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Program:
#include<stdio.h>
#include<string.h>
void main()
{
int cust_no, unit_con;
float charge,surcharge=0, amt, total_amt;
char nm[25];
printf("Enter the customer IDNO :t");
scanf("%d",&cust_no);
printf("Enter the customer Name :t");
scanf("%s",nm);
printf("Enter the unit consumed by customer :t");
scanf("%d",&unit_con);
if (unit_con <200 )
charge = 0.80;
else if (unit_con>=200 && unit_con<300)
charge = 0.90;
else
charge = 1.00;
amt = unit_con*charge;
if (amt>400)
surcharge = amt*15/100.0;
total_amt = amt+surcharge;
printf("tttnElectricity Billnn");
printf("Customer IDNO :t%d",cust_no);
printf("nCustomer Name :t%s",nm);
printf("nunit Consumed :t%d",unit_con);
printf("nAmount Charges @Rs. %4.2f per unit :t%0.2f",charge,amt);
printf("nSurchage Amount :t%.2f",surcharge);
printf("nMinimum meter charge Rs :t%d",100);
printf("nNet Amount Paid By the Customer :t%.2f",total_amt+100);
}](https://image.slidesharecdn.com/labmanualcpl21scheme-2-221016024033-45d0dea8/75/LAB_MANUAL_cpl_21scheme-2-pdf-10-2048.jpg)
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Output
:
4. Implement binary search on Integers/Numbers
Step 1: Start
Step 2: Read size of the array n
Step 3: Read the list of elements in sorted order a[i].
Step 4: Read the key element in key
Step 5: initialize low=0 and high= n-1
Step 6: Check if low is less than or equal to high. If condition is true, goto step
7, other wise goto Step 11
Step 7: find the middle element.i.e. mid=(low+high)/2;
Step 8: if key element is equal to mid element, then initialize loc value, loc =
mid+1; otherwise goto step 9
Step 9: Check if key element is less than mid element is true, then initialize
high=mid-1 then goto step 6, if it false goto step10.
Step 10: Check if key element is greater than mid element is true, then](https://image.slidesharecdn.com/labmanualcpl21scheme-2-221016024033-45d0dea8/75/LAB_MANUAL_cpl_21scheme-2-pdf-11-2048.jpg)
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initialize low=mid+1 then goto step 6.
Step 11: Check if loc value is greater then zero then print the search is
successful then goto step 13, otherwise goto step 12.
Step 12: print search is unsuccessful, then goto step 13.
Step 13: Stop
Program:
#include<stdio.h>
void main()
{
int n, a[100], i, key, high, low, mid, loc=-1;
printf("Enter the size of the arrayn"); scanf("%d",&n);
printf("Enter the elements of array in sorted ordern");
for(i=0;i<n;i++)
scanf("%d",&a[i]);
printf("Enter the key element to be searchedn");
scanf("%d",&key);
low=0; high=n-1;
while(low<=high)
{
mid=(low+high)/2;
if(key<a[mid])
{
loc = mid+1;
break}
if(key<a[mid])
high=mid-1;
else}
}
5. Implement Matrix multiplication and validate the rules of multiplication.
Algorithm
Step-1: Start](https://image.slidesharecdn.com/labmanualcpl21scheme-2-221016024033-45d0dea8/75/LAB_MANUAL_cpl_21scheme-2-pdf-12-2048.jpg)
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Step-2: [Read the order of both matrices M, N, P, Q]
Step-3: Check for i = 0 to M and j = 0 to N for the matrix A, read A[i][j]
Step-4: Check for i = 0 to P and j = 0 to Q for the matrix B, read B[i][j]
Step-5: If in case (N=P) then only multiplication is possible then goto step 6
otherwise goto step 7
Step-6: Initially set matrix C[i][j] as 0
For k=0 to n
C[i][j] = C[i][j] + A[i][k]*B[i][k] than goto step 8
Step-7: Multiplication is not possible
Step-8: Prints the multiplication of the two matrix
Step-9: Stop](https://image.slidesharecdn.com/labmanualcpl21scheme-2-221016024033-45d0dea8/75/LAB_MANUAL_cpl_21scheme-2-pdf-13-2048.jpg)
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Program:
#include<stdio.h>
int main()
{
int a[20][20],b[20][20],c[20][20];
int m,n,p,q,i,j,k;
printf("Enter rows and columns of matrix An");
scanf("%d%d",&m,&n);
printf("Enter rows and columns of matrix Bn");
scanf("%d%d",&p,&q);
if(n!=p)
{
printf("Matrix multiplication not possiblen");
return 0;
}
printf("Enter elements of matrix An");
for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
{
scanf("%d",&a[i][j]);
}
}
printf("Enter elements of matrix Bn");
for(i=0;i<p;i++)
{
for(j=0;j<q;j++)
{
scanf("%d",&b[i][j]);
}
}
for(i=0;i<p;i++)
{
for(j=0;j<q;j++)
{
c[i][j]=0;](https://image.slidesharecdn.com/labmanualcpl21scheme-2-221016024033-45d0dea8/75/LAB_MANUAL_cpl_21scheme-2-pdf-17-2048.jpg)
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for(k=0;k<n;k++)
{
c[i][j]=c[i][j]+a[i][k]*b[k][j];
}
}
}
printf("Product of two matrices isn");
for(i=0;i<m;i++)
{
for(j=0;j<q;j++)
{
printf("%dn",c[i][j]);
}
}
}
Output:
Enter rows and columns of matrix A
2 2
Enter rows and columns of matrix B
2 2
Enter elements of matrix A
6 6
6 6
Enter elements of matrix B
2 2
2 2
Product of two matrices is
24 24
24 24
6. Compute sin(x)/cos(x) using Taylor series approximation. Compare your
result with the built-in library function. Print both the results with appropriate
inferences.
Algorithm
Step-1: Start](https://image.slidesharecdn.com/labmanualcpl21scheme-2-221016024033-45d0dea8/75/LAB_MANUAL_cpl_21scheme-2-pdf-18-2048.jpg)
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Step-2: [Read the value of x in degree]- Read x
Step-3: [Initialization and Radians Conversion]
Temp = x
x=x*(3.142/180.0)
Term = x
sinx = term
n=1
Step-4: [Computer sin value]
Repeat while (term>FLT_EPSILON)
Fact = 2*n*(2*n+1)
Term = term * x * x /fact Sinx
sinx +term = n + 1n
Step-5: [Output]- Print sin(x) without using library function of Print sin(x) and
with using library function
Step-6: Stop](https://image.slidesharecdn.com/labmanualcpl21scheme-2-221016024033-45d0dea8/75/LAB_MANUAL_cpl_21scheme-2-pdf-19-2048.jpg)
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7. Sort the given set of N numbers using Bubble sort.
Algorithm
Step-1: Start
Step-2: Read un-sorted array of n elements into a[i], where i is the index value
Step-3: Initialize index i=0
Step-4: Check for the i if it is less than n. if ture, than goto step-5. Other wise
goto step output array
Step-5: Initialize the index J to zero
Step-6: Check if j is less than (n-i-1). If it is true than goto step-7, otherwise
increment j and goto step-4
Step-7: Check if a[j] is greater than a[j+1](i.e-adjacent elements are compared)
inside the for loop. if true swap the elements using temporary variables,
otherwise goto
step-6
Step-8: Using the for loop output the sorted array elements
Step-9: Stop](https://image.slidesharecdn.com/labmanualcpl21scheme-2-221016024033-45d0dea8/75/LAB_MANUAL_cpl_21scheme-2-pdf-22-2048.jpg)
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Program:
#include<stdio.h>
int main()
{
int i,j,n,temp;
int a[20];
printf("enter the value of n");
scanf("%d",&n);
printf("Enter the numbers in unsorted order:n");
for(i=0;i<n;i++)
scanf("%d", &a[i]);
// bubble sort logic
for(i=0;i<n;i++)
{
for(j=0;j<(n-i)-1;j++)
{
if( a[j]>a[j+1])
{
temp=a[j];
a[j]=a[j+1];
a[j+1]=temp;
}
}
}
printf("The sorted array isn");
for(i=0;i<n;i++)
{
printf("%dn",a[i]);
}
}](https://image.slidesharecdn.com/labmanualcpl21scheme-2-221016024033-45d0dea8/75/LAB_MANUAL_cpl_21scheme-2-pdf-24-2048.jpg)
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Program:
#include<stdio.h>
#include<string.h>
void compare(char [ ],char [ ]); .
void concat(char [ ],char [ ]);
void length(char *[ ]);
void main( )
{
int n,digit;
char str1[10],str2[10];
do
{
printf("press 1-compare 2-concatenate 3-length of string");
printf("n enter your choice=");](https://image.slidesharecdn.com/labmanualcpl21scheme-2-221016024033-45d0dea8/75/LAB_MANUAL_cpl_21scheme-2-pdf-27-2048.jpg)
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scanf("%d",&n);
switch(n)
{
case 1:printf("enter first string=");
scanf("%s",str1);
printf("enter second string=");
scanf("%s",str2);
compare(str1,str2);
break;
case 2: printf("enter first string=");
scanf("%s",str1);
printf("enter second string=");
scanf("%s",str2);
concat(str1,str2);
break;
case 3:printf("enter string=");
scanf("%s",str1);
length(&str1);
break;
default: printf("wrong choice");
break;
}
printf("n Do you want to continue(1/0)? ");
scanf("%d", &digit);
}while(digit==1);
}
void compare(char str1[ ],char str2[ ])
{
int i;
i=strcmp(str1,str2);
if(i==0)
printf("strings are equaln ");
else
printf("string are not equaln");
}
void concat(char str1[ ],char str2[ ])
{
strcat(str1,str2);
printf("concatenate string=%s",str1);](https://image.slidesharecdn.com/labmanualcpl21scheme-2-221016024033-45d0dea8/75/LAB_MANUAL_cpl_21scheme-2-pdf-28-2048.jpg)
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}
void length(char *str1[ ])
{
int len;
len=strlen(str1);
printf("the length of string=%d",len);
}
Output:
9.Implement structures to read, write and compute average- marks and the students scoring
above and below the average marks for a class of N students.
Algorithm:
Step-1: Start
Step-2: Read number of students
Step-3: For every student, read the student id, name , marks for all the subjects
Step-4: Calculate the avarage marks and store it in the avg field
Step-5: Print the results
Step-6: Initialise loop
Step-7: Read teh average of every student
Step-8: Check for if avg>35.00
Step-9: If yes than print the result else goto next interation
Step-10: Initialise the loop
Step-11: Read average of every student
Step-12: Check if avg<35.00
Step-13: If yes than print result else goto next iteration
Step-14: Stop
Flowchart:](https://image.slidesharecdn.com/labmanualcpl21scheme-2-221016024033-45d0dea8/75/LAB_MANUAL_cpl_21scheme-2-pdf-29-2048.jpg)
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ra
am
mC
Co
ol
ll
le
eg
ge
eo
of
fE
En
ng
gi
in
ne
ee
er
ri
in
ng
g,
,A
An
ne
ek
ka
al
l,
,B
Be
en
ng
ga
al
lu
ur
ru
u.
.
3
30
0|
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P a
a g
g e
e
Program:
#include<stdio.h>
struct student
{
char usn[10];
char name[10];
float m1,m2,m3;
float avg,total;
};
void main()
{
struct student s[20];
int n,i;
float tavg,sum=0.0;
printf("Enter the number of student=");](https://image.slidesharecdn.com/labmanualcpl21scheme-2-221016024033-45d0dea8/75/LAB_MANUAL_cpl_21scheme-2-pdf-30-2048.jpg)
![C
Co
om
mp
pu
ut
te
er
rP
Pr
ro
og
gr
ra
am
mm
mi
in
ng
gL
La
ab
bo
or
ra
at
to
or
ry
y2
21
1C
CP
PL
L2
27
7
S
Sr
ri
iS
Sa
ai
ir
ra
am
mC
Co
ol
ll
le
eg
ge
eo
of
fE
En
ng
gi
in
ne
ee
er
ri
in
ng
g,
,A
An
ne
ek
ka
al
l,
,B
Be
en
ng
ga
al
lu
ur
ru
u.
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3
31
1|
|P
P a
a g
g e
e
scanf("%d",&n);
for(i=0;i<n;i++)
{
printf("Enter the detail of %d studentsn",i+1);
printf("n Enter USN=");
scanf("%s",s[i].usn);
printf("n Enter Name=");
scanf("%s",s[i].name);
printf("Enter the three subject scoren");
scanf("%f%f%f",&s[i].m1,&s[i].m2,&s[i].m3);
s[i].total=s[i].m1+s[i].m2+s[i].m3;
s[i].avg=s[i].total/3;
}
for(i=0;i<n;i++)
{
if(s[i].avg>=35)
printf("n %s has scored above the average marks",s[i].name);
else
printf("n %s has scored below the average marks",s[i].name);
}
}
Output](https://image.slidesharecdn.com/labmanualcpl21scheme-2-221016024033-45d0dea8/75/LAB_MANUAL_cpl_21scheme-2-pdf-31-2048.jpg)
![C
Co
om
mp
pu
ut
te
er
rP
Pr
ro
og
gr
ra
am
mm
mi
in
ng
gL
La
ab
bo
or
ra
at
to
or
ry
y2
21
1C
CP
PL
L2
27
7
S
Sr
ri
iS
Sa
ai
ir
ra
am
mC
Co
ol
ll
le
eg
ge
eo
of
fE
En
ng
gi
in
ne
ee
er
ri
in
ng
g,
,A
An
ne
ek
ka
al
l,
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Be
en
ng
ga
al
lu
ur
ru
u.
.
3
33
3|
|P
P a
a g
g e
e
Program:
#include<stdio.h>
#include<math.h>
int main()
{
int n , i;
float x[20],sum,mean;
float variance , deviation;
printf("Enter the value of n n");
scanf("%d",&n);
printf("enter %d real values n",n);
for (i=0;i<n;i++)
{](https://image.slidesharecdn.com/labmanualcpl21scheme-2-221016024033-45d0dea8/75/LAB_MANUAL_cpl_21scheme-2-pdf-33-2048.jpg)
LAB_MANUAL_cpl_21scheme-2.pdf
- 1. C Co om mp pu ut te er rP Pr ro og gr ra am mm mi in ng gL La ab bo or ra at to or ry y2 21 1C CP PL L2 27 7 S Sr ri iS Sa ai ir ra am mC Co ol ll le eg ge eo of fE En ng gi in ne ee er ri in ng g, ,A An ne ek ka al l, ,B Be en ng ga al lu ur ru u. . 1 1| |P P a a g ge e Familiarization with programming environment, the concept of naming the program files, storing, compilation, execution, and debugging. Taking any simple C- code Program: #include<stdio.h> { int l, b, area, peri; printf("Enter the length and the breadthn"); scanf("%d%d", &l, &b); area = l * b; peri = 2 * (l + b); printf("The area is %dn", area); printf("The perimeter is %dn', peri); } Output Enter the length and the breadth 5 4
- 2. C Co om mp pu ut te er rP Pr ro og gr ra am mm mi in ng gL La ab bo or ra at to or ry y2 21 1C CP PL L2 27 7 S Sr ri iS Sa ai ir ra am mC Co ol ll le eg ge eo of fE En ng gi in ne ee er ri in ng g, ,A An ne ek ka al l, ,B Be en ng ga al lu ur ru u. . 2 2| |P P a a g ge e The area is 20 The perimeter is 18 1. Develop a Program to solve simple computational problems using arithmetic expressions and use of each operator leading to the simulation of a Commercial calculator Algorithm: Step-1: Start Step-2: Read the variables a and b Step-3: Options Display Step-4: Read the input choice Step-5: If Choice is 1 then execute step 6 2 then execute step 7 3 then execute step 8 4 then execute step 9 Other than above mentioned choice than execute step 10 Step-6: Addition of a and b Step-7: Subtraction of a and b Step-8: Multiplication of a and b Step-9: Division of a and b Step-10: Invalid choice Step-11: Print the result Step-12: Stop
- 3. C Co om mp pu ut te er rP Pr ro og gr ra am mm mi in ng gL La ab bo or ra at to or ry y2 21 1C CP PL L2 27 7 S Sr ri iS Sa ai ir ra am mC Co ol ll le eg ge eo of fE En ng gi in ne ee er ri in ng g, ,A An ne ek ka al l, ,B Be en ng ga al lu ur ru u. . 3 3| |P P a a g ge e Program: #include<stdio.h> void main() { int a,b; int op; printf(" 1.Additionn 2.Subtractionn 3.Multiplicationn 4.Divisionn 5.Remaindern"); printf("Enter the values of a & b: "); scanf("%d %d",&a,&b); printf("Enter your Choice : "); scanf("%d",&op); switch(op) { case 1 : printf("Sum of %d and %d is : %d",a,b,a+b); break; case 2 :
- 4. C Co om mp pu ut te er rP Pr ro og gr ra am mm mi in ng gL La ab bo or ra at to or ry y2 21 1C CP PL L2 27 7 S Sr ri iS Sa ai ir ra am mC Co ol ll le eg ge eo of fE En ng gi in ne ee er ri in ng g, ,A An ne ek ka al l, ,B Be en ng ga al lu ur ru u. . 4 4| |P P a a g ge e printf("Difference of %d and %d is : %d",a,b,a-b); break; case 3 : printf("Multiplication of %d and %d is : %d",a,b,a*b); break; case 4 : printf("Division of Two Numbers is %d : ",a/b); break; case 5: printf("Remainder of %d and %d is : %d",a,b,a%b); break; default : printf(" Enter Your Correct Choice."); break; } } Output: Enter the values of a & b: 5 5 Enter your choice 1 Sum of a and b is : 10 2. Develop a program to compute the roots of a quadratic equation by accepting the coefficients. Print the appropriate messages Algorithm Step-1: [Read values of A, B, C]-Read A, B, C. Step-2: [Check for Co-efficient]-If A=B=C=0 then roots are non-determinant Step-3:[Compute the value Disc]- Disc=B*B-4*A*C; Step-4:[Different roots are obtained depending on the value of disc]- If Disc is lesser than 0 than goto step 5 If Disc is equals to 0 than goto step 6 If Disc is greater than 0 than goto step 7.
- 5. C Co om mp pu ut te er rP Pr ro og gr ra am mm mi in ng gL La ab bo or ra at to or ry y2 21 1C CP PL L2 27 7 S Sr ri iS Sa ai ir ra am mC Co ol ll le eg ge eo of fE En ng gi in ne ee er ri in ng g, ,A An ne ek ka al l, ,B Be en ng ga al lu ur ru u. . 5 5| |P P a a g ge e Step-5: Print Imaginary roots. Realp=-B/2*A; Imagp=sqrt(Disc)/2*A; Root1=realp + imagep; Root2 = realp – imagep; Step-6:Roots are real and equal Root1=B/2*A; Root2 = root1; Step-7: roots are real and distinct Root1= – B +sqrt(Disc)/2*A; Root2 = -B -sqrt(Disc)/2*A; Step-8: Stop Flowchart: Program:
- 6. C Co om mp pu ut te er rP Pr ro og gr ra am mm mi in ng gL La ab bo or ra at to or ry y2 21 1C CP PL L2 27 7 S Sr ri iS Sa ai ir ra am mC Co ol ll le eg ge eo of fE En ng gi in ne ee er ri in ng g, ,A An ne ek ka al l, ,B Be en ng ga al lu ur ru u. . 6 6| |P P a a g ge e #include<stdio.h> #include<stdlib.h> #include<math.h> void main() { float a,b,c,disc; float root1,root2,realp,imagp; printf("Enter values of a,b,cn"); scanf("%f%f%f",&a,&b,&c); if(a==0 && b==0 && c==0) { printf("roots cannot be determinedn"); exit(0); } else { disc=b*b-4*a*c; if(disc>0) { root1=(-b+sqrt(disc))/(2*a); root2=(-b-sqrt(disc))/(2*a); printf("roots are real and distinctn"); printf("root1=%fn",root1); printf("root2=%fn",root2); } else if(disc==0) { root1=-b/(2*a); root2=root1; printf("roots are real and equaln"); printf("root1=%fn",root1); printf("root2=%fn",root2); } else if(disc<0) { realp=-b/(2*a); imagp=sqrt(abs(disc)/(2*a)); printf("roots are complexn");
- 7. C Co om mp pu ut te er rP Pr ro og gr ra am mm mi in ng gL La ab bo or ra at to or ry y2 21 1C CP PL L2 27 7 S Sr ri iS Sa ai ir ra am mC Co ol ll le eg ge eo of fE En ng gi in ne ee er ri in ng g, ,A An ne ek ka al l, ,B Be en ng ga al lu ur ru u. . 7 7| |P P a a g ge e printf("root1=%f+i%fn",realp,imagp); printf("root2=%f-i%fn",realp,imagp); } } } Output: 3. An electricity board charges the following rates for the use of electricity: for the first 200 units 80 paise per unit: for the next 100 units 90 paise per unit: beyond 300 units rupees 1 per unit. All users are charged a minimum of rupees 100 as meter charge. If the total amount is more than Rs 400, then an additional Surcharge of 15% of total amount is charged. Write a program to read the name of the user, number of units consumed and print out the charges. Algorithm: Input: Customer name and unit consumed Output: Customer name, units consumed and total amount to be paid Step 1: Start Step 2: Read the name of customer and the unit consumed by the customer. Step 3: Check if the unit consumed is greater than 1 and less than 200,if true goto step 4 else goto step 5. Step 4: Compute: amt=100+(0.8*units). Step 5: if unit is greater than 200 and less than 300,if true goto step 6 else goto step7 Step 6: Compute amt=100+(200*0.8)+((units-200)*0.9) Step 7:Compute amt=100+(200*0.8)+(100*0.9)+((units-300)*1), then goto step 8
- 8. C Co om mp pu ut te er rP Pr ro og gr ra am mm mi in ng gL La ab bo or ra at to or ry y2 21 1C CP PL L2 27 7 S Sr ri iS Sa ai ir ra am mC Co ol ll le eg ge eo of fE En ng gi in ne ee er ri in ng g, ,A An ne ek ka al l, ,B Be en ng ga al lu ur ru u. . 8 8| |P P a a g ge e Step 8: Check if the amt is less than or equal to 400, if true goto step 9 otherwise goto step 10. Step 9: Print the amount charged and goto step 11. Step 10: compute amt=amt*1.15,and print the amount charged. Step 11: Stop
- 9.
- 10. C Co om mp pu ut te er rP Pr ro og gr ra am mm mi in ng gL La ab bo or ra at to or ry y2 21 1C CP PL L2 27 7 S Sr ri iS Sa ai ir ra am mC Co ol ll le eg ge eo of fE En ng gi in ne ee er ri in ng g, ,A An ne ek ka al l, ,B Be en ng ga al lu ur ru u. . 1 10 0| |P P a a g ge e Program: #include<stdio.h> #include<string.h> void main() { int cust_no, unit_con; float charge,surcharge=0, amt, total_amt; char nm[25]; printf("Enter the customer IDNO :t"); scanf("%d",&cust_no); printf("Enter the customer Name :t"); scanf("%s",nm); printf("Enter the unit consumed by customer :t"); scanf("%d",&unit_con); if (unit_con <200 ) charge = 0.80; else if (unit_con>=200 && unit_con<300) charge = 0.90; else charge = 1.00; amt = unit_con*charge; if (amt>400) surcharge = amt*15/100.0; total_amt = amt+surcharge; printf("tttnElectricity Billnn"); printf("Customer IDNO :t%d",cust_no); printf("nCustomer Name :t%s",nm); printf("nunit Consumed :t%d",unit_con); printf("nAmount Charges @Rs. %4.2f per unit :t%0.2f",charge,amt); printf("nSurchage Amount :t%.2f",surcharge); printf("nMinimum meter charge Rs :t%d",100); printf("nNet Amount Paid By the Customer :t%.2f",total_amt+100); }
- 11. C Co om mp pu ut te er rP Pr ro og gr ra am mm mi in ng gL La ab bo or ra at to or ry y2 21 1C CP PL L2 27 7 S Sr ri iS Sa ai ir ra am mC Co ol ll le eg ge eo of fE En ng gi in ne ee er ri in ng g, ,A An ne ek ka al l, ,B Be en ng ga al lu ur ru u. . 1 11 1| |P P a a g ge e Output : 4. Implement binary search on Integers/Numbers Step 1: Start Step 2: Read size of the array n Step 3: Read the list of elements in sorted order a[i]. Step 4: Read the key element in key Step 5: initialize low=0 and high= n-1 Step 6: Check if low is less than or equal to high. If condition is true, goto step 7, other wise goto Step 11 Step 7: find the middle element.i.e. mid=(low+high)/2; Step 8: if key element is equal to mid element, then initialize loc value, loc = mid+1; otherwise goto step 9 Step 9: Check if key element is less than mid element is true, then initialize high=mid-1 then goto step 6, if it false goto step10. Step 10: Check if key element is greater than mid element is true, then
- 12. C Co om mp pu ut te er rP Pr ro og gr ra am mm mi in ng gL La ab bo or ra at to or ry y2 21 1C CP PL L2 27 7 S Sr ri iS Sa ai ir ra am mC Co ol ll le eg ge eo of fE En ng gi in ne ee er ri in ng g, ,A An ne ek ka al l, ,B Be en ng ga al lu ur ru u. . 1 12 2| |P P a a g ge e initialize low=mid+1 then goto step 6. Step 11: Check if loc value is greater then zero then print the search is successful then goto step 13, otherwise goto step 12. Step 12: print search is unsuccessful, then goto step 13. Step 13: Stop Program: #include<stdio.h> void main() { int n, a[100], i, key, high, low, mid, loc=-1; printf("Enter the size of the arrayn"); scanf("%d",&n); printf("Enter the elements of array in sorted ordern"); for(i=0;i<n;i++) scanf("%d",&a[i]); printf("Enter the key element to be searchedn"); scanf("%d",&key); low=0; high=n-1; while(low<=high) { mid=(low+high)/2; if(key<a[mid]) { loc = mid+1; break} if(key<a[mid]) high=mid-1; else} } 5. Implement Matrix multiplication and validate the rules of multiplication. Algorithm Step-1: Start
- 13. C Co om mp pu ut te er rP Pr ro og gr ra am mm mi in ng gL La ab bo or ra at to or ry y2 21 1C CP PL L2 27 7 S Sr ri iS Sa ai ir ra am mC Co ol ll le eg ge eo of fE En ng gi in ne ee er ri in ng g, ,A An ne ek ka al l, ,B Be en ng ga al lu ur ru u. . 1 13 3| |P P a a g ge e Step-2: [Read the order of both matrices M, N, P, Q] Step-3: Check for i = 0 to M and j = 0 to N for the matrix A, read A[i][j] Step-4: Check for i = 0 to P and j = 0 to Q for the matrix B, read B[i][j] Step-5: If in case (N=P) then only multiplication is possible then goto step 6 otherwise goto step 7 Step-6: Initially set matrix C[i][j] as 0 For k=0 to n C[i][j] = C[i][j] + A[i][k]*B[i][k] than goto step 8 Step-7: Multiplication is not possible Step-8: Prints the multiplication of the two matrix Step-9: Stop
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- 16.
- 17. C Co om mp pu ut te er rP Pr ro og gr ra am mm mi in ng gL La ab bo or ra at to or ry y2 21 1C CP PL L2 27 7 S Sr ri iS Sa ai ir ra am mC Co ol ll le eg ge eo of fE En ng gi in ne ee er ri in ng g, ,A An ne ek ka al l, ,B Be en ng ga al lu ur ru u. . 1 17 7| |P P a a g ge e Program: #include<stdio.h> int main() { int a[20][20],b[20][20],c[20][20]; int m,n,p,q,i,j,k; printf("Enter rows and columns of matrix An"); scanf("%d%d",&m,&n); printf("Enter rows and columns of matrix Bn"); scanf("%d%d",&p,&q); if(n!=p) { printf("Matrix multiplication not possiblen"); return 0; } printf("Enter elements of matrix An"); for(i=0;i<m;i++) { for(j=0;j<n;j++) { scanf("%d",&a[i][j]); } } printf("Enter elements of matrix Bn"); for(i=0;i<p;i++) { for(j=0;j<q;j++) { scanf("%d",&b[i][j]); } } for(i=0;i<p;i++) { for(j=0;j<q;j++) { c[i][j]=0;
- 18. C Co om mp pu ut te er rP Pr ro og gr ra am mm mi in ng gL La ab bo or ra at to or ry y2 21 1C CP PL L2 27 7 S Sr ri iS Sa ai ir ra am mC Co ol ll le eg ge eo of fE En ng gi in ne ee er ri in ng g, ,A An ne ek ka al l, ,B Be en ng ga al lu ur ru u. . 1 18 8| |P P a a g ge e for(k=0;k<n;k++) { c[i][j]=c[i][j]+a[i][k]*b[k][j]; } } } printf("Product of two matrices isn"); for(i=0;i<m;i++) { for(j=0;j<q;j++) { printf("%dn",c[i][j]); } } } Output: Enter rows and columns of matrix A 2 2 Enter rows and columns of matrix B 2 2 Enter elements of matrix A 6 6 6 6 Enter elements of matrix B 2 2 2 2 Product of two matrices is 24 24 24 24 6. Compute sin(x)/cos(x) using Taylor series approximation. Compare your result with the built-in library function. Print both the results with appropriate inferences. Algorithm Step-1: Start
- 19. C Co om mp pu ut te er rP Pr ro og gr ra am mm mi in ng gL La ab bo or ra at to or ry y2 21 1C CP PL L2 27 7 S Sr ri iS Sa ai ir ra am mC Co ol ll le eg ge eo of fE En ng gi in ne ee er ri in ng g, ,A An ne ek ka al l, ,B Be en ng ga al lu ur ru u. . 1 19 9| |P P a a g ge e Step-2: [Read the value of x in degree]- Read x Step-3: [Initialization and Radians Conversion] Temp = x x=x*(3.142/180.0) Term = x sinx = term n=1 Step-4: [Computer sin value] Repeat while (term>FLT_EPSILON) Fact = 2*n*(2*n+1) Term = term * x * x /fact Sinx sinx +term = n + 1n Step-5: [Output]- Print sin(x) without using library function of Print sin(x) and with using library function Step-6: Stop
- 20.
- 21. C Co om mp pu ut te er rP Pr ro og gr ra am mm mi in ng gL La ab bo or ra at to or ry y2 21 1C CP PL L2 27 7 S Sr ri iS Sa ai ir ra am mC Co ol ll le eg ge eo of fE En ng gi in ne ee er ri in ng g, ,A An ne ek ka al l, ,B Be en ng ga al lu ur ru u. . 2 21 1| |P P a a g ge e Program: #include<stdio.h> #include<stdlib.h> #include<math.h> void main() { int x,n,i; float rad, res, sum=0; printf("Enter degreen"); scanf("%d",&x); printf("Enter number of termsn"); scanf("%d",&n); rad=x*3.14/180; for(i=1;i<=n;i+=2) { if ((i-1)%4==0) sum=sum+pow(rad,i)/fact(i); else sum=sum-pow(rad,i)/fact(i); } printf("Calculate sin(%d) = %f", x,sum); printf("nLibrary sin(%d) = %f", x,sin(rad)); } int fact(int m) { int i,f=1; for(i=1;i<=m;i++) { f=f*i; } return 1; } Output
- 22. C Co om mp pu ut te er rP Pr ro og gr ra am mm mi in ng gL La ab bo or ra at to or ry y2 21 1C CP PL L2 27 7 S Sr ri iS Sa ai ir ra am mC Co ol ll le eg ge eo of fE En ng gi in ne ee er ri in ng g, ,A An ne ek ka al l, ,B Be en ng ga al lu ur ru u. . 2 22 2| |P P a a g ge e 7. Sort the given set of N numbers using Bubble sort. Algorithm Step-1: Start Step-2: Read un-sorted array of n elements into a[i], where i is the index value Step-3: Initialize index i=0 Step-4: Check for the i if it is less than n. if ture, than goto step-5. Other wise goto step output array Step-5: Initialize the index J to zero Step-6: Check if j is less than (n-i-1). If it is true than goto step-7, otherwise increment j and goto step-4 Step-7: Check if a[j] is greater than a[j+1](i.e-adjacent elements are compared) inside the for loop. if true swap the elements using temporary variables, otherwise goto step-6 Step-8: Using the for loop output the sorted array elements Step-9: Stop
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- 24. C Co om mp pu ut te er rP Pr ro og gr ra am mm mi in ng gL La ab bo or ra at to or ry y2 21 1C CP PL L2 27 7 S Sr ri iS Sa ai ir ra am mC Co ol ll le eg ge eo of fE En ng gi in ne ee er ri in ng g, ,A An ne ek ka al l, ,B Be en ng ga al lu ur ru u. . 2 24 4| |P P a a g ge e Program: #include<stdio.h> int main() { int i,j,n,temp; int a[20]; printf("enter the value of n"); scanf("%d",&n); printf("Enter the numbers in unsorted order:n"); for(i=0;i<n;i++) scanf("%d", &a[i]); // bubble sort logic for(i=0;i<n;i++) { for(j=0;j<(n-i)-1;j++) { if( a[j]>a[j+1]) { temp=a[j]; a[j]=a[j+1]; a[j+1]=temp; } } } printf("The sorted array isn"); for(i=0;i<n;i++) { printf("%dn",a[i]); } }
- 25. C Co om mp pu ut te er rP Pr ro og gr ra am mm mi in ng gL La ab bo or ra at to or ry y2 21 1C CP PL L2 27 7 S Sr ri iS Sa ai ir ra am mC Co ol ll le eg ge eo of fE En ng gi in ne ee er ri in ng g, ,A An ne ek ka al l, ,B Be en ng ga al lu ur ru u. . 2 25 5| |P P a a g ge e Output: 8. Write functions to implement string operations such as compare, concatenate, string length. Convince the parameter passing techniques. Step-1: Start Step-2: Press 1 – Compare, 2 – Concatenate, 3-length of string, if press 1 than goto step 3, if press step 2 then goto step 4, if press 3 then goto step 5 Step-3: Read the string1 and string2 and cojmparsion function by using strcmp built in unction is used. If res = 0 then print both strings are equal, otherwise print strints are not equal and than goto the step 4 Step-4: Read String1 and sring2 and find concatenation of two strings using string handling function strcat() and display sring and return back to main fuctions. Step-5: Read string1 and call function to find the length of string by calling function length ( *string) Step-6: if digit=1 then gtoto step 2 else goto step 7 Step-7: Stop
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- 27. C Co om mp pu ut te er rP Pr ro og gr ra am mm mi in ng gL La ab bo or ra at to or ry y2 21 1C CP PL L2 27 7 S Sr ri iS Sa ai ir ra am mC Co ol ll le eg ge eo of fE En ng gi in ne ee er ri in ng g, ,A An ne ek ka al l, ,B Be en ng ga al lu ur ru u. . 2 27 7| |P P a a g ge e Program: #include<stdio.h> #include<string.h> void compare(char [ ],char [ ]); . void concat(char [ ],char [ ]); void length(char *[ ]); void main( ) { int n,digit; char str1[10],str2[10]; do { printf("press 1-compare 2-concatenate 3-length of string"); printf("n enter your choice=");
- 28. C Co om mp pu ut te er rP Pr ro og gr ra am mm mi in ng gL La ab bo or ra at to or ry y2 21 1C CP PL L2 27 7 S Sr ri iS Sa ai ir ra am mC Co ol ll le eg ge eo of fE En ng gi in ne ee er ri in ng g, ,A An ne ek ka al l, ,B Be en ng ga al lu ur ru u. . 2 28 8| |P P a a g ge e scanf("%d",&n); switch(n) { case 1:printf("enter first string="); scanf("%s",str1); printf("enter second string="); scanf("%s",str2); compare(str1,str2); break; case 2: printf("enter first string="); scanf("%s",str1); printf("enter second string="); scanf("%s",str2); concat(str1,str2); break; case 3:printf("enter string="); scanf("%s",str1); length(&str1); break; default: printf("wrong choice"); break; } printf("n Do you want to continue(1/0)? "); scanf("%d", &digit); }while(digit==1); } void compare(char str1[ ],char str2[ ]) { int i; i=strcmp(str1,str2); if(i==0) printf("strings are equaln "); else printf("string are not equaln"); } void concat(char str1[ ],char str2[ ]) { strcat(str1,str2); printf("concatenate string=%s",str1);
- 29. C Co om mp pu ut te er rP Pr ro og gr ra am mm mi in ng gL La ab bo or ra at to or ry y2 21 1C CP PL L2 27 7 S Sr ri iS Sa ai ir ra am mC Co ol ll le eg ge eo of fE En ng gi in ne ee er ri in ng g, ,A An ne ek ka al l, ,B Be en ng ga al lu ur ru u. . 2 29 9| |P P a a g ge e } void length(char *str1[ ]) { int len; len=strlen(str1); printf("the length of string=%d",len); } Output: 9.Implement structures to read, write and compute average- marks and the students scoring above and below the average marks for a class of N students. Algorithm: Step-1: Start Step-2: Read number of students Step-3: For every student, read the student id, name , marks for all the subjects Step-4: Calculate the avarage marks and store it in the avg field Step-5: Print the results Step-6: Initialise loop Step-7: Read teh average of every student Step-8: Check for if avg>35.00 Step-9: If yes than print the result else goto next interation Step-10: Initialise the loop Step-11: Read average of every student Step-12: Check if avg<35.00 Step-13: If yes than print result else goto next iteration Step-14: Stop Flowchart:
- 30. C Co om mp pu ut te er rP Pr ro og gr ra am mm mi in ng gL La ab bo or ra at to or ry y2 21 1C CP PL L2 27 7 S Sr ri iS Sa ai ir ra am mC Co ol ll le eg ge eo of fE En ng gi in ne ee er ri in ng g, ,A An ne ek ka al l, ,B Be en ng ga al lu ur ru u. . 3 30 0| |P P a a g ge e Program: #include<stdio.h> struct student { char usn[10]; char name[10]; float m1,m2,m3; float avg,total; }; void main() { struct student s[20]; int n,i; float tavg,sum=0.0; printf("Enter the number of student=");
- 31. C Co om mp pu ut te er rP Pr ro og gr ra am mm mi in ng gL La ab bo or ra at to or ry y2 21 1C CP PL L2 27 7 S Sr ri iS Sa ai ir ra am mC Co ol ll le eg ge eo of fE En ng gi in ne ee er ri in ng g, ,A An ne ek ka al l, ,B Be en ng ga al lu ur ru u. . 3 31 1| |P P a a g ge e scanf("%d",&n); for(i=0;i<n;i++) { printf("Enter the detail of %d studentsn",i+1); printf("n Enter USN="); scanf("%s",s[i].usn); printf("n Enter Name="); scanf("%s",s[i].name); printf("Enter the three subject scoren"); scanf("%f%f%f",&s[i].m1,&s[i].m2,&s[i].m3); s[i].total=s[i].m1+s[i].m2+s[i].m3; s[i].avg=s[i].total/3; } for(i=0;i<n;i++) { if(s[i].avg>=35) printf("n %s has scored above the average marks",s[i].name); else printf("n %s has scored below the average marks",s[i].name); } } Output
- 32. C Co om mp pu ut te er rP Pr ro og gr ra am mm mi in ng gL La ab bo or ra at to or ry y2 21 1C CP PL L2 27 7 S Sr ri iS Sa ai ir ra am mC Co ol ll le eg ge eo of fE En ng gi in ne ee er ri in ng g, ,A An ne ek ka al l, ,B Be en ng ga al lu ur ru u. . 3 32 2| |P P a a g ge e 10.Develop a program using pointers to compute the sum, mean and standard deviation of all elements stored in an array of N real numbers. Algorithm Step-1: Start Step-2: Read n Step-3: For every value of n read the x Step-4: Initialize sum=0 and i=0 Step-5: For every value of n and i, comuter sum using sum = sum + (*(x+i) – mean) * ( * ( x+i) – mean) Step-6: Using the sum value computer variance = sum / n and deviation = sqrt ( variance ) Step-7: Display mean, variance, deviation Step-8: Stop Flowchart:
- 33. C Co om mp pu ut te er rP Pr ro og gr ra am mm mi in ng gL La ab bo or ra at to or ry y2 21 1C CP PL L2 27 7 S Sr ri iS Sa ai ir ra am mC Co ol ll le eg ge eo of fE En ng gi in ne ee er ri in ng g, ,A An ne ek ka al l, ,B Be en ng ga al lu ur ru u. . 3 33 3| |P P a a g ge e Program: #include<stdio.h> #include<math.h> int main() { int n , i; float x[20],sum,mean; float variance , deviation; printf("Enter the value of n n"); scanf("%d",&n); printf("enter %d real values n",n); for (i=0;i<n;i++) {
- 34. C Co om mp pu ut te er rP Pr ro og gr ra am mm mi in ng gL La ab bo or ra at to or ry y2 21 1C CP PL L2 27 7 S Sr ri iS Sa ai ir ra am mC Co ol ll le eg ge eo of fE En ng gi in ne ee er ri in ng g, ,A An ne ek ka al l, ,B Be en ng ga al lu ur ru u. . 3 34 4| |P P a a g ge e scanf("%f",(x+i)); } sum=0; for(i=0;i<n;i++) { sum= sum+*(x+i); } printf("sum=%fn",sum); mean=sum/n; sum=0; for(i=0;i<n;i++) { sum=sum+(*(x+i)-mean)*(*(x+i)-mean); } variance=sum/n; deviation=sqrt(variance); printf("mean(Average)=%fn",mean); printf("variance=%fn",variance); printf("standard deviation=%fn",deviation); } Output 11.Implement Recursive functions for Binary to Decimal Conversion Algorithm Step-1: Start Step-2: Read the binary number value Step-3: Call the function find by passing arguments
- 35. C Co om mp pu ut te er rP Pr ro og gr ra am mm mi in ng gL La ab bo or ra at to or ry y2 21 1C CP PL L2 27 7 S Sr ri iS Sa ai ir ra am mC Co ol ll le eg ge eo of fE En ng gi in ne ee er ri in ng g, ,A An ne ek ka al l, ,B Be en ng ga al lu ur ru u. . 3 35 5| |P P a a g ge e Step-4: Check if number entered is zero than return 0 otherwise compute using formula- ( pow ( 2, count ) * ( x % 10 ) + find ( x /10 , count + 1 ) ) Step-5: Repeat the step-4 until recursive condition find becomes false Step-6: Return the result of converted number Step-7: Stop Flowchart: Program: #include <stdio.h> #include <math.h> int find(int x, int count) { if (x == 0) return 0; else return(pow(2, count) * (x%10 ) + find(x/10 ,count+1)); } void main() { int binary; printf("Enter a Binary number"); scanf ("%d",&binary); printf(" The decimal value for binary %d is",binary); printf("%d", find(binary,0));
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