1)Program find sum of digits of a number
#include<stdio.h>
#include<conio.h>
void main()
{
int n,temp,rem,ans=0;
clrscr();
printf("enter a no");
scanf("%d",&n);
temp=n;
while(n!=0)
{
rem=n%10;
ans+=rem;
n=n/10;
}
printf("the sum of digits of %d is %d\n",temp,ans);
getch();
}
OUTPUT:
enter a no234
the sum of digits of 234 is 9
2)Program to print Fibonacci series
void main()
{
int i,n,f,f1,f2;
printf("enter the range");
scanf("%d",&n);
f=0;
f1=1;
f2=1;
do
{
i++;
printf("%d\n",f);
f1=f2;
f2=f;
f=f1+f2;
}
while(i<=n);
}
OUTPUT:
Enter the range 9
0 1 1 2 3 5 8 13 21
3)Program to generate prime number till nth number
void main()
{
int n,i,fact,j;
printf("enter the range");
scanf("%d",&n);
printf(“Prime numbers are\n”);
for(i=1;i<=n;i++)
{
fact=0;
for(j=1;j<=n;j++)
{
if(i%j==0)
fact++;
if(fact==2)
printf("%d “,i);
}
getch();
}
OUTPUT:
Enter the range 10
Prime numbers are
3 5 7
4)To find the sum of the terms of the given series by the input value x.
#include<stdio.h>
4)To find the sum of the terms of the given series by the input value x.
The given series is 1-X2/2! + X4/4! –X6/6! +X8/8! –X10/10!
#include<stdio.h>
#include<conio.h>
#include<math.h>
void main()
{
int p,i,j;
float x,sum=0,fact=1;
clrscr();
printf("enter the value of x\t");
scanf("%f",&x);
for(i=0,p=0;p<=10;i++,p+=2)
{
for(j=1;j<=p;j++)
{
fact=fact*j;
}
sum=sum+(pow(-1,i)*pow(x,p))/fact;
}
printf("\nThe sum of series 1-x2/2!+x4/4!-x6/6!+x8/8!-x10/10!");
printf("\n when x value %f is %f",x,sum);
getch();
}
OUTPUT:
Input: enter the value of x 2
Input: enter the value of x 2
The sum of series 1-x2/2!+x4/4!-x6/6!+x8/8!-x10/10!
when x value 2.000000 is – 0.668518
5)Program to calculate the roots of a Quadratic equation
#include<stdio.h>
#include<conio.h>
#include<math.h>
void main()
{
float a,b,c,d,r1,r2,rp,ip;
clrscr();
printf("\n enter a b c values");
scanf("%f%f%f",&a,&b,&c);
if(a==0)
printf("not a QE");
else
{
d=b*b-4*a*c;
if(d>=0)
{
printf("\n roots are real");
r1=(-b+sqrt(d))/(2*a);
r2=(b-sqrt(d))/(2*a);
printf("\n roots are %f and %f",r1,r2);
}
else
{
printf("\n roots are imaginary");
rp=-b/(2*a);
ip=sqrt(-d)/(2*a);
printf("\n roots are (%f,%f) and (%f,%f)",rp,ip,rp,-ip);
}
}
getch();
}
OUTPUT:
enter a b c values3 5 2
roots are real
roots are -0.666667 and 0.666667
6)Program to calulate the factorial of the given number
#include<stdio.h>
#include<conio.h>
void main()
{
int i,n;
long fact=1;
clrscr();
printf("\n enter n integer");
scanf("%d",&n);
for(i=1;i<=n;i++)
fact=fact*i;
printf("factorial is %ld\n",fact);
getch();
}
OUTPUT:
enter n integer5
factorial is 120
7)Program to find factorial using recursive and non-recursive functions
C-Program: using non – recursive technique.
#include<stdio.h>
#include<conio.h>
void main()
{
void ff(int);
int n;
clrscr();
printf("enter the value of n\t");
scanf("%d",&n);
ff(n);
getch();
}
void ff(int n)
{
int i, fact=1;
for(i=1;i<=n;i++)
{
fact=fact*i;
}
printf("\nThe %d! is %d",n,fact);
}
8)To find the sum of the terms of the given series by the input value x.
The given series is 1-X2/2! + X4/4! –X6/6! +X8/8! –X10/10!
#include<stdio.h>
#include<conio.h>
#include<math.h>
void main()
{
int p,i,j;
float x,sum=0,fact=1;
clrscr();
printf("enter the value of x\t");
scanf("%f",&x);
for(i=0,p=0;p<=10;i++,p+=2)
{
for(j=1;j<=p;j++)
{
fact=fact*j;
}
sum=sum+(pow(-1,i)*pow(x,p))/fact;
}
printf("\nThe sum of series 1-x2/2!+x4/4!-x6/6!+x8/8!-x10/10!");
printf("\n when x value %f is %f",x,sum);
getch();
}
OUTPUT:
Input: enter the value of x 2
Input: enter the value of x 2
The sum of series 1-x2/2!+x4/4!-x6/6!+x8/8!-x10/10!
when x value 2.000000 is – 0.668518
9)Program to calculate the roots of a Quadratic equation
10)Program to calulate the factorial of the given number
OUTPUT:
11)Program to find factorial using recursive and non-recursive functions
9)Program to calculate the roots of a Quadratic equation
#include<stdio.h>
#include<conio.h>
#include<math.h>
void main()
{
float a,b,c,d,r1,r2,rp,ip;
clrscr();
printf("\n enter a b c values");
scanf("%f%f%f",&a,&b,&c);
if(a==0)
printf("not a QE");
else
{
d=b*b-4*a*c;
if(d>=0)
{
printf("\n roots are real");
r1=(-b+sqrt(d))/(2*a);
r2=(b-sqrt(d))/(2*a);
printf("\n roots are %f and %f",r1,r2);
}
else
{
printf("\n roots are imaginary");
rp=-b/(2*a);
ip=sqrt(-d)/(2*a);
printf("\n roots are (%f,%f) and (%f,%f)",rp,ip,rp,-ip);
}
}
getch();
}
OUTPUT:
enter a b c values3 5 2
roots are real
roots are -0.666667 and 0.666667
10)Program to calulate the factorial of the given number
#include<stdio.h>
#include<conio.h>
void main()
{
int i,n;
long fact=1;
clrscr();
printf("\n enter n integer");
scanf("%d",&n);
for(i=1;i<=n;i++)
fact=fact*i;
printf("factorial is %ld\n",fact);
getch();
}
OUTPUT:
enter n integer5
factorial is 120
11)Program to find factorial using recursive and non-recursive functions
C-Program: using non – recursive technique.
#include<stdio.h>
#include<conio.h>
void main()
{
void ff(int);
int n;
clrscr();
printf("enter the value of n\t");
scanf("%d",&n);
ff(n);
getch();
}
void ff(int n)
{
int i, fact=1;
for(i=1;i<=n;i++)
}
OUTPUT:
enter the value of n 6
the 6! Is 720
12)Using recursive technique.
#include<stdio.h>
#include<conio.h>
void main()
{
int n,f;
clrscr();
printf("enter the value of n\t");
scanf("%d",&n);
f=factorial(n);
printf("\nThe %d! is %d",n,f);
getch();
}
factorial(int x)
{
if(x==1)
return 1;
else
return (x*factorial(x-1));
}
enter the value of n 6
the 6! Is 720
13)Program to find GCD using recursive and non-recursive functions .C-Program using non – recursive technique.
#include<stdio.h>
void main ()
{
int x,y,z;
printf("enter x,y");
scanf("%d%d",&x,&y);
z=gcdr(x,y);
printf("recursive-z=%d",z);
z=gcdn(x,y);
printf("non-recursive-z=%d",z);
}
int gcdr(int a,int b)
{
int r;
r=a%b;
if(r==0)
return(b);
else
return(a);
}
int gcdn(int a,int b)
{
int r;
if(a>b)
{
r=a%b;
while(r!=0)
{
a=b;
b=r;
r=a%b;
}
return(b);
}
else
{
r=b%a;
while(r!=0)
{
b=a;
a=r;
r=b%a;
}
return(a);
}
}
OUTPUT:
enter x,y 69
13
recursive-z=69non-recursive-z=1
16) The total distance traveled by vehicle in ‘t’ seconds is given by
OUTPUT:
enter initial velocity u(mts/sec);3
enter n value 5
19)Addition of two matrices
OUTPUT:
#include<stdio.h>
int i=0;
void hanoi(int n,char source,char by,char dest)
{
if(n==1)
printf("\n %d more disk from %c to %c \n",++i,source,dest);
else
{
hanoi(n-1,source,dest,by);
hanoi(1,source,by,dest);
hanoi(n-1,by,source,dest);
}
}
void main()
{
char s='a',b='b',d='c';
int n;
clrscr();
printf("enter n value");
scanf("%d",&n);
hanoi(n,s,b,d);
getch();
}
OUTPUT:
enter n value3
OUTPUT:
enter n value3
1 move disk from a to c
2 move disk from a to b
3 move disk from c to b
4 move disk from a to c
5 move disk from b to a
6 move disk from b to c
7 move disk from a to c
16) The total distance traveled by vehicle in ‘t’ seconds is given by
distance = ut+1/2at2 where ‘u’ and ‘a’ are the initial velocity (m/sec.)4 and
acceleration (m/sec2). Write C program to find the distance trav*-` 0eled at
regular intervals of time given the values of ‘u’ and ‘a’. The program
should provide the flexibility to the user to select his own time intervals
and repeat the calculations for different values of ‘u’ and ‘a’.
#include<stdio.h>
/* to complete distance */
void main()
{
float s,u,t,a,time,interval;
char c;
clrscr();
while(1)
{
printf("\n enter initial velocity u(mts/sec);");
scanf("%f",&u);
printf("\n enter accleration a(mts/sec2);");
scanf("%f",&a);
printf("\n enter time(secs);");
scanf("%f",&time);
printf("\n enter time interval(secs);");
scanf("%f",interval);
printf("\n time distance \n");
for(t=interval;t<=time;t+=interval)
{
s=u*t+0.5*a*t*t;
printf("\n %10.4f %10.4f \n",t,s);
}
printf("\n press any key to continue / space bar to quit");
c=getch();
if(c==" ")
break;
}
}
OUTPUT:
enter initial velocity u(mts/sec);3
enter accleration a(mts/sec2);2
enter time(secs);5.
enter time interval(secs);1
time distance
1.0000 4.0000
2.0000 10.0000
3.0000 18.0000
4.0000 28.0000
5.0000 40.0000
press any key to continue / space bar to quit
17)Using switch-case statement, write a C program that takes two operands
and one operator from the user, performs the operation and then prints
the answer. (consider operators +,-,/,* and %).
#include<stdio.h>
#include<conio.h>
void main()
{
int a,b,c,ch;
printf("enter a , b values \n");
scanf("%d%d",&a,&b);
printf("enter choice\n");
scanf(" %d",&ch);
switch(ch)
{
case 1:
printf("addtion of a,b is%d",a+b);
break;
case 2:
printf("subsraction of a,b is%d",a-b);
break;
case 3:
printf("multiplication of a,b is %d",a*b);
break;
case 4:
printf("division of a,b is%d",a/b);
break;
case 5:
printf("modulas of a,b is%d",a%b);
break;
default:
printf("enter right choice:");
}
}
OUTPUT:
enter a , b values
5
6
enter choice
3
multiplication of a,b is 30
enter a , b values
4
2
enter choice
4
division of a,b is2
enter a , b values
2
2
enter choice
5
modulas of a,b is0
enter a , b values
2
3
enter choice
6
enter right choice:
OUTPUT:
enter n value 5
enter elements 5 2 3 1 6
the max value is 6 and the min value is 1
19)Addition of two matrices
#include<stdio.h>
#include<conio.h>
int main()
{
int a[5][5],b[5][5],c[5][5],i,j,n,m,p,q;
clrscr();
printf("enter first matrix size\n");
scanf("%d%d",&m,&n);
printf("enter second matrix size\n");
scanf("%d%d",&p,&q);
if(m!=p||n!=q)
printf("size mismatch,Addition is not possible");
else
{
printf("enter first matrix elements\n");
for(i=0;i<m;i++)
for(j=0;j<n;j++)
scanf("%d",&a[i][j]);
printf("enter second matrix elements\n");
for(i=0;i<p;i++)
for(j=0;j<q;j++)
scanf("%d",&b[i][j]);
printf("addtion of two matrices is\n");
for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
c[i][j]=a[i][j]+b[i][j];
}
for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
printf("%4d",c[i][j]);
printf("\n");
}
}
getch();
return 0;
}
OUTPUT:
enter first matrix size
2
3
enter second matrix size
2
3
enter first matrix elements
1
2
3
4
5
6
enter second matrix elements
6
5
4
3
2
1
addtion of two matrices is
7 7 7
7 7 7
20)Multiplecation of two matrices
#include<stdio.h>
#include<conio.h>
int main()
{
int a[5][5],b[5][5],c[5][5],i,j,n,k,m,p,q;
clrscr();
printf("enter first matrix size\n");
scanf("%d%d",&m,&n);
printf("enter second matrix size\n");
scanf("%d%d",&p,&q);
if(n!=p)
printf("Not compatible,Multiplication is not possible");
else
{
printf("enter first matrix elements\n");
for(i=0;i<m;i++)
for(j=0;j<n;j++)
scanf("%d",&a[i][j]);
printf("enter second matrix elements\n");
for(i=0;i<p;i++)
for(j=0;j<q;j++)
scanf("%d",&b[i][j]);
printf("product of two matrices is\n");
for(i=0;i<m;i++)
{
for(j=0;j<q;j++)
c[i][j]=0;
for(k=0;k<n;k++)
c[i][j]+=a[i][k]*b[k][j];
}
for(i=0;i<m;i++)
{
for(j=0;j<q;j++)
printf("%4d",c[i][j]);
printf("\n");
}
}
getch();
return 0;
}
OUTPUT:
enter first matrix size
2
3
enter second matrix size
4
3
Not compatible,Multiplication is not possible
enter first matrix size
2
3
enter second matrix size
3
1 enter first matrix elements
1
2
3
4
5
6
enter second matrix elements
1
2
3
product of two matrices is
14
32
27)Program to generate pascal Triangle.
#include<stdio.h>
void main()
{
int binom=1,p,q=0,r,x;
clrscr();
printf("enter the no of rows");
scanf("%d",&p);
while(q<p)
{
for(r=40-3*q;r>0;--r)
{
printf(" ");
}
for(x=0;x<=q;++x)
{
if(x==0||q==0)
binom=1;
else
binom=(binom*(q-x+1)/x);
printf("%6d",binom);
}
printf("\n");
++q;
}
printf("\n");
}
OUTPUT:
enter the no of rows5
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
27)Program to construct a pyramid of numbers
#include<stdio.h>
void main()
{
int p,m,q,n;
clrscr();
printf("enter the no of lines");
scanf("%d",&n);
for(p=1;p<=n;p++)
{
for(q=1;q<=n-p;q++)
printf(" ");
m=p;
for(q=1;q<=p;q++)
printf("%d",m++);
m-=2;
for(q=1;q<p;q++)
printf("%d",m--);
printf("\n\n");
}
}
OUTPUT:
enter the no. of lines5
1
232
34543
4567654
567898765
28)Program to read two numbers x&n and then sompute sum of this Geometric Expression
1+x+x2+x3+-------+xn
#include<stdio.h>
#include<conio.h>
#define out 0
#define in 1
void main()
{
int c,x,n,i,sum,term;
while(1)
{
clrscr();
sum=1;
term=1;
printf("\n enter n");
scanf("%d",&n);
if(n<0)
{
printf("\n error press any key to come out");
getch();
continue;
}
printf("\n enter x");
scanf("%d",&x);
for(i=0;i<n;i++)
{
term*=x;
sum+=term;
}
printf("\n x=%d,n=%d,sum=%d\n",x,n,sum);
printf("\n press any key to continue / spacebar to quick");
c=getch();
if(c==' ')
break;
}
}
OUTPUT:
enter n
3
enter x5
x=5,n=3,sum=156
press any key to continue / spacebar to quick
29)Program to find the 2’s complement of a binary number
#include<stdio.h>
#include<conio.h>
#include<string.h>
void complement(char bn[],int c)
{
if(bn[i]=='1')
bn[i]='0';
else
bn[i]='1';
}
void tows_complement(char st[])
{
int i,flg=0;
i=strlen(st)-1;
while(i>=0)
{
if(flg)
complement(st,i);
if((!flg)&&(st[i]=='1'))
flg=1;
i--;
}
}
void main()
char s[25],c;
while(1)
{
clrscr();
printf("\n enter the binary no");
gets(s);
tows_complement(s);
printf("\n the 2s complement is %s",s);
printf("\n press anykey to continue/space bar to quit...");
c=getch();
if(c==' ')
break;
}
}
OUTPUT:
Enter a binary no:101
The 2s complement is:011
Pres any key to continye/space bar to quit
30)write a c program to convert a roman numeral to its decimal equilent
#include<string.h>
#include<stdlib.h>
void main()
{
int *a,len,i,j,k;
char *rom;
clrscr();
printf("Enter the Roman Numeral:");
scanf("%s",rom);
len=strlen(rom);
for(i=0;i<len;i++)
{
if(rom[i]=='I')
a[i]=1;
else if(rom[i]=='V')
a[i]=5;
else if(rom[i]=='X')
a[i]=10;
else if(rom[i]=='L')
a[i]=50;
else if(rom[i]=='C')
a[i]=100;
else if(rom[i]=='D')
a[i]=500;
else if(rom[i]=='M')
a[i]=1000;
else
{
printf("\nInvalid Value");
getch();
exit(0);
}
}
k=a[len-1];
for(i=len-1;i>0;i--)
{
if(a[i]>a[i-1])
k=k-a[i-1];
else if(a[i]==a[i-1] || a[i]<a[i-1])
k=k+a[i-1];
}
printf("\nIts Decimal Equivalent is:");
printf("%d",k);
getch();
OUTPUT:
Enter the Roman Numeral:III
Its Decimal Equivalent is:3
31)Write a C program that uses functions to perform the following operations:
i) Reading a complex number
ii) Writing a complex numbe
iii) Addition of two complex number
iv) Multiplication of two complex numbers
(Note: represent complex number using a structure.)
i) Reading a complex number
ii) Writing a complex numbe
iii) Addition of two complex number
iv) Multiplication of two complex numbers
(Note: represent complex number using a structure.)
#include<stdio.h>
#include<math.h>
#include<math.h>
void arithmetic(int opern);
struct comp
{
double realpart;
double imgpart;
};
{
double realpart;
double imgpart;
};
void main()
{
int opern;
clrscr();
printf("\n\n \t\t\t***** MAIN MENU *****");
printf("\n\n Select your option: \n 1 : ADD\n 2 : MULTIPLY\n 0 : EXIT \n\n\t\t Enter your Option [ ]\b\b");
{
int opern;
clrscr();
printf("\n\n \t\t\t***** MAIN MENU *****");
printf("\n\n Select your option: \n 1 : ADD\n 2 : MULTIPLY\n 0 : EXIT \n\n\t\t Enter your Option [ ]\b\b");
scanf("%d",&opern);
switch(opern)
{
case 0:
exit(0);
case 1:
case 2:
arithmetic(opern);
default:
printf("Invalid choice\n");
exit(0);
}
getch();
}
{
case 0:
exit(0);
case 1:
case 2:
arithmetic(opern);
default:
printf("Invalid choice\n");
exit(0);
}
getch();
}
void arithmetic(int opern)
{
{
struct comp w1, w2, w;
printf("\n Enter two Complex Numbers (x+iy)\n Real Part(x) of First Number:");
scanf("%lf",&w1.realpart);
printf("\n Imaginary Part(y) of First Number:");
scanf("%lf",&w1.imgpart);
printf("You Entered complex number %2lf + %2lfi",w1.realpart,w1.imgpart);
printf("\n Real Part(x) of Second Number:");
scanf("%lf",&w2.realpart);
printf("\n Imaginary Part(y) of Second Number:");
scanf("%lf",&w2.imgpart);
printf("You Entered complex number %2f+%2fi",w2.realpart,w2.imgpart);
scanf("%lf",&w1.realpart);
printf("\n Imaginary Part(y) of First Number:");
scanf("%lf",&w1.imgpart);
printf("You Entered complex number %2lf + %2lfi",w1.realpart,w1.imgpart);
printf("\n Real Part(x) of Second Number:");
scanf("%lf",&w2.realpart);
printf("\n Imaginary Part(y) of Second Number:");
scanf("%lf",&w2.imgpart);
printf("You Entered complex number %2f+%2fi",w2.realpart,w2.imgpart);
switch(opern)
{
{
/*addition of complex number*/
case 1:
w.realpart = w1.realpart+w2.realpart;
w.imgpart = w1.imgpart+w2.imgpart;
break;
case 1:
w.realpart = w1.realpart+w2.realpart;
w.imgpart = w1.imgpart+w2.imgpart;
break;
/*multiplication of complex number*/
case 2:
w.realpart=(w1.realpart*w2.realpart)-(w1.imgpart*w2.imgpart);
w.imgpart=(w1.realpart*w2.imgpart)+(w1.imgpart*w2.realpart);
break;
}
case 2:
w.realpart=(w1.realpart*w2.realpart)-(w1.imgpart*w2.imgpart);
w.imgpart=(w1.realpart*w2.imgpart)+(w1.imgpart*w2.realpart);
break;
}
if (w.imgpart>0)
printf("\n Answer = %lf+%lfi",w.realpart,w.imgpart);
else
printf("\n Answer = %lf%lfi",w.realpart,w.imgpart);
getch();
printf("\n Answer = %lf+%lfi",w.realpart,w.imgpart);
else
printf("\n Answer = %lf%lfi",w.realpart,w.imgpart);
getch();
}
OUTPUT:
***** MAIN MENU *****
***** MAIN MENU *****
Select your option:
1 : ADD
2 : MULTIPLY
0 : EXIT
1 : ADD
2 : MULTIPLY
0 : EXIT
Enter your Option [ 1]
Enter two Complex Numbers (x+iy)
Real Part(x) of First Number:2
Real Part(x) of First Number:2
Imaginary Part(y) of First Number:3
You Entered complex number 2.000000 + 3.000000i
Real Part(x) of Second Number:3
You Entered complex number 2.000000 + 3.000000i
Real Part(x) of Second Number:3
Imaginary Part(y) of Second Number:4
You Entered complex number 3.000000+4.000000i
Answer = 5.000000+7.000000iInvalid choice
Program:
You Entered complex number 3.000000+4.000000i
Answer = 5.000000+7.000000iInvalid choice
32)Name of the experiment:
Area of the circle
Program:
#include<stdio.h>
#define PIE 3.14
main()
{
float r, area;
Printf(“enter r value”);
Scanf(“%d”,&r);
area=PIE*r*r;
printf(“area of the circle %f”,area);
}
Output:
enter r value
3
area of the circle is 28.285
33)Name of the experiment:
Area of the triangle
Program:
#include<math.h>
void main()
{
int a,b,c;
float s,area;
printf("enter the values of a,b,c");
scanf("%d%d%d",&a,&b,&c);
s=(a+b+c)/2;
area=sqrt(s*(s-a)*(s-b)*(s-c));
printf("the area of a trangle is =%f",area);
}
Output:
enter the values of a,b,c 10 20 30
The area of a triangle is = 0.000000
34)Name of the experiment:
Binary to decimal
Program:
#include<stdio.h>
#include<conio.h>
void main()
{
#include<conio.h>
void main()
{
int bin,binary,decimal=0,digits,base=1;
printf(" Enter input binary number ");
scanf("%d",&binary);
scanf("%d",&binary);
bin=binary
while(binary)
{
digit=binary%10;
decimal=digit*base;
base*=2;
binary/=10;
}
printf(“decimal equalant of binary number%d=%d\n”,bin,decimal);
}
Output:
input a binary no
1001
decimal equalent of binary no 1001= 9
35)Name of the experiment:
Matrix transpose
Program:
#include< stdio.h>
int main()
{
int m, n, c, d, matrix[10][10], transpose[10][10];
printf("Enter the number of rows and columns of matrix ");
scanf("%d%d",&m,&n);
printf("Enter the elements of matrix \n");
for( c = 0 ; c < m ; c++ )
{
for( d = 0 ; d < n ; d++ )
{
scanf("%d",&matrix[c][d]);
}
}
for( c = 0 ; c < m ; c++ )
{
for( d = 0 ; d < n ; d++ )
{
transpose[d][c] = matrix[c][d];
}
}
printf("Transpose of entered matrix :-\n");
for( c = 0 ; c < n ; c++ )
{
for( d = 0 ; d < m ; d++ )
{
printf("%d\t",transpose[c][d]);
}
printf("\n");
}
return 0;
}
Output:
Eneter the number of rows and coloums:2 3
Enter the elements of the matrix:
1 2 3
4 5 6
Transpose of entered matrix:
1 4
2 5
3 6
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