Since the last few postings we are in C Language array. Today’s topic in our C programming tutorialis also c array . We’ll solve some programs on array using C Language.
Program on identity matrices
#include< stdio.h >
void main ()
{
int mat1[4][4],mat2[4][4],i,j,x=3;
clrscr();
printf("Program on identity matrix.\n");
for(i=0;i< 4;i++)
{
for(j=0;j< 4;j++)
{
if(i==j)
mat1[i][j]=1;
else
mat1[i][j]=0;
}
}
printf("The identity matrix is :-\n");
for(i=0;i< 4;i++)
{
for(j=0;j< 4;j++)
{
printf("%4d",mat1[i][j]);
}
printf("\n");
}
printf("\n AND\n");
for(i=0;i< 4;i++)
{
for(j=0;j< 4;j++)
{
if(j==x)
mat1[i][j]=1;
else
mat1[i][j]=0;
}
x--;
}
for(i=0;i< 4;i++)
{
for(j=0;j< 4;j++)
{
printf("%4d",mat1[i][j]);
}
printf("\n");
}
getch ();
}
In this C Language program, we’ll work on matrix. Given a 2-D matrix of order 3*3 .Search an element x if the element in the matrix, display that the search is successful and display the position at which the element occurs. If the search is not successful-display that element is not in the matrix.
#include< stdio.h>
void main ()
{
int mat[3][3],i,j,x;
clrscr();
printf("\nNow we will enter the values(integer) for the matrix.");
for(i=0;i< 3;i++)
{
for(j=0;j< 3;j++)
{
printf("\nValue:-");
scanf("%d",&mat[i][j]);
}
}
printf("\nEnter the specific value to be searched:-");
scanf("%d",&x);
clrscr();
printf("\nThe matrix:-\n");
for(i=0;i< 3;i++)
{
for(j=0;j< 3;j++)
{
printf("%4d",mat[i][j]);
}
printf("\n");
}
for(i=0;i< 3;i++)
{
for(j=0;j< 3;j++)
{
if(x==mat[i][j])
break;
}
if(j!=3)
break;
}
if(i!=3)
{
printf("Your input value exists in row no %d and column no %d",i+1,j+1);
}
else
{
printf("Value not available.");
}
getch();
}
Same array program with some modification
#include< stdio.h>
void main ()
{
int mat[3][3],i,j,n,x[3],y[3],a=0;
clrscr();
printf("\nNow we will enter the values(integer) for the matrix.");
for(i=0;i< 3;i++)
{
for(j=0;j< 3;j++)
{
printf("\nValue:-");
scanf("%d",&mat[i][j]);
}
}
printf("\nEnter the specific value to be searched:-");
scanf("%d",&n);
clrscr();
printf("\nThe matrix:-\n");
for(i=0;i< 3;i++)
{
for(j=0;j< 3;j++)
{
printf("%4d",mat[i][j]);
}
printf("\n");
}
for(i=0;i< 3;i++)
{
for(j=0;j< 3;j++)
{
if(n==mat[i][j])
{
x[a]=i+1;
y[a]=j+1;
a++;
}
}
}
if(a==0)
printf("Value does not exists in the matrix.");
else
{
printf("The value exists in the matrix and locations are:-\n");
for(i=0;i< a;i++)
{
printf("%d and %d\n",x[i],y[i]);
}
}
getch();
}
*
ReplyDelete* *
* * *
* * * *
* * * * * *
any body can make this programme
#include< iostream.h >
ReplyDelete# include < conio.h >
class Pattern
{
public:
void show()
{
for(int i=0; i< 5; i++0
{
for( int j=0;j < =i;j++0
{
cout<< "*";
}
cout<< endl;}
}
}
};
void main ()
{
Pattern ob;
ob.show();
getch();
}
1 0 0 0
ReplyDelete0 1 0 0
0 0 1 0
0 0 0 1
please give me solution of this system.
* * * *
ReplyDelete* * *
* *
*
please give me solution of this problem
#include< stdio.h >
ReplyDeletevoid main ()
{
int i,j;
for(i=0;i< 4;i++)
{
for(j=0; j< 4;j++)
{
if(i==j)
printf("1");
else
printf("0");
}
printf("\n");
}
getch();
}
#include< stdio.h>
ReplyDeletevoid main()
{
int i,j;
for ( i=0; i < 4;i++)
{
for( j=i;j < 4; j++)
{
printf("*");
}
printf("\n");
}
getch();
}