C Program To find equivalent resistance of Parallel combination of resistive circuits

C Program To find equivalent resistance of Parallel combination of resistive circuits | C assignment to find equivalent resistance in C language | Program to find equivalent resistance when value of Resistance is given

#include
#include

int main()
{
int r[10],num,i,Rs=0;
clrscr();

printf("Enter the number of Resistances : ");
scanf("%d",&num);

printf("\nEnter Value of Each Resistance : \n");
for(i=0;i<num;i++)
   {
   printf("\nR%d : ",i+1);
   scanf("%d",&r[i]);
   }

for(i=0;i<num;i++)
   {
   Rs = Rs + r[i];
   }

printf("\nEquivalent Series Resistance : %d Kohm",Rs);

getch();
}

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#include

void main(int argc , char * argv[])
{
    int i,sum=0;

    if(argc!=3)
      {
      printf("Oops....!! \n you have forgot to type numbers. \n");
      exit(1);
      }

    printf("The sum is : ");

    for(i=1;i<argc; i++)
        sum = sum + atoi(argv[i]);

    printf("%d",sum);

}

How Run Program?

Step 1 : Write a Program

Step 2 : Open Command Prompt inside Borland C/C++.

Step 3 : Click on DOS Shell.

Step 4 : Inside Command Prompt type this command.

C:\TC\BIN>add 10 20

Step 5 : Hit Enter , You will get following Output.



C:\TC\BIN>add 10 20

The sum is : 30

C:\TC\BIN>

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#include

int main(){

float var[5]={1.1f,2.2f,3.3f};
float(*ptr)[5];

ptr=&var;
printf("Value inside ptr : %u",ptr);

ptr=ptr+1;
printf("Value inside ptr : %u",ptr);

return 0;
}

Output :

Value inside ptr : 1000
Value inside ptr : 1020

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1
2 1
1 2 3 2 1
2 3 4 3 2 1
1 2 3 4 5 4 3 2 1

# include 
# include 
void main ( )
{
clrscr ( );
int n,i,j,k;
printf ("\n Enter no. of rows :");
scanf ("%d",&n);
printf ("\n\n");
for (i =1;i < = n;i++)
{
for (j = 1;j < = n-i;j++)
printf (" ");
for (j = i;j > =1;j--)
printf ("%d",j);
for (j = 2;j < = i;j++)
printf ("%d",j);
printf ("\n");
}
getch ( );
}

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#include
#include
#include

void dec_oct(long int num)   // Function Decl.
{
long int rem[50],i=0,length=0;
while(num>0)
 {
 rem[i]=num%8;
 num=num/8;
 i++;
 length++;
 }
printf("\n Converted Octal number is : ");
     for(i=length-1;i>=0;i--)
             printf("%ld",rem[i]);
}





void main()
{
long int num;
clrscr();
 
 printf("Enter the decimal number : ");
 scanf("%ld",&num);

    dec_oct(num);   // Calling function

 getch();
}

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#include
#include
#include

void dec_hex(long int num)   // UDF
{
long int rem[50],i=0,length=0;

while(num>0)
   {
      rem[i]=num%16;
      num=num/16;
      i++;
      length++;
   }

printf("\n Hexadecimal number : ");
for(i=length-1;i>=0;i--)
  {
    switch(rem[i])
    {
      case 10:
          printf("A");
          break;
      case 11:
          printf("B");
          break;
      case 12:
          printf("C");
          break;
      case 13:
          printf("D");
          break;
      case 14:
          printf("E");
          break;
      case 15:
          printf("F");
          break;
      default :
         printf("%ld",rem[i]);
    }
  }
}
//================================================
void main()
{
long int num;
clrscr();

 printf("Enter the decimal number : ");
 scanf("%ld",&num);

    dec_hex(num);   // Calling function

 getch();
}

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#include
#include
#include

void bin_dec(long int num)   // Function declaration
{
long int rem,sum=0,power=0;
while(num>0)
 {
 rem = num%10;
 num = num/10;
 sum = sum + rem * pow(2,power);
 power++;
 }

printf("\n Converted Decimal number is : %d",sum);
}



// END of Function




void main()
{
long int num;
clrscr();

printf(" Please Enter the Binary number (0 and 1 only): \n");
scanf("%ld",&num);

bin_dec(num);

getch();
}

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#include
#include

void binary(unsigned int); 

void main()
{
unsigned int num;
printf(" Please Enter Any Decimal Number : ");
scanf("%u",&num);
binary(num);   // Function Call
getch();
}




// Function Starts 
void binary(unsigned int num)
{
unsigned int mask=32768;   //mask = [1000 0000 0000 0000]
printf("\n The Binary Eqivalent is : ");

while(mask > 0)
   {
   if((num & mask) == 0 )
         printf("0");
   else
         printf("1");
  mask = mask >> 1 ;  // Right Shift
   }
}

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C program to accept a string from user and reverse all words but not string using pointer.

/*c program for reverse all words but not string using pointer*/

#include
#include
int main()
{
 char str[30];
 char *p,*tmp;
 printf(" Enter any string : \n");
 gets(str);
 for(p=str; *p!='\0'; p++)
 {
  if(*(p+1)==' ' || *(p+1)=='\0')
  {
   for(tmp=p; *tmp!=' ' && tmp>=str; tmp--)
     printf("%c",*tmp);
  }
  printf(" ");
 }
 getch();
 return 0;
}

OUTPUT :

Enter Any String : 
Life is Good


efiL si dooG

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#include
#include
int main()
{
 int i;
 printf("ASCII table & its equivalent values with numbering: \n");
 for(i=1; i<=255; i++)
 printf("\nValue: %d -> ASCII character: %c",i,i);
 getch();
 return 0;
}

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#include
#include

void main()
{
int i,j,a[10][10],sum,m,n;

/* m - Number of rows 
   n - Number of Columns   */

printf("\n Enter the number of Rows : ");
scanf ("%d",&m);

printf("\n Enter the number of Columns : ");
scanf ("%d",&n);

/* Accept the Elements in m x n Matrix */

for(i=0;i<m;i++ )
       for(j=0;j<n;j++)
       {
       printf("Enter the Element a[%d][%d] : ", i , j);
       scanf("%d",&a[i][j]);
       } 

/* Addition of all Diagonal Elements */

sum = 0;

for(i=0;i<m;i++ )
       for(j=0;j<n;j++)
       {
        if ( i == j )   
        sum = sum + a[i][j]; 
       } 

/*  Print out the Result */

printf("\nSum of All Diagonal Elements in Matrix : %d",sum); 

getch();

}

Enter the number of Rows : 2

Enter the number of Columns : 2

Enter the Element a[0][0] : 1
Enter the Element a[0][1] : 1
Enter the Element a[1][0] : 1
Enter the Element a[1][1] : 1

The Addition of All Elements in the Matrix : 2

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#include
#include

void main()
{
int i,j,a[10][10],sum,m,n;

/* m - Number of rows 
   n - Number of Columns */

printf("\n Enter the number of Rows : ");
scanf ("%d",&m);

printf("\n Enter the number of Columns : ");
scanf ("%d",&n);

/* Accept the Elements in m x n Matrix */

for(i=0;i<m;i++)
       for(j=0;j<n;j++ )
       {
       printf(" \n Enter the Element a[%d][%d] : ", i , j);
       scanf("%d",&a[i][j]);
       } 

/* Addition of all Elements */

sum = 0;

for(i=0;i<m;i++ )
       for(j=0;j<n;j++ )
       {
       sum = sum + a[i][j]; 
       } 

/* Print out the Result */
printf("\nThe Addition of All Elements in the Matrix : %d",sum); 
getch();
}

Enter the number of Rows : 2
Enter the number of Columns : 2

Enter the Element a[0][0] : 1
Enter the Element a[0][1] : 1
Enter the Element a[1][0] : 1
Enter the Element a[1][1] : 1

The Addition of All Elements in the Matrix : 4

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#include
#include
void main()
{
int i,j,a[3][3];

/* i = For Counting Rows
j = For Counting Columns */

/* two loops will Accept 9 elements from user */

for(i=0;i<3;i++)
   for(j=0;j<3;j++)
   {
   printf("\n Enter the a[%d][%d] = ",i,j);
   scanf("%d",&a[i][j]);
   }

/* Print 9 elements */

for(i=0;i<3;i++)
{
   for(j=0;j<3;j++)
      printf("%d\t",a[i][j]);
printf("\n");
}

getch();
}

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#include < stdio.h >
#include < conio.h > 

void main()
{
	int a[20];
	int i,j,n;
	clrscr();
	printf("\n Enter value for n: ");
	scanf("%d",&n);
	printf("\n Enter %d values \n",n);
	for(i=0; i < n ; i++)
	{
		scanf("%d", &a[i]);
	}
	printf(" %d values before sorting are \n",n);
	for(i=0; i < n; i++)
	{
		printf("%4d",a[i]);
	}
	simple_sort(a,n);
	printf("\n %d values after sorting are \n",n);
	for(i=0;in;i++)
	{
		printf("%4d",a[i]);
	}
}
int simple_sort(int a[],int n)
{
	int i,j,t;
	for(i = 0;i < n-1;i++)
	{
		for(j= i + 1;j < n;j++)
		{
			if(a[i]>a[j])
			{
				t=a[i];
				a[i]=a[j];
				a[j]=t;
			}
		}
	}
 return;
}

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#include
#include
struct tree_node
{
 int data;
 struct tree_node *left,*right,*parent;
};
typedef struct tree_node node;
node * new_node(int x);
node * insert_node(node *root, int x);
int count_nodes(node *root);

int main()
{
 node *root=NULL;
 int i,x,max;
 printf("how many numbers do you want to enter in tree?\n");
 scanf("%d",&max);
 printf("Enter %d numbers \n",max);
 for(i=0;i<max;i++)
 {
  scanf("%d",&x);
  root = insert_node(root, x);
 }
 printf("all numbers inserted into the tree\t press enter to count");
 fflush(stdin);
 getchar();
 printf("\nThere are total %d nodes in this tree\n",count_nodes(root));
    return 0;
}

node * insert_node(node *root, int x)
{
 if(!root)
 {
  root = new_node(x);
  return root;
 }
 if(root->data > x)
  root->left = insert_node(root->left,x);
 else
  root->right = insert_node(root->right,x);
 return root;
}

int count_nodes(node *root)
{
 if(!root)
 return 0;
 else
 return(count_nodes(root->left) + 1 + count_nodes(root->right));
}

node * new_node(int x)
{
 node *furnished;
 furnished = (node*)malloc(sizeof(node));
 furnished->data=x;
 furnished->left=NULL;
 furnished->right=NULL;
 furnished->parent=NULL;
 return furnished;
}

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