cpp | Student Projects

C++ class program to perform rational number arithmetic

Editorial Team — Sun, 04 Apr 2010 13:34:26 +0000

Write a program to perform rational number arithmetic.

 #include 	
 #include
 #include
 class rational 
 {
	 int numer;
	 int denom;
	 public:
	 void getdata()
	 {
		 cout<<"\n enter the numerator part of the rational no.";
		 cin>>numer;
		 cout<<"\n enter the denominator part of the rational no.";
		 cin>>denom;
	 }
	 void operator+(rational);
	 void operator-(rational);
	 void operator *(rational);
	 void operator /(rational);
 };
 void rational ::operator+(rational c1)
 {
	 rational temp;
	 temp.numer=(numer*c1.denom)+(c1.numer*denom);
	 temp.denom=denom*c1.denom;
	 cout<<"\nrational no. after addition";
	 cout<<"\n numerator="<>n;
		 switch(n)
		 {
			 case 1:
			 cout<The post C++ class program to perform rational number arithmetic first appeared on Student Projects.

C++ program using class to multiply by 10 to every member of a list

Editorial Team — Sun, 04 Apr 2010 13:19:02 +0000

Write a program using class which reads a list of N number of integer type in an array. It modifies the list by multiplying 10 to every number of the list. The modified list is displayed.

#include
#include
class array
{
 public: void readarray();
         void multiply();
};
void array::readarray()
{
	int a[10];
	cout<<"Enter the elements of the Array"<>a[i];
}
void array::multiply()
{
	int i,j,a[10],temp[10];
	for(i=0;i<=9;i++)
	{
		for(j=0;j<=9;j++)
		temp[i]=(a[i]*10);
	}
	cout<<"Result are as follows"<The post C++ program using class to multiply by 10 to every member of a list first appeared on Student Projects.

C++ program to multiply two numbers without using multiplication operator

Editorial Team — Sun, 04 Apr 2010 10:11:00 +0000

Write a c++ program to multiply two numbers without using multiplication operator.

#include
#include
void main ()
{
int a,b,i,temp=0;
clrscr();
cout<<"Enter Two numbers for multiplication";
cin>>a>>b;
for(i=1;i<=b;i++)
{
temp=temp+a;
}
cout<The post C++ program to multiply two numbers without using multiplication operator first appeared on Student Projects.

C++ program to illustrate Hybrid Inheritance.

Editorial Team — Sat, 13 Mar 2010 16:34:11 +0000

C++ program to illustrate Hybrid Inheritance.

#include
#include
class stu
{
protected:
  	int rno;
public:
 	void get_no(int a)
{
rno=a;
 	}
void put_no(void)
 	{
 		out<<"Roll no"<
OUTPUT
	Roll no  123

	Marks obtained : part1=27.5

	Part2=33

	Sports=6

	Total score = 66.5
The post C++ program to illustrate Hybrid Inheritance. first appeared on Student Projects.

C++ program to implement three classes using multiple inheritance

Editorial Team — Sat, 13 Mar 2010 15:11:45 +0000

MULTIPLE INHERITANCES

AIM:
Write a program to illustrating how all the three classes are implemented in multiple inheritance mode

ALGORITHM
• Start the process
• Invoke the class M
• Invoke the another class N
• Invoke one more class,ie class P,which is inherited by both M and N
• Create an object for class p,ie P p
• Call p.get_m(),assign the value in to ‘m’
• Call p.get_n(),assign the value in to ‘n’
• Call display(), for dispay the result
• Stop the result

PROGRAM:

	#include
	#include
	
	Class M
        {
		Protected:
		Int m;
		Public :
		Void get_M();
	};
	Class N
        {		
		Protected:
		Int n;
		Public:
		Void get_N();
	};
	Class p: public M, public N 		
	{
		Public:
		Void disply(void);
	};
	Void M ::get_m(int x)
	{
		m=x;
	}
	Void N::get_n(int y)
	{
		n=y;
	}
	Void P:: disply(void)
	{
	Cout<<”m=”<
OUTPUT
	m=10

	n=20

	m*n=200
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C++ program for overloading the unary operator ++.

Editorial Team — Sat, 13 Mar 2010 12:42:46 +0000

ALGORITHAM:
• Start the process
• Invoke the class counter
• Crate two objects c1 and c2
• Assign values to c1 an c2
o Call c1.get_count()
o Call c2.get_count()
• Increment the values
o C1++
o C2++
o ++c2
• Print c1 and c2
• Stop the process

PROGRAM

#include
#include
class counter
{
int count;
public:
counter()
{
	count=0;
}
int get_count()
{
return count;
}
void operator++()
{
count++;
}
};

void main()
{
counter c1,c2;
cout<<"\nC1 ="<C++ program for overloading the unary operator ++. first appeared on Student Projects.

c++ Program to illustrate the use of difference operators using friend function

Editorial Team — Sat, 13 Mar 2010 04:29:18 +0000

AIM:
A program to illustrate the use of difference operators to access the class member using friend function

ALGORITHM:

1) Start the process
2) Invoke the classes
3) Call the set_xy() first
a) Assign the value of x and y
b) Print the value of x and y
4) Call the sum() for second(friend)
a) Again assign the temp value of x and y
5) Print the value of x and y
6) Stop the process

PROGRAM

#include
class M
{
int x;
int y;
public:
void set_xy(int a,int b)
{
x=a;
y=b;
}
friend int sum(M m);
};
int sum(M m)
{
int M ::* px=&M :: x;
int M ::* py=&M :: y;
M *pm=&m;
int s=m.*px + pm->*py;
return s;
}
main()
{
M n;
void (M :: *pf)(int,int)=&M :: set_xy;
(n.*pf)(10,20);
cout<<"Sum="<*pf)(30,40);
cout<<"Sum="<
Output:
	Sum=30

	Sum=70
The post c++ Program to illustrate the use of difference operators using friend function first appeared on Student Projects.

C++ program that uses dynamic programming algorithm to solve the optimal binary search tree problem

Editorial Team — Thu, 11 Mar 2010 17:39:44 +0000

/* Write a C++ program that uses dynamic programming algorithm to solve the optimal binary search tree problem */

#include
#include
#include
using namespace std;
#define MAX 10
int find(int i,int j);
void print(int,int);
int p[MAX],q[MAX],w[10][10],c[10][10],r[10][10],i,j,k,n,m;
char idnt[7][10];

main()
{
	cout << "enter the no, of identifiers";
	cin >>n;
	cout <<"enter identifiers";
	for(i=1;i<=n;i++)
	gets(idnt[i]);
	cout <<"enter success propability for identifiers";
	for(i=1;i<=n;i++)
		cin >>p[i];
	cout << "enter failure propability for identifiers";
	for(i=0;i<=n;i++)
		cin >> q[i];
	for(i=0;i<=n;i++)
	{
		w[i][i]=q[i];
		c[i][i]=r[i][i]=0;
		w[i][i+1]=q[i]+q[i+1]+p[i+1];
		r[i][i+1]=i+1;
		c[i][i+1]=q[i]+q[i+1]+p[i+1];
	}
	w[n][n]=q[n];
	r[n][n]=c[n][n]=0;
	for(m=2;m<=n;m++)
	{
		for(i=0;i<=n-m;i++)
		{
		     j=i+m;
		     w[i][j]=w[i][j-1]+p[j]+q[j];
		     k=find(i,j);
		     r[i][j]=k;
		     c[i][j]=w[i][j]+c[i][k-1]+c[k][j];
		}
	}
       cout <<"\n";
       print(0,n); }

int find(int i,int j)
{
int min=2000,m,l;
for(m=i+1;m<=j;m++)
if(c[i][m-1]+c[m][j]
OUTPUT
enter the no, of identifiers4

enter identifiers

do

if

int

while

enter success propability for identifiers3 3 1 1

enter failure propability for identifiers2 3 1 1 1
tree in preorder form

if

do

int

while
The post C++ program that uses dynamic programming algorithm to solve the optimal binary search tree problem first appeared on Student Projects.

C++ programs for sorting numbers in ascending order using Quick sort method

Editorial Team — Thu, 11 Mar 2010 17:27:51 +0000

/* Write C++ programs for sorting a given list of elements in ascending order using Quick sort sorting methods */

#include
#include
int a[10],l,u,i,j;
void quick(int *,int,int);
void main()
{
clrscr();
cout <<"enter 10 elements";
for(i=0;i<10;i++)
cin >> a[i];
l=0;
u=9;
quick(a,l,u);
cout <<"sorted elements";
for(i=0;i<10;i++)
cout << a[i] << " ";
getch();
}

void quick(int a[],int l,int u)
{
   int p,temp;
   if(lp && i<=j )
	   j--;
      if(i<=j)
      {
      temp=a[i];
      a[i]=a[j];
      a[j]=temp;}
  }
  temp=a[j];
  a[j]=a[l];
  a[l]=temp;
  cout <<"\n";
  for(i=0;i<10;i++)
  cout <
OUTPUT
enter 10 elements5 2 3 16 25 1 20 7 8 61 14
1 2 3 5 25 16 20 7 8 61

1 2 3 5 25 16 20 7 8 61

1 2 3 5 25 16 20 7 8 61

1 2 3 5 25 16 20 7 8 61

1 2 3 5 25 16 20 7 8 61

1 2 3 5 8 16 20 7 25 61

1 2 3 5 7 8 20 16 25 61

1 2 3 5 7 8 16 20 25 61

1 2 3 5 7 8 16 20 25 61 
sorted elements1 2 3 5 7 8 16 20 25 61
The post C++ programs for sorting numbers in ascending order using Quick sort method first appeared on Student Projects.

C++ program that uses non-recursive functions to traverse a binary tree in Post-order

Editorial Team — Thu, 11 Mar 2010 17:22:47 +0000

/* Write C++ program that uses non-recursive functions to traverse a binary tree in Post-order */

#include
#include
#include

class node
{
public:
class node *left;
class node *right;
int data;
};

class tree: public node
{
public:
int stk[50],top;
node *root;
tree()
{
root=NULL;
top=0;
}
void insert(int ch)
{
	node *temp,*temp1;
	if(root== NULL)
	{
		root=new node;
		root->data=ch;
		root->left=NULL;
		root->right=NULL;
		return;
	}
	temp1=new node;
	temp1->data=ch;
	temp1->right=temp1->left=NULL;
	temp=search(root,ch);
	if(temp->data>ch)
		temp->left=temp1;
	else
		temp->right=temp1;

}
node *search(node *temp,int ch)
{
	if(root== NULL)
	{
		cout <<"no node present";
		return NULL;
	}
	if(temp->left==NULL && temp->right== NULL)
		return temp;

	if(temp->data>ch)
	     {  if(temp->left==NULL) return temp;
		search(temp->left,ch);}
	else
	      { if(temp->right==NULL) return temp;
	      search(temp->right,ch);

}              }

void display(node *temp)
{
	if(temp==NULL)
	 return ;
	display(temp->left);
       cout<data << " ";
	display(temp->right);
}
void postorder( node *root)
{
	node *p;
	p=root;
	top=0;

	while(1)
	{
	while(p!=NULL)
	{
		stk[top]=p->data;
		top++;
		if(p->right!=NULL)
			stk[top++]=-p->right->data;
		p=p->left; 
	}
	while(stk[top-1] > 0 || top==0)
	{
	   if(top==0) return;
	   cout << stk[top-1] <<" ";
	   p=pop(root);
	}
	if(stk[top-1]<0)
	{
	  stk[top-1]=-stk[top-1];
	  p=pop(root);
      }	}

}
node * pop(node *p)
{
int ch;
ch=stk[top-1];
if(p->data==ch)
{
top--;
return p;
}
if(p->data>ch)
pop(p->left);
else
pop(p->right);
}
};
void main()
{
	tree t1;
	int ch,n,i;
	clrscr();
	while(1)
	{
		cout <<"\n1.INSERT\n2.DISPLAY 3.POSTORDER TRAVERSE\n4.EXIT\nEnter your choice:";
		cin >> ch;
		switch(ch)
		{
		case 1:   cout <<"enter no of elements to insert:";
			  cout<<"\n enter the elements";
			  cin >> n;
			  for(i=1;i<=n;i++)
			  {  cin >> ch;
			     t1.insert(ch);
			  }
			   break;
		case 2:   t1.display(t1.root);break;
		case 3:   t1.postorder(t1.root); break;
		case 4:   exit(1);
		}
	}
}

OUTPUT

1.INSERT
2.DISPLAY 3.POSTORDER TRAVERSE
4.EXIT
Enter your choice:1
enter no of elements to insert:
enter the elements7
5 24 36 11 44 2 21

1.INSERT
2.DISPLAY 3.POSTORDER TRAVERSE
4.EXIT
Enter your choice:2
2 5 11 21 24 36 44

1.INSERT
2.DISPLAY 3.POSTORDER TRAVERSE
4.EXIT
Enter your choice:3
2 21 11 44 36 24 5

1.INSERT
2.DISPLAY 3.POSTORDER TRAVERSE
4.EXIT
Enter your choice:4

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