A C++ program to generate a Pascal’s triangle

A C++ program to generate a Pascal’s triangle which is as follows:

      
      1
     1 1
    1 2 1
   1 3 3 1
  1 4 6 4 1
<pre>#include<iostream>
using namespace std;
int fact(int);
main()
{
	int rows,i,j,k;
	cout<<"Enter the numbe of rows you want in the triangle:";
	cin>>rows;
	for(i=0;i<rows;i++)
	{
		//Moving each row by rows-i spaces to get a triangular shape
		for(k=0;k<(rows-i);k++) 
		cout<<" ";
		//Loop for printing each row
		for(j=0;j<=i;j++)       
		cout<<" "<<fact(i)/(fact(j)*fact(i-j)); //nCr=n!/(r!*(n-r)!)
		cout<<endl;
	}
}

int fact(int i)
{
	int value=1;	
	while(i!=0)
	{
		value=value*i;
		i--;
	}
	return value;
}</pre>

Download the program Pascals-Triangle

Program to convert infix expression to postfix in C | Shunting yard algorithm

This program is a implementation of shunting yard algorithm to convert an infix expression to post fix expression, This is a extension of the stack program published earlier

Algorithm:(Taken from wikipedia)

  • While there are tokens to be read:
  • Read a token .
  • If the token is a number, then add it to the output queue.
  • If the token is a function token, then push it onto the stack.
  • If the token is a function argument separator (e.g., a comma):
  • Until the topmost element of the stack is a left parenthesis, pop the element from the stack and push it onto the output queue. If no left parentheses are encountered, either the separator was misplaced or parentheses were mismatched.
  • If the token is an operator, o1, then:
  • while there is an operator, o2, at the top of the stack, and either
o1 is associative or left-associative and its precedence is less than (lower precedence) or equal to that of o2, or
o1 is right-associative and its precedence is less than (lower precedence) that of o2,
pop o2 off the stack, onto the output queue;
  • push o1 onto the stack.
  • If the token is a left parenthesis, then push it onto the stack.
  • If the token is a right parenthesis:
  • Until the token at the top of the stack is a left parenthesis, pop operators off the stack onto the output queue.
  • Pop the left parenthesis from the stack, but not onto the output queue.
  • If the token at the top of the stack is a function token, pop it and onto the output queue.
  • If the stack runs out without finding a left parenthesis, then there are mismatched parentheses.
  • When there are no more tokens to read:
  • While there are still operator tokens in the stack:
  • If the operator token on the top of the stack is a parenthesis, then there are mismatched parenthesis.
  • Pop the operator onto the output queue.
  • Exit.
#include <stdio.h>
#define size 10
char stack[size];
int tos=0,ele;
void push();
char pop();
void show();
int isempty();
int isfull();
char infix[30],output[30];
int prec(char);
int main()
{
                int i=0,j=0,k=0,length;
                char temp;
                printf("\nEnter an infix expression:");
                scanf("%s",infix);
                printf("\nThe infix expresson is %s",infix);
                length=strlen(infix);
                for(i=0;i<= prec(stack[tos-1])  )
						{
						temp=pop();
                                                printf("\n the poped element is :%c",temp);
                                                output[j++]=temp;
						push(infix[i]);
						printf("\n The pushed element is :%c",infix[i]);
                                                show();
						}
						else
						{
						push(infix[i]);
						printf("\nThe pushed element is:%c",infix[i]);
                                                show();
						}
					}
					else
					{
						if(infix[i]=='(')
						{
                                                push(infix[i]);
                                                printf("\nThe pushed-- element is:%c",infix[i]);
                                                }
						if(infix[i]==')')
						{
						temp=pop();
                                                while(temp!='(')
                                               {output[j++]=temp;
                                                printf("\nThe element added to Q is:%c",temp);
                                                //temp=pop();
						printf("\n the poped element is :%c",temp);
						temp=pop();}
						}
					}
				}
			}
 printf("\nthe infix expression is: %s",output);
		}
		while(tos!=0)
		{
			output[j++]=pop();
		}
printf("the infix expression is: %s\n",output);
}
//Functions for operations on stack
void push(int ele)
{
	stack[tos]=ele;
	tos++;
}
char pop()
{
	tos--;
	return(stack[tos]);
}
void show()
{
	int x=tos;
	printf("--The Stack elements are.....");
	while(x!=0)
	printf("%c, ",stack[--x]);
}

//Function to get the precedence of an operator
int prec(char symbol)
{
if(symbol== '(')
return 0;
if(symbol== ')')
return 0;
if(symbol=='+' || symbol=='-')
return 1;
if(symbol=='*' || symbol=='/')
return 2;
if(symbol=='^')
return 3;
return 0;
}




Download the Program Infix-to-postfix.c

Implementaion of Stack as an Array in C language

/////////////////////////////////////////////////////

// Implementation of stack(containing names) as an //

// array, with its basic operations PUSH,POP and   //

// also the IS_EMPTY and IS_FULL operations.       //

// Program written by Satish Gandham on 01-09-08   //

// Feel free to modify or use it directly          //

// Program compiled with GCC Compiler              //

/////////////////////////////////////////////////////

#include<stdio.h>
#define size 5
char stack[size][15],ele[20];
int tos;
void push();
char* pop();
void show();
int isempty();
int isfull();

int main()
{
	int choice;
	tos=0;
	do
	{
		printf("\tEnter 1 for push,2 for pop,3 to show,and 4 to exit\n");
		scanf("%d",&choice);
		switch(choice)
		{
		case 1:
		if (isfull())
		printf("\nThe stack is full");
		else
			{
		printf("\n Enter element to insert");
		scanf("%s",ele);
		push(ele);
			}
		break;
		case 2:
		if(isempty())
		printf("\n The stack is empty");
		else
		printf("\nThe removed element is:%s",pop());
		break;	
		case 3:
		if(isempty())
		printf("\nThe stack is empty, I cant show you any elements");
		else
		show();
		break;
		case 4:
		exit(1);
		default:
		printf("\nDo you understand english and numbers??");
		}
	}while(1);
}
int isempty()
{
	return(tos==0);
}
int isfull()
{
	return(tos==size);
}
void push(ele)
{
	//stack[tos]=ele;
	strcpy(stack[tos],ele);
	tos++;
}
char* pop()
{
	tos--;
	return(stack[tos]);
}
void show()
{
	int x=tos;	
	printf("\nThe Stack elements are.....\n");
	while(x!=0)
	printf("\t%s\n",stack[--x]);
}



Download the Program
Page 3 of 4«1234»