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Program to arrange the string in alphabetical order

chitra — Tue, 30 Aug 2011 14:24:39 +0000

Here is the program to arrange the string in alphabetical order.

  #include "stdio.h"
  #include "conio.h"
  #include "string.h"
  void main()
  {
     char str[10],temp;
     int i,j;
     clrscr();
     printf("\nEnter the string : ");
     scanf("%s",str);
     for(i=0;i<strlen(str);i++)
       for(j=0;j<strlen(str);j++)
	 if(str[i]<str[j])
	 {
	   temp = str[i];
	   str[i] = str[j];
	   str[j] = temp;
	 }
     printf("Reversed String is : %s",str);
     getch();
  }

Binary Search

chitra — Mon, 15 Aug 2011 17:43:26 +0000

A binary search program locates the position of an item in a sorted array.

Logic: The Binary search starts by testing the element at the middle of the array. Let us consider an array ‘a’ with all elements arranged in ascending order. Let low and high are the lower and upper indices of the array ‘a’, respectively. We want to search an item K in it. To do this, we calculate a middle index with the values of low and high, let it be mid. Then we compare if K=a[mid] or not.

If it is true, then search is done. Otherwise, the search is to be repeated on left or right half depending on the Ka[mid], respectively. In case of low =high and K a[mid], the search terminates unsuccessfully.

Example 1:

Consider the following elements stored in an array,

Example 2:

Consider the following elements stored in an array,

Program:

#include"stdio.h"
#include"conio.h"
void main()
{
	int a[25], i, n, K, flag = 0, low, high, mid;
	clrscr();
	printf("Enter the number of elementsn");
	scanf("%d", &n);
	printf("Enter the elementsn");
	for(i = 0; i<n; i++)
	{
		scanf("%d",&a[i]);
	}
	printf("Enter the key to be searchedn");
	scanf("%d",&K);
	low = 0;
	high = n - 1;
	while(low <= high)
	{
		mid = (low+high)/2;
		if(a[mid] == K)
		{
			flag = 1;
			break;
		}
		else if(K<a[mid])
			high = mid-1;
		else
			low = mid + 1;
	}
	if(flag == 1)
	{
		printf("Key element is found");
	}
	else
	{
		printf("Key element not found");
	}
	getch();
}

C program to implement tower of Hanoi problem

Ranjith — Sun, 04 Apr 2010 17:28:33 +0000

To implement tower of Hanoi problem.

#include 	  
void hanoi_tower(char,char,char,int);
void hanoi_tower(char peg1,char peg2,char pege3,int n)
{
	if(n<=0)
        printf("\n Illegal Entry");
        if(n==1)
			printf ("\n Move disk from %c to %c", pege1,pege3);
		else
		{
		   hanoi_tower(peg1,peg3,peg2,n-1);
		   hanoi_tower(peg1,peg2,peg3,1);
		   hanoi_tower(peg2,peg1,peg3,n-1);
		}
}
Void main ()
{
    int n;
    printf("\n Input the number of dise:);
    scanf("%d", &n);
    printf("\n Tower of Hanoi for 5th DISC", n);
    hanoi_tower('x','y','z',n);
}

C program to Implement Morse code to text conversion and vice-versa.

Ranjith — Sun, 04 Apr 2010 15:52:44 +0000

C program to Implement Morse code to text conversion and vice-versa.

#include
#include
#include
#include
 
void main()
{
	char str[25],str1[100];
	clrscr();
	fflush(stdin);
	printf("enter the String");
	gets(str);
 
	int j=0;
	for(int i=0;i<=strlen(str);++)
	{
		switch(toupper(str[i]))
		{
		  case 'A':
		  str1[j++]='.';
		  str1[j]='.';
		  break;
 
		  case 'b':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='.';
		  break;
 
		  case 'c':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='.';
		  break;
 
		  case 'D':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='.';
		  break;
 
		  case 'E':
		  str1[j]='.';
		  break;
 
		  case 'F':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='.';
		    break;
 
		  case 'G':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='.';
		  break;
 
		  case 'H':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='.';
		    break;
 
		   case 'I':
		  str1[j++]='.';
		  str1[j]='.';
		    break;
 
		case 'J':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='.';
		    break;
 
		case 'K':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='.';
		    break;
 
		case 'L':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='.';
		  break;
 
		case 'M':
		  str1[j++]='-';
		    str1[j]='-';
		    break;
 
		case 'N':
		  str1[j++]='-';
		  str1[j]='.';
		    break;
 
		case 'O':
		  str1[j++]='-';
		  str1[j++]='-';
		  str1[j]='.';
		    break;
 
		case 'P':
		  str1[j++]='.';
		  str1[j++]='-';
		  str1[j++]='-';
		  str1[j]='.';
		    break;
 
		case 'Q':
		  str1[j++]='-';
		  str1[j++]='-';
		  str1[j++]='.';
		  str1[j]='.';
		    break;
 
		case 'R':
		  str1[j++]='.';
		  str1[j++]='-';
		  str1[j]='.';
		    break;
 
		case 'S':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='.';
		    break;
 
		case 'T':
		  str1[j]='-';
		    break;
 
		case 'U':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='.';
		    break;
 
		case 'V':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='-';
		    break;
 
		case 'W':
		  str1[j++]='.';
		  str1[j++]='-';
		  str1[j]='-';
		    break;
 
		case 'X':
		  str1[j++]='-';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='-';
		    break;
 
		case 'y':
		  str1[j++]='-';
		  str1[j++]='.';
		  str1[j++]='-';
		  str1[j]='.';
		    break;
 
		case 'Z':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='.';
		    break;
 
		case '0':
		  str1[j++]='-';
		  str1[j++]='-';
		  str1[j++]='-';
		  str1[j++]='-';
		  str1[j]='-';
		    break;
 
		case '1':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='-';
		  str1[j++]='-'; 
		  str1[j]='-';
		    break;
 
		case '2':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='.';
		    break;
 
		case 'F':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='.';
		    break;
 
		case '3':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='-';
		  str1[j]='.';
		    break;
 
		case '4':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='-';
		    break;
 
		case '5':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='.';
		    break;
 
		case '6':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='-';
		    break;
 
		case '7':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='-';
		    break;
 
		case '8':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='-';
		    break;
 
		case '9':
		  str1[j++]='-';
		  str1[j++]='-';
		  str1[j++]='-';
		  str1[j++]='-';
		  str1[j]='.';
		    break;
 
		case '.':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='-';
		    break;
 
		case ',':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='-';
		    break;
 
		case ':':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='-';
		    break;
 
		case '?':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='-';
		    break;
 
		/* case '\';
		  str1[j++]='.';
		  str1[j++]='.'; 
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='-';
		    break; */
 
		case '-':
		  str1[j++]='-';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='-';
		    break;
 
		case ';':
		  str1[j++]='-';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='-';
		  str1[j]='.';
		    break;
 
		case '(':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='-';
		    break;
 
		case ')':
		  str1[j++]='.';
		  str1[j++]='.'; 
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='-';
		    break;
 
		case ']':
		  str1[j++]='-';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='-';
		  str1[j++]='-';
		  str1[j]='-';
		    break;
 
		case '[':
		  str1[j++]='-';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='-';
		  str1[j++]='-';
		  str1[j]='.';
		    break;
 
		case '}':
		  str1[j++]='-';
		  str1[j++]='.';
		  str1[j++]='-';
		  str1[j++]='-';
		  str1[j++]='.';
		  str1[j]='-';
		    break;
 
		case '{':
		  str1[j++]='-';
		  str1[j++]='.';
		  str1[j++]='-';
		  str1[j++]='-';
		  str1[j++]='.';
		  str1[j]='-';
		    break;
 
		case '"':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='-';
		    break;
 
		case '+':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='-';
		    break;
 
		case '/':
		  str1[j++]='.';
		  str1[j++]='.';  
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='-';
		    break;
 
		case '%':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='-';
		    break;
 
		case '&':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='-';
		    break;
 
		case '$':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='-';
		    break;
 
		case '*':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='-';
		    break;
 
		case '^':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='-';
		    break;
 
		case '#':
		  str1[j++]='.';
		  str1[j++]='.';  
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='-';
		    break;
 
		case '@':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='-';
		    break;
 
		case '=':
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j++]='.';
		  str1[j]='';
		    break;
		}
	    j++;
	}
	str1[j-1]='\0';
	puts(str1);
	getch();
 }

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

Ranjith — 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"<<endl;
	for(int i=0;i<=9;i++)
	  cin>>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"<<endl;
	for(j=0;j<=9;j++)
		cout<<temp[j]<<endl;
}
void main()
{
	array mul;
	clrscr();
	mul.readarray();
	mul.multiply();
	getch();
}

C program to calculate the GCD of a number

Ranjith — Sun, 04 Apr 2010 13:04:32 +0000

Write a program to calculate the Greatest Common Divisor(GCD) of a number

#include
#include
#include
int gcd(int m,int n)
{
	int rem;
	while(n!=0)
	{
		rem=m%n;
		m=n;
		n=rem;
	}
	return(m);
}
main()
{
	int num1,num2,num3,gcd1,gcd2;
	clrscr();
	printf("Enter three positive integers");
	scanf("%d%d%d",&num1,&num2,&num3);
	if(num1==0 && num2==0 && num3==0)
	{
		printf("\n Invalid number");
		exit(0);
	}
	gcd1=gcd(num1,num2);
	gcd2=gcd(num3,gcd1);
	printf("\n GCD of %d %d %d is : %d\n",num1,num2,num3,gcd2);
	getch();
}

C++ program to calculate a to the power b without using power function

Ranjith — Sun, 04 Apr 2010 12:51:05 +0000

Write a C++ program to calculate a to the power b without using power function.

#include
#include
void main()
{
	int a,b,i,temp=1;
	clrscr();
	cout<<"Enter number for operation";
	cin>>a>>b;
	for(i=1;i<=b;i++)
	{
		temp=temp*a;
	}
	cout<<endl<<"Result are:: "<<temp;
	getch();
}

C program for Syntax Analyzer

Ranjith — Mon, 29 Mar 2010 16:54:13 +0000

C program for Syntax Analyzer

#include
#include
#include
#include
void main()
 {
    int i,j,k=0,count,inc=0,n;
   char name[30],open[30],ch,chh,o[30];
   char op[20]={'=','+','-','*','/','%','^','&','|'};
   clrscr();
   textcolor(3);
   cprintf("--Syntax Analyser--");
   printf("\n");
   printf("\n Enter Syntax");
   printf("\n");
   scanf("%s",name);
   n=strlen(name);
   for(i=0;i<n;i++)
    {
      ch=tolower(name[i]);
      for(j=0;j<9;j++)
      {
        if(ch==op[j])
         {
           open[k]=i;
           o[k]=ch;
           k++;
         }
      }
   }
   for(i=0;i<k;i++)
    {
      count=open[i];
      ch=tolower(name[count-1]);
      chh=tolower(name[count+1]);
      if(isalpha(ch)&&isalpha(chh)||isdigit(chh))
       ++inc;
     }
   if(k==inc)
  printf("\n %s is a valid syntax",name);
 else
  printf("\n %s is an invalid syntax",name);
  getch();
}

C program for Shortest Job Next

Ranjith — Mon, 29 Mar 2010 16:51:08 +0000

C program for Shortest Job Next

 #include
 #include
 #include
 void main()
 {
   int s,t[20],temp,n,x[20],w[20],i,j,c,d,b[20];
   float avgw,avgt;
   clrscr();
   printf("\n Enter no.of job:");
   scanf("%d",&n);
   printf("enter the job:");
   for(i=1;i<=n;i++)
    {
      scanf("%d",&b[i]);
      x[i]=i;
    }
   for(i=1;i<=n-1;i++)
   {
    for(j=i+1;j<=n;j++)
    {
      if(b[i]>b[j])
      {
        temp=b[i];
        b[i]=b[j];
        b[j]=temp;
        s=x[i];
        x[i]=x[j];
        x[j]=s;
      }
   }
 }
   c=0;
   d=0;
   for(i=1;i<=n;i++)
    {
       w[i]=w[i-1]+b[i-1];
       t[i]=b[i]+t[i-1];
       w[1]=0;
       t[1]=b[1];
       c+=w[i];
       d+=t[i];
    }
  avgw=(float)c/n;
  avgt=(float)d/n;
 printf("job \t waiting time \t service time \t turn around time \t \n :");
 for(i=1;i<=n;i++)
  {
     printf("%d \t %d \t ",x[i],b[i]);
     printf("%d \t %d \t",w[i],t[i]);
     printf("\n");
  }
     printf("average wait time %f \n",avgw);
     printf("average turn around %f \n",avgt);
     getch();
}

C program for Round Robin

Ranjith — Mon, 29 Mar 2010 16:48:08 +0000

C program for Round Robin

 
#include
#include
#include
int t,n,s,bt[10],ct[10],ta[10],w[10],lat[10],wav,taav;
int allover()
 {
   for(int i=0;i<n;i++)
   if(ct[i]>0)
   return 0;
   return 1;
 }
void select(int p)
 {
   w[p]+=t-lat[p];
   if(ct[p]>=s)
   {
     ct[p]-=s;
     t+=s;
   }
 else
  {
    t+=ct[p];
    ct[p]=0;
  }
   if(ct[p]==0)
    ta[p]=t;
   lat[p]=t;
 }
void main()
 {
   int p=0;t=0;taav=0;wav=0;
   clrscr();
   printf("Enter the number of process : ");
   scanf("%d",&n);
   printf("\n Enter the time slice : ");
   scanf("%d",&s);
   printf("\nEnter the burst time of processes \n ");
   for(int i=0;i<n;i++)
   {
     printf("\n process %d : ",i+1);
     scanf("%d",&bt[i]);
     ct[i]=bt[i];
     lat[i]=0;
   }
 while(!allover())
  {
    if(ct[p]!=0)
   select(p);
   p=(p+1)%n;
  }
 printf("\n\n Process    Burst time   Wait time    Turn around \n\n");
 for(i=0;i<n;i++)
  {
    printf("\n %d\t\t%d\t\t%d\t\t%d",i+1,bt[i],w[i],ta[i]);
    wav+=w[i];
    taav+=ta[i];
  }
 wav/=n;
 taav/=n;
 printf("\n\nAverage waiting time : %d ",wav);
 printf("\n\nAverage turn around time : %d ",taav);
 getch();
}