it is wrong .

1. Divide the coins in 3 groups of 3 coins each
2. Put 2 groups in the weighing scale
3a. If both the groups weigh equal
    1. Pick 2 coins from the 3rd group and put in weighing scale
    2. If they weigh equal, the 3rd coin is counterfeit.
Else the coin which weigh more is counterfeit
3b. If they do not weigh equal, pick the group with more
weight and do steps 3a1 and 3a2.  

1.t(n)=10*n*n*(n+1)/2
2.t(n)={n/4*(n/4)*[(n/4)+1]}/2

a-   (n*n)*log(n).
b-  log(n)*log(n).
c-log(n)*log(n)*log(n).

The reader and writer problem is IMPLEMENTED THROUGH 
SEMAPHORES.

THE EFFECTIVE SOLUTION TO SOLVE THE READER AND WRITER 
SOLUTION IS :


A)    If a writer is waiting in the ready queue to enter 
into the critical section, then the current reading process 
should be completed without delay.

B)    Only one reader can be put into the ready queue 
because the writer is waiting.

c)    The semaphores such as ReadCount-to count the number 
of reader in ready queue,
Reader and Writer, these are the important semaphores to 
solve the reader and writer problem.

i << 3

/*      PROGRAM TO FIND WHETHER A NUMBER IS A MULTIPLE OF 2 
AND 8 WITHOUT
		USING ARITHMETIC OPERATOR  */

#include<stdio.h>
#include<conio.h>
void main()
 {
   int num;
   clrscr();
   printf("\n enter a  number");
   scanf("%d",&num);
   if(num & 1)
    printf("\n not a multiple of 2");
   else
    printf("\n a multiple of 2");
   if(num & 7)
    printf("\n not a multiple of 8");
   else
    printf("\n a multiple of 8");

   getch();
 }

i=n<<3;

boolean div8(x){
  return (x & 7)
}

if return is 0 = divisible by 8

int i,n=1;
 i=n<<3

#include <stdio.h>
main()
{
      int x,n;
      printf("Enter the X number:");
      scanf("%d",&x);
      n=(x<<3)-x;
      printf("The Answer is : %d",n);
      getch();
}

#include <stdio.h>
main()
{
      int x,n;
      printf("Enter the X number:");
      scanf("%d",&x);
      n=(x<<3)-x;
      printf("The Answer is : %d",n);
      getch();
}

log of base 2 means given variable is power of 2. So 

void main(){
     int x;
     if(x&(x-1){ 
        printf("Given Number is not log base 2\n");
     }
 
}

insertion sort took more time, because it done after 
complete selection sort...

recursive reverse(ptr)
  if(ptr->next==NULL)
  return ptr;
  temp=reverse(ptr->next);
  ptr=ptr->next;
  return ptr;
  end

node *reverse(node *first)


{
node *cur,*temp;


cur=NULL;

while(first!=NULL)


{temp=first;

first=first->link;

temp->link=cur;

cur=temp;
}

return cur;

}

/*
the structure is as follows:
struct node
{
   int data;
   struct node *next;
}
*/

struct node * reverse (struct node *home , struct node *rev)
{  
  struct node temp , *p;

  if(home != NULL)
  {
    temp = home;
    while(temp != NULL)
    {
       //this part will create a new node with the name p.
       p = myalloc;
       p -> data = temp -> data;
       p -> next = NULL;
          
       if(rev == NULL)
          rev = p;
       else
       {
              p -> next = rev;
              rev = p;
       }
       temp = temp -> next;
      }
   }
return rev;
}

**Liked list as a stack = linked list in reverse order.

/*
the structure is as follows:
struct node
{
   int data;
   struct node *next;
}
*/

struct node * reverse (struct node *home , struct node *rev)
{  
  struct node temp , *p;

  if(home != NULL)
  {
    temp = home;
    while(temp != NULL)
    {
       //this part will create a new node with the name p.
       p = myalloc;
       p -> data = temp -> data;
       p -> next = NULL;
          
       if(rev == NULL)
          rev = p;
       else
       {
              p -> next = rev;
              rev = p;
       }
       temp = temp -> next;
      }
   }
return rev;
}

write a link list program to insert integer number and then 
display its sum as well. Your program should display the 
address as well as the result
write a program to reverse a link list so that the last 
element becomes the first one and so on
copy one link list to another list

//Go through the code, incase any issues feel free to 
revert.

Copying a link list:

//home is the stationary pointer of the main linked list 
which is to be copied.

//head is the stationary pointer of the new linked list 
which has to be created..

//temp and temp1 are the moving pointers attached to the 
respective linked list...


struct node *copy(struct node *home , struct node *head)
{
  struct node *temp , *temp1 , *p;
  
  temp = home;
  while(temp != NULL)
  {
       p = (struct node *) malloc (sizeof(struct node));
       p -> data = temp -> data;
      
       if(head == NULL)
             head = p;
        
        else
        {
             temp1 = head;
             while(temp1 -> next != NULL)
                   temp1 = temp1 -> next;
               
              temp1 -> next = p;
         }
    temp = temp -> next;
}
return head;
}

#include <stdio.h>

typedef struct list
{
 int data;
 struct list * next;
} LIST;

LIST* reverse(LIST *head)
{
  LIST* temp;
  LIST* temp1;

  if (head == NULL || head->next == NULL)
    return head;
  else
  {
    temp = head->next;
    head->next = NULL;
    while(temp)
    {
      temp1 = temp->next;
      temp->next = head;
      head = temp;
      temp = temp1;
    }

    return head;
  }
}

int main(int argc, char ** argv)
{
 int i=0;
 LIST* head = NULL;
 LIST* node = NULL;
 int count;

 if (argc < 2) printf("usage: a.out <count>");
 else count = atoi(argv[1]);

 for (i=0;i<count;i++)
 {
   node = (LIST*)calloc(sizeof(LIST));
   node->data = i;
   node->next = head;

   head = node;
 }

 node = head;
 while(node)
 {
   printf("before %d\n", node->data);
   node = node->next;
 }

 head = reverse(head);
 printf("after head=%d\n", head->data);
 node = head;
 while(node)
 {
   printf("after %d\n", node->data);
   node = node->next;
 }
}

#include<iostream.h>

#include<string.h>//complexity-n/2

int main()
{
int i,l,l1;
char str[100];
cout<<"enter string:";
gets(str);
l=strlen(str);
if(l%2==0)
l1=(l/2-1);
else
l1=l/2;
for(i=0;i<=l1;i++)/*swap elements from 2 ends till u reach
middle part of array*/
{
char t=str[i];
str[i]=str[l-i-1];
str[l-i-1]=t;
}
str[l]=0;
cout<<"\n\nreversed string is:"<<str;
getch();
return 0;
}

#include<stdio.h>
#include<string.h>
main(int argc, char *argv[])
{
  char  *string = argv[1];
  int len = strlen(string);
  int i = 0;
  int j = len - 1;
  printf("string before is %s\n", string);
  printf("len is %d\n", len);
  while(i <= j)
  {
    *(string+i) += *(string+j);
    *(string+j) = *(string+i) - *(string+j);
    *(string+i) = *(string+i) - *(string+j);
    i++;
    j--;
    if(len % 2)
     if(i == j) break;
  }
  printf("reversed string is %s\n", string);
}

#include<stdio.h>

void reverse(char *);

void main()
{
	char str[]="Hello";

	reverse(str);
	printf("Reverse String is %s",str);
	
}

void reverse(char *p)
{

	char *q=p;
	while(*++q!='\0');
	q--;

	while(p<q)
	{
		*p=*p+*q;
		*q=*p-*q;
		*p=*p-*q;
		p++;
		q--;
	}

}

refer c book!
    or
contact me

#inclue<stdio.h>
       #include<string.h>
       main(){
           char a[10],i;
           int len=1;
           printf(" Enter string ");
           fgets(a,9,stdin);
     
           len = strlen(a);
     
      
          for(i=0 ; i<(len/2) ; i++){
                  a[i]=a[i]+a[len-2];
                  a[len-2]=a[i]-a[len-2];
                  a[i]=a[i]-a[len-2];
                  len--;
          }
          printf("\n Reverse string with n/2 complexity
 %s",a);
          return 0;
      }

It's all O(n). You're finding the length of the string,
which itself is an O(n) operation.
So, O(n + n/2) = O(n).

#include <iostream>
#include <string>
using std::string;
using std::cout;
using std::endl;
using std::cin;

int main()
{
    string s("abcdefghijklmnopqrstuvwxyz");
    string::size_type s_size = s.size();

    for (string::size_type x = 0; x != s_size; x++){

        if ( (s_size -x -1) > x ){

            s[x] ^= s[s_size - x -1];
            s[s_size - x -1] ^= s[x];
            s[x] ^= s[s_size - x -1];

        }else{
            break;
        }

    }

    cout << s << endl;
}

strpy(n1,n2)
here copy of string n2 in string n1

while(str1[i]!=0) str2[j++]=str1[i++];

for(i=0;(str2[i]=str1[i])!='\0';i++);

while ((*target++ = *source++));

for(i=0; str2[i] = str1[i]; i++);

strcpy(s1,s2); //copies string s1 to s2

void cpy_user(char *s, char *t)
{
while ((*s++ = *t++) != '\0');
}

If your document store in database in binary format then
code for retrive the document is

string sqlQuery="write the select statement"
datareaderObject =cmd.executereader(sqlQuery,connectionname);
if(datareaderObject.read())
{
 Response.Buffer = false;
            Response.ClearHeaders();
            Response.ContentType = "application/octet-stream";
            Response.AddHeader("Content-Disposition",
"attachment; filename=" +
datareaderObject["docname"].ToString());

            //Code for streaming the object while writing
            const int ChunkSize = 1024;
            byte[] buffer = new byte[ChunkSize];
            byte[] binary = (datareaderObject["doc"]) as byte[];
            MemoryStream ms = new MemoryStream(binary);
            int SizeToWrite = ChunkSize;
            for (int i = 0; i < binary.GetUpperBound(0) - 1;
i = i + ChunkSize)
            {
                if (!Response.IsClientConnected) return;
                if (i + ChunkSize >= binary.Length)
SizeToWrite = binary.Length - i;
                byte[] chunk = new byte[SizeToWrite];
                ms.Read(chunk, 0, SizeToWrite);
                Response.BinaryWrite(chunk);
                Response.Flush();
            }
            Response.Close();
        }
    }

if document store in virtual path


 if (datareadear.Read())
        {
            FileInfo file = new
System.IO.FileInfo(Server.MapPath("datareadear["filename"].ToString());
            if (file.Exists)
            {
                Response.Clear();
                Response.AddHeader("content-disposition",
"attachment; filename=" + file.Name);
                Response.AddHeader("Content-Length",
file.Length.ToString());
                Response.ContentType =
"application/octet-stream";
                Response.WriteFile(file.FullName);
                Response.End();
            }
            else
            {
                Response.Redirect("../error.aspx?error=" +
"File dose not exits");
                Response.End();
            }
        }

dim con as sqlconncetion
dim cmd as sqlcommand
dim parm as sqlparameter
con=new sqlconnection
("server=localhost;database=Database1;username=sa;password=s
a")
cmd=new sqlcommand
con.open()
cmd.commandtype=commandtype.storedprocedure
cmd.commandtext="StoredprocedureName"
cmd.parmaeters.add("@Name",txtname.txt)
cmd.parameters.add("@sal",txtname.txt)
cmd.excutenonquery()
con.close()

dim con as sqlconncetion
dim cmd as sqlcommand
dim parm as sqlparameter
con=new sqlconnection
("server=localhost;database=Database1;username=sa;password=s
a")
cmd=new sqlcommand
con.open()
cmd.commandtype=commandtype.storedprocedure
cmd.commandtext="StoredprocedureName"
cmd.parmaeters.add("@Name",txtname.txt)
cmd.parameters.add("@sal",txtname.txt)
cmd.excutenonquery()
con.close()

SqlConnection conn = new SqlConnection(strconn);
SqlCommand comm = new 
SqlCommand"[StoreProcedureName]",conn);
comm.CommandType = CommandType.StoredProcedure;
comm.Parameters.AddWithValue("@UserId", Id);
sqldataadapter da= new sqldataadapter;
Dataset ds = new dataset;
try
{
conn.open();
da.fill(ds);
conn.fill;
}
catch
{
throw(ex);
}
dlFilldatalist.datasourse = ds;
dlFilldatalist.databind();