int m_plus(string a, char c)
{
if (a[0] != c)
return -1;
int i=0;
while(a[i++] == c);
return i-1;
}

int m_star(string a, char c)
{
if (a[0] != c)
return 0;
int i=0;
while(a[i++] == c);
return i-1;
}

int m_dot(string a, char c)
{
if (a[0] != c)
return -1;
else
return 1;
}

int main()
{
string s,reg;
int i,j,p;
cin >> reg>>s;

while (reg!=”" )
{
if (reg.length()>=2)
{
if (reg[1] == ‘+’)
p = m_plus(s,reg[0]);
else if (reg[1] == ‘*’)
p = m_star(s,reg[0]);
else
p = m_dot(s,reg[0]);
if ( p == -1)
{
cout<<”Error”;
return 0;
}

s = s.substr(p,s.length());

if (reg[1] == ‘+’ || reg[1] == ‘*’ || reg[1] == ‘.’)

reg = reg.substr(2,reg.length());
else
reg = reg.substr(1,reg.length());
}
else
{
p = m_dot(s,reg[0]);
if ( p == -1)
{cout<<”Error”;
return 0;
}
s = s.substr(p,s.length());
reg = reg.substr(1,reg.length());
}
}

if (s!=”")
{cout<<”Error”;

}
else
cout<<”Matched”;
return 0;
}

if a is smaller , c becomes negetive . thus when right shifting by 31 bits , c becomes 1 and k becomes 1.
In calculation for max -k * negative c becomes positive assigning b as max value .

public static int numZeros(int num) {

int count = 0;

if (num < 0) {

System.out.println(“Factorial is not defined for 0; i *= 5) {

count += num / i;

}

return count;
}
Let’s walk through an example to see how this works: Suppose num = 26. In the first loop, we
count how many multiples of five there are by doing 26 / 5 = 5 (these multiples are 5, 10, 15,
20, and 25). In the next loop, we count how many multiples of 25 there are: 26 / 25 = 1 (this
multiple is 25). Thus, we see that we get one zero from 5, 10, 15 and 20, and two zeros from
25 (note how it was counted twice in the loops). Therefore, 26! has six zeros.

ll factorial (int x)
{
ll f = 1;
for(int i=2;i<=x; i++)
f*= i;
return f;
}

ll count(int c1, int c2, int c3)
{
ll c = factorial(c1+c2+c3)/ (factorial(c1) * factorial(c2) * factorial(c3));
return c;
}

int main()
{

int s = 47, p1= 2, p2 = 3, p3 = 7;

int m1,m2,m3;

int i,j,k;
ll c= 0;
cout<<s/p2<<endl;
for(i=0;i<=s/p1;i++)
for(j=0;j<=s/p2;j++)
for(k=0;k<=s/p3; k++)
if (i * p1 + j * p2 + k*p3 == s)
{
//cout<<i<<” “<<j<<” “<<k<<endl;
c += count(i,j,k);
}
cout<<c<<endl;
return 0;
}

using namespace std;

int main()
{
vector<int> maxi,mini;

int a[10] = {3,2,1,0,9,4,-1,6,8,9};
int i,s,m;

for ( i=0; i<9; i+=2) // n/2 comparisons
{
s = a[i]+a[i+1];
m = max(a[i],a[i+1]);
maxi.push_back(m);
mini.push_back(s-m);

}

cout<<*max_element(maxi.begin(), maxi.end())<<endl; // n/2 comparisons
;cout<<*min_element(mini.begin(), mini.end())<<endl; // n/2 comparisons

return 0;
}

using namespace std;

int main()
{
int a[100]={0},n;
int i;
int maxy[10]={0};
cout<<”\nEnter number of elements in array”;
cin >> n;
for(i=0; i<n;i++)
cin >> a[i];

for(i=n-1; i>=0 ; i–)
maxy[i] = a[i] +max( maxy[i+2],maxy[i+3]);

cout<<max(maxy[0], maxy[1]);
return 0;
}

Snake

Hi everyone ,
In this blogpost I would like to share my OpenGl game project which I had developed for my semester course on Computer Graphics . It took almost a week to complete this project .

Project Description

The project’s aim was to create a 2d snake game . This snake keeps on growing with the consumption of fruit . The game ends with the snake either colliding with itself or with the blocks . The game has a number of features associated with it like specular lighting effect, score and high score, play/pause the game and 3 exciting game levels :- worm, snake and kobra .

Keys

Up – moves up
Down – moves down
Left – moves left
Right – moves right
p/P – pauses game
any other key – resumes game

Platform :- Linux / Ubuntu

How to Run
Change the .doc extension to .zip
Unzip the file
run the file using ./snake in the terminal
compile the files using gl -o snake *.cpp

Source code can be downloaded from here :- snake

main()
{
int ret;
ret = fork();
ret = fork();
ret = fork();
ret = fork();

if(!ret)
printf(“zero\n”);
else
printf(“one\n”);

}

fork creates a child process. So, 2^4 times the if statement gets executed out of which 8 are ones and other 8 are zero’s. The order of output is compiler dependent.

#include<stdio.h>
#include<stdlib.h>
//#define NULL 0
/* STRUCTURE CONTANING A DATA PART AND A LINK PART */

struct node
{
int data;
struct node *next;
}*p;

/* P IS A GLOBAL POINTER CONTAINS THE ADRESS OF THE FIRST NODE IN
LIST
*/

/*THIS FUNCTION DELETES A NODE */

void delnode(int num)
{
struct node *temp, *m;
temp=p;
while(temp!=NULL)
{
if(temp->data==num)
{
if(temp==p)
{
p=temp->next;
free(temp);
return;
}
else
{
m->next=temp->next;
free(temp);
return;
}
}else
{
m=temp;
temp= temp->next;
}

}
printf(” ELEMENT %d NOT FOUND “, num);
}/*THIS FUNCTION ADDS A NODE AT THE LAST OF LINKED LIST */

void append( int num )
{
struct node *temp,*r;
/* CREATING A NODE AND ASSIGNING A VALUE TO IT */

temp= (struct node *)malloc(sizeof(struct node));
temp->data=num;
r=p;

if (p == NULL) /* IF LIST IS EMPTY CREATE FIRST NODE */
{
p=temp;
p->next =NULL;
}
else
{ /* GO TO LAST AND ADD*/

while( r->next != NULL)
r=r->next;
r->next =temp;
r=temp;
r->next=NULL;
}
}/* ADD A NEW NODE AT BEGINNING */

addbeg( int num )
{
/* CREATING A NODE AND INSERTING VALUE TO IT */

struct node *temp;
temp=(struct node *)malloc(sizeof(struct node));
temp->data=num;

/* IF LIST IS NULL ADD AT BEGINNING */
if ( p== NULL)
{
p=temp;
p->next=NULL;
}

else
{
temp->next=p;
p=temp;
}
}

/* ADD A NEW NODE AFTER A SPECIFIED NO OF NODES */

addafter(int num, int loc)
{
int i;
struct node *temp,*t,*r;
r=p; /* here r stores the first location */
if(loc > count() + 1 || loc <= 0)
{
printf(” insertion is not possible : “);
return;
}
if (loc == 1)/* if list is null then add at beginning */
{
addbeg(num);
return;
}
else
{
for(i=1;i<loc;i++)
{
t=r; /* t will be holding previous value */
r=r->next;
}
temp=(struct node *)malloc(sizeof(struct node));
temp->data=num;
t->next=temp;
t=temp;
t->next=r;
return;
}
}/* THIS FUNCTION DISPLAYS THE CONTENTS OF THE LINKED LIST */

display(struct node *r)
{
r=p;
if(r==NULL)
{
printf(“NO ELEMENT IN THE LIST :”);
return;
}
/* traverse the entire linked list */
while(r!=NULL)
{
printf(” -> %d “,r->data);
r=r->next;
}
printf(“”);
}
//THIS FUNCTION COUNTS THE NUMBER OF ELEMENTS IN THE LIST
count()
{
struct node *n;
int c=0;
n=p;
while(n!=NULL)
{
n=n->next;
c++;
}
return(c);
}
//THIS FUNCTION REVERSES A LINKED LIST
reverse(struct node *q)
{
struct node *m, *n,*l,*s;
m=q;
n=NULL;
while(m!=NULL)
{
s=n;
n=m;
m=m->next;
n->next=s;
}
p=n;
}

/* THIS IS THE MAIN PROGRAM */

main()
{
int i;
p=NULL;
while(1) /* this is an indefinite loop */
{
printf(“\n1.INSERT A NUMBER AT BEGINNING;\n”);
printf(” 2.INSERT A NUMBER AT LAST:\n”);
printf(” 3.INSERT A NUMBER AT A PARTICULAR LOCATION INlIST:\n”);
printf(” 4.PRINT THE ELEMENTS IN THE LIST :\n”);
printf(” 5.PRINT THE NUMBER OF ELEMENTS IN THE LIST\n”);
printf(” 6.DELETE A NODE IN THE LINKED LIST:\n”);
printf(” 7.REVERSE A LINKED LIST :\n”);
printf(” 8.GET OUT OF LINKED LIST (BYEE BYEE):\n”);
printf(” PLEASE, ENTER THE NUMBER:\n”);

scanf(“%d”,&i); /* ENTER A VALUE FOR SWITCH */

switch(i)
{
case 1:
{
int num;
printf(” PLEASE ENTER THE NUMBER :-”);
scanf(“%d”,&num);
addbeg(num);
break;
}
case 2:
{
int num;
printf(” PLEASE ENTER THE NUMBER :-”);
scanf(“%d”,&num);
append(num);
break;
}

case 3:
{
int num, loc,k;
printf(” PLEASE ENTER THE NUMBER :-”);
scanf(“%d”,&num);
printf(“PLEASE ENTER THE LOCATION NUMBER :-”);
scanf(“%d”,&loc);
addafter(num,loc);
break;
} case 4:
{
struct node *n;
printf(” THE ELEMENTS IN THE LIST ARE : “);
display(n);
break;
}

case 5:
{
struct node *n;
//display(n);
printf(” TOTAL NO OF ELEMENTS IN THE LSIT ARE %d”,count());
break;
} case 6:
{
int num;
printf(” PLEASE ENTER A NUMBER FROM THE LIST :”);
scanf(“%d”,&num);
delnode(num);
break;
}
case 7:
{
reverse(p);
//display(p);
break;
}
case 8:
{
exit(1);
}
}/* end if switch */
}/* end of while */
}/* end of main */

#include<stdio.h>
#include<stdlib.h>

int nol=0,non=0;

struct tree
{
int no;
struct tree *l,*r;
};

struct tree *root=NULL;

struct tree * search(int val,int *found)
{
struct tree *p=root,*par=NULL;
while(p!=NULL)
{
if(val==p->no)
{
*found=1;
break;
}
else if(val > p->no)
{
par=p;
p=p->r;
}
else
{
par=p;
p=p->l;
}
}
return par;
}

void insert(int val)
{
int f=0;
struct tree *n,*parent;
n=(struct tree*)malloc(sizeof(struct tree));
n->no=val;
n->l=n->r=NULL;
if (root==NULL)
{
root=n;
return;
}
parent=search(val ,&f);
if(f==1)
{
printf(“\n\n\n DUPlicate number”);
free(n);
return;
}

else if(val>parent->no)
parent->r=n;
else
parent->l=n;
}

void countnoofleaf(struct tree *p)
{
if(p!=NULL)
{
if((p->l==NULL)&&(p->r==NULL))
nol++;
else
countnoofleaf(p->l);
countnoofleaf(p->r);
}
}

int level(struct tree *p,int val)
{
int f=0;
search(val,&f);
if(f==1)
{
while(val!=p->no)
{
if(val>p->no)
{
p=p->r;
nol++;
}
else
{
p=p->l;
nol++;
}
}
return nol;
}
else
{
printf(“\n\nThe Number u entered is not Available”);
return -1;
}
}

void inorder(struct tree *p)
{
if(p!=NULL)
{
inorder(p->l);
printf(” %d”,p->no);
inorder(p->r);
}
}

void countnode(struct tree *p )
{
if(p!=NULL)
{
if(p->l!=NULL)
{
non++;
countnode(p->l);
}
if(p->r!=NULL)
{non++;
countnode(p->r);
}
}
}

void countsubnode(struct tree *p,int val)
{
int f=0;
search(val,&f);
if(f==1)
{
while(val!=p->no)
{
if(val>p->no)
{
p=p->r;
}
else
p=p->l;
}
countnode(p);
}
}

void plevelno(struct tree *p,int lvl)
{ int i=0;
if(p!=NULL)
{
i=level(root,p->no);
nol=0;

if(i==lvl)
printf(“\n %d”,p->no);
else
plevelno(p->l,lvl);
plevelno(p->r,lvl);
}
}

void main()
{

int val,ch,i;
do
{
printf(“\n\n\n 1 To Insert”);
printf(“\n 2 To INORDER traversing of TREE”);
printf(“\n 3 To Count no of leaf”);
printf(“\n 4 To Find The LEVEL of a particular NUMber”);
printf(“\n 5 To find the subnode”);
printf(“\n 6 To print element of the same LEVEL”);
printf(“\n Enter ur choice”);
scanf(“%d”,&ch);

switch(ch)
{
case 1:printf(“\n\n Enter NUMBER”);
scanf(“%d”,&val);
insert( val);
break;
case 2: inorder(root);
break;
case 3: countnoofleaf(root);
printf(“\n\n Number of Leaf is %d”,nol);
nol=0;
break;
case 4: printf(“\n\n Enter the number”);
scanf(“%d”,&val);
i=level(root,val);
if(i!=-1)
printf(“\n\n The level of %d is %d”,val,i);
nol=0;
break;
case 5: printf(“\n\n Enter the NUMBER”);
scanf(“%d”,&val);
countsubnode(root,val);
printf(“\n\nNumber of subnode is %d”,non);
non=0;
break;
case 6: printf(“\n\n Enter the LEVEL no.”);
scanf(“%d”,&val);
plevelno(root,val);
break;
case 0: break;
default:printf(“\n\nInvalid Choice”);
}
}while(ch!=0) ;

}