program of baal game catch circles coming down with use left right arrow key

program of baal game catch circles coming down with use left right arrow key


Program
#include
#include
#include
#include
#include
#include
int a,b,ch;
int c=2,d=3;
static char msg[1];
static int chance,pts,x1,x2,y1,y2;
over()
{
int i=0,j,x=185,y=240;
while(i<1000) { j=1+rand()%15; setcolor(j); settextstyle(0,0,4); outtextxy(x,y,"GAME OVER"); delay(100); i++; } exit(1); return; } move(int x1,int x2,int y1,int y2,int k) { int i=0; i=i+10; if(k==1) //left {i=i; bar1(x2-i,x2,y1,y2,2); } else { i=-i; bar1(x1,x1-i,y1,y2,3); } bar1(x1-i,x2-i,y1,y2,1); return; } bar1(int x1,int x2,int y1,int y2,int k) { if(k==1) {c=2; d=3;} else {c=d=0;} setfillstyle(2,d); bar(x1,479-y1,x2,479-y2); return; } keyin() { int i; ch=getch(); switch(ch) { case 75: if(x1>24)
{
i=10; //left
move(x1,x2,y1,y2,1);
x2=x2-i;
x1=x1-i;
}
break;
case 77: //right
if(x2<614) { i=10; move(x1,x2,y1,y2,2); x1=x1+i; x2=x2+i; } break; }//end switch if(ch==27) over(); return; } ball() { int a,b,t=0; a=639-rand()%639; b=350; while(b>100)
{
b=b-1;
setcolor(RED);
if((a-5)<=x2&&(a+5)>=x1&&(479-(b-5)>=(479-y1))&&(479-(b+5)<=(479-y2)))
{score();
t=1;
break;}
if(kbhit())
keyin();
circle(a,479-b,5);
delay(10);
setcolor(BLACK);
circle(a,479-b,5);
}
if(t==0)
chance++;
if(chance==2)
over();
return;
}
score()
{
setcolor(0);
settextstyle(0,0,5);
sprintf(msg,"%d",pts);
outtextxy(110,20,msg);
setcolor(4);
sound(1100);
delay(50);
nosound();
pts++;
settextstyle(0,0,5);
sprintf(msg,"%d",pts);
outtextxy(110,20,msg);
return;
}
play()
{
while(ch!=27)
ball();
return;
}

program of 2 player game in c language

2 player game









Program
#include
#include
#include
#include
#include
static float a,b,h1,h2,x1=20,x2=25,y1=250,y2=200,x3=20,x4=25,y3=190,y4=140;
static int ch=3,ch1=3,q=1,t,s=1,s1,s2,c=0,c1=0,i=0,k=1,t1,t2,p1=0,p2=0;
gameover()
{
int j=0;
while(j<100) p="="1)" t="getpixel(x1+2,479-(y1+y2)/2);" t1="getpixel(a-6,479-b);">(479-(y1+y2)/2)&&(t1==2))
c=1; //down
else if((479-b)<(479-(y1+y2)/2)&&(t1==2)) c=2; //up return c; } else if(p==2) { t=getpixel(639-(x3+2),479-(y3+y4)/2); t2=getpixel(a+6,479-b); if((479-b)>(479-(y3+y4)/2)&&(t2==2))
{c1=1; //down
//printf("%2.f %2.f",479-(y3+y4)/2,479-b);
//printf("yes ");
}
else if((479-b)<(479-(y3+y4)/2)&&(t2==2)) { //printf("%2.f %2.f",479-(y3+y4)/2,479-b); c1=2; //up //printf("no "); } return c1; } return 0; } ball() { ch1=getch(); while(ch1!=13) { move(); ch1=getch(); if(ch1==27) break; } //ch1=getch(); if(ch1==13) { q=2; sound(1100); delay(50); nosound(); while(ch1!=27) { setfillstyle(1,5); setcolor(5); circle(a,479-b,5); floodfill(a,479-b,5); c=check(s1); c1=check(s2); delay(5); setfillstyle(1,0); setcolor(0); circle(a,479-b,5); floodfill(a,479-b,0); if(kbhit()) move(); if(k==1) { t=getpixel(639-(x3+2),479-(y3+y4)/2); t2=getpixel(a+7,479-b); if(t==t2) { sound(1100); delay(50); nosound(); h2=a; c=0;c1=0;//printf("h2=%2.f,yes",h2); k=2; } a++; } else if(k==2) { t=getpixel(x1,479-(y1+y2)/2); t1=getpixel(a-7,479-(b)); if(t==t1) { sound(1100); delay(50); nosound(); h1=a; c=0;c1=0; //printf("yes %2.f =h1",h1); k=1; } a--; } if((479-b)==(479-50)) { //printf("s=%d",s); if(c==1) c=2; else if(c1==1) c1=2; } if(a==639||(a+25)==639) { ch1=3; p1++; printf("%d %d",p1,p2); s=1; q=1; c=0; c1=0; start(1,5); } if(a==0||(a-25)==0) { p2++; s=2; c=0;c1=0; q=1; ch1=3; printf("%d %d",p1,p2); start(2,5); } if((479-b)==(479-429)) { //printf("s=%d",s); if(c1==2) c1=1; else if(c==2) c=1; } if(c==1) //down frm 1st { a++; b--; } else if(c==2) //up frm 1st { a++; b++; } else if(c1==1) { a--; b--; } else if(c1==2) { a--; b++; } if(p1==5||p2==5) gameover(); }//end while } //end ch if return; } move() { fflush(stdin); ch1=getch(); if(ch1==119||ch1==115) move1(); else move2(); return; } move1() { //ch1=getch(); switch(ch1) { case 119://up if(y1<=420) { if(q==1&&s==1) { star(1,0); //delay(1); } setfillstyle(1,2); bar(x1,479-(y1+10),x2,479-(y2+10)); //delay(1); setfillstyle(1,1); bar(x1,479-(y2),x2,479-(y2+10)); y1=y1+10; y2=y2+10; setfillstyle(1,2); if(q==1&&s==1) star(1,5); } break; case 115://down if(y2>=50)
{
if(q==1&&s==1)
{
star(1,0);
//delay(1);
}
setfillstyle(1,2);
bar(x1,479-(y1-10),x2,479-(y2-10));
//delay(1);
setfillstyle(1,1);
bar(x1,479-(y1),x2,479-(y1-10));
y1=y1-10;
y2=y2-10;
setfillstyle(1,2);
if(q==1&&s==1)
star(1,5);
}
break;
}
return;
}
move2()
{
switch(ch1)
{
case 112://up
if(y3<=420) { if(q==1&&s==2) star(2,0); setfillstyle(1,2); bar(639-x3,479-(y3+10),639-x4,479-(y4+10)); setfillstyle(1,1); bar(639-x3,479-(y4),639-x4,479-(y4+10)); y3=y3+10; y4=y4+10; setfillstyle(1,2); if(q==1&&s==2) star(2,5); } break; case 108://down if(y4>=(50))
{
if(q==1&&s==2)
star(2,0);
setfillstyle(1,2);
bar(639-x3,479-(y3-10),639-x4,479-(y4-10));
setfillstyle(1,1);
bar(639-x3,479-(y3),639-x4,479-(y3-10));
y3=y3-10;
y4=y4-10;
setfillstyle(1,2);
if(q==1&&s==2)
star(2,5);
}
break;
}
return;
}
star(int o,int c)
{
s=o;
if(s==1)
{
a=x2+7;
b=(y2+y1)/2;
i=608;
k=1;
s1=1;
s2=2;
setfillstyle(1,c);
setcolor(c);
circle(a,479-b,5);
floodfill(a,479-b,c);
}
if(s==2)
{
a=620-(x3-7);
b=(y3+y4)/2;
i=608;
k=2;
s1=1;
s2=2;
setfillstyle(1,c);
setcolor(c);
circle(a,479-b,5);
floodfill(a,479-b,c);
}
return;
}
start(int o,int c)
{
s=o;
if(s==1)
{
a=x2+7;
b=(y2+y1)/2;
i=608;
k=1;
s1=1;
s2=2;
setfillstyle(1,c);
setcolor(c);
circle(a,479-b,5);
floodfill(a,479-b,c);
}
if(s==2)
{
a=620-(x3-7);
b=(y3+y4)/2;
i=608;
k=2;
s1=1;
s2=2;
setfillstyle(1,c);
setcolor(c);
circle(a,479-b,5);
floodfill(a,479-b,c);
}
ball();
return;
}
play()
{
setfillstyle(1,2); //solid color
bar(x1,479-y1,x2,479-y2);
bar(639-x3,479-y3,639-x4,479-y4);
line(0,479-435,639,479-435);
line(0,440,639,440);
start(s,5);
return;
}
main()
{
int k,a=DETECT,b,ch;
initgraph(&a,&b,"D:\\TC\\BGI");
play();
getch();
return 0;
}

Program a menu driven program showing several operations on string.

Program
#include
#include
#include
void stringlen(char s1);
void stringcopy(char *s1);
void stringcon(char *s1);
void stringrev(char *s1);
void stringcompare(char *s1);
void wordcount(char *s1);
void lettercount(char *s1);
void main()
{
char s[40],x;
clrscr();
printf("\n Enter a string");
gets(s1);
printf("\n Enter 'a' for string length");
printf("\n Enter 'b' for string copy");
printf("\n Enter 'c' for string concatanation");
printf("\n Enter 'd' for string rev");
printf("\n Enter 'e' for string compare");
printf("\n Enter 'f' for word count within a string");
printf("\n Enter 'g' for letter count within a string\n");
scanf("%2c",&x);
switch(x)
{
case 'a':
stringlen(s1);
break;
case 'b':
stringcopy(s1);
break;
case 'c':
stringcon(s1);
break;
case 'd':
stringrev(s1);
break;
case 'e':
stringcompare(s1);
break;
case 'f':
wordcount(s1);
break;
case 'g':
lettercount(s1);
break;
}
}
void stringlen(char *s1)
{
int n=0;i=0;
while(s[i]!=NULL)
{
n++;
i++;
}
printf("length of the string is:-\n%d",n);
getch();
}
void stringcopy(char *s1)
{
int i;
char s2[40];
for(i=0;s1[i]=!NULL;i++)
{
s2=s1[i];
}
s2[i]=NULL;
printf("NEW STRING IS;-\n%s",s2);
getch();
}
void stringconcatanation(char *s1)
{
int i,j;
char str[40];
printf("ENTER ANOTHER STRING:-\n");
fflush(stdin);
gets(str);
for(i=0;s1[i]!=NULL;i++);
for(j=0;s1[j]!=NULL;j++)
{
s1[i]=str[j];
i++;
s1[i]=NULL;
}
printf("NEW STRING IS;-\n%s",s1);
getch();
}
void stringrev(char *s1)
{
int i,c=0,l;
char s2[100];
l=strlen(s1);
for(i=l-1;i>=0;i--)
s2[c++]=s1[i];
printf("NEW STRING IS \n");
for(i=0;i
printf("%c",s2[i]);
getch();
}
void stringcompare(char *s1)
{
int i,flag=1;
char s2[40];
printf("ENTER ANOTHER STRING");
fflush(stdin);
gets(*s2);
for(i=0;s2[i]!=NULL;i+)
{
if(s1[i]!=s2[i])
{
printf("\n not equal");
flag=0;break;
}
}
if(flag==1)
printf("equal");
getch();
}

program to check how many 1 there on binary code of a number

prg to chk how many 1 there on binary code of a number.......(popular q in interviews)




Program
#include

int main()
{
int i=10;
int count=0;


while(i)
{
if(i &1==1)
{
count++;
}
i>>=1;
}
printf("%d",count);
return 0;
}

program converts an infix string to postfix expression

program converts an infix string to postfix expression....i have kept no place for input..please change the char a[] in void main() and it will act as the input.....if any new symbols are to be added please add them to op[] and also its corresponding order of precedence in prec[]




















#include
#include
#include
#include
int top=0,opn=5;
char op[]={'^','*','/','+','-'};
int prec[]={1,2,2,3,3};
char pop(char s[]);
int isoperand(char);
void push(char s[],char);
int prcdnc(char,char);

void main()
{

char a[]="a+(b*c-(d/e^f)*g)*h",stk[50],c[50],temp,symb;
int l,i=0,r=0;
l=strlen(a);
a[l]=')';
a[l+1]=NULL;
stk[0]='(';
while(a[r]!=NULL)
{
symb=a[r];
if(isoperand(symb))
{
c[i]=symb;
i++;
}
else if(symb==')')
{
while(1)
{
temp=pop(stk);
if(temp=='(')
break;
else
{
c[i]=temp;
i++;
}
}
}
else if(symb=='(')
push(stk,symb);
else
{
while(1)
{
if(stk[top]=='(')
{
push(stk,symb);
break;
}
else if(prcdnc(stk[top],symb))
{
c[i]=pop(stk);
i++;
}
else
{ push(stk,symb);
break;
}
}
}
r++;
}
c[i]=NULL;
printf("\nThe post fix expression is= %s",c);
getch();
}

int prcdnc(char a,char b)
{
int num1,num2,i;
for(i=0;i
{
if(op[i]==a)

num1=prec[i];

if(op[i]==b)
num2=prec[i];
}
if(num2<=num1) return 0; else return 1; } void push(char s[],char a) { top=top+1; s[top]=a; } char pop(char s[]) { char val; val= s[top]; top=top-1; return val; } int isoperand(char a) { int i; if(a=='(' || a==')') return 0; for(i=0;i
{
if(op[i]==a)
return 0;
}
return 1;
}

To find saddle point in a triangle with c programming

Program
#include
#include
void main()
{
int a[5][5],i,j,min,max,m,n,flag=1,p,q;
printf("Enter size of the matrix");
scanf("%d%d",&m,&n);
printf("\nEnter the elements of the matrix");
for(i=0;ia[i][j])
min=a[i][j];
p=i;
q=j;
}
}
for(j=0;ja[p][q])
flag=0;
}
if(flag)
printf("Saddle point %d",a[p][q]);
else
printf("no saddle point");
flag=1;
getch();
}

how to use signature of printf in c programming language

how to use signature of printf... according to our needs.....




Program
#include
main()
{
char p[]={"gaurav kapoor"};
int *name;
name=&p;
printf(name);
return 0;}

Make a Binary tree from Infix expression

Program
Make a Binary tree from Infix expression.


Infix :-> 4 $ 2 * 3 - 3 + 8 / 4 / (1 + 1)


Solution: First separate the operator and operand
___________________________________________________________


Operator:-> $ * - + / / (+)


Operand:-> 4 2 3 3 8 4 1 1
___________________________________________________________


Now separate the operator according to their priority


First priority :-> ( + ) // Because ( ) has greater priority


Second priority :-> $


Third priority :-> * / /


Fourth priority :-> - +
_____________________________________________________
Step 1:
_____________________________________________________


Root will be fixed according to lowest priority


So, In Foruth priority - +


Here + will be the first root of tree (right to left)


+
_________________________________________________
Step 2:
_________________________________________________




Now - will be the left child of root of this tree


Because - is existing in left side of +.


+
/
/
-


_________________________________________________
Step 3:
_________________________________________________


Now in Third priority :-> * / /


Here / will be right child of +.


Because / is existing in right side of +.
+
/ \
/ \
- /


_________________________________________________
Step 4:
_________________________________________________


Here / will be right child of /.


Because / is existing in left side of /.
+
/ \
/ \
- (/) Devide sign (no 1)
/
/
(/) devide sign (no 2)


_________________________________________________
Step 5:
_________________________________________________
Here * will be left child of -.


Because, In infix expression, * is existing in left side of -.


+
/ \
/ \
- (/) Devide sign (no 1)
/ /
/ /
* (/) devide sign (no 2)


_________________________________________________
Step 6:
_________________________________________________


In Second priority:-> $


Here $ will be left child of *.


Because, In infix expression,


$ is existing in left side of *.


+
/ \
/ \
- (/) Devide sign (no 1)
/ /
/ /
* (/) devide sign (no 2)
/
/
$
_________________________________________________
Step 7:
_________________________________________________


In First priority:-> (+)


Here (+) will be left child of / (devide sign no 1).


Because, In infix expression,


$ is existing in left side of *.


+
/ \
/ \
- (/)
/ / \
/ / \
* (/) +
/
/
$


_________________________________________________
Step 8:
_________________________________________________


Now As we know the expression is


Infix :-> 4 $ 2 * 3 - 3 + 8 / 4 / (1 + 1)


Operand:-> 4 2 3 3 8 4 1 1


Now put the operand only


_____________________________________________________


First of all take 4, Here 4 will be left child of $.


Because 4 is left side of $ in expression


+
/ \
/ \
- (/)
/ / \
/ / \
* (/) +
/
/
$
/
/
4
_____________________________________________________


Now take 2 , Here 2 will be right child of $.


Because 2 is right side of $ in expression.


+
/ \
/ \
- (/)
/ / \
/ / \
* (/) +
/
/
$
/ \
/ \
4 2
_____________________________________________________


Now take 3 , Here 3 will be right child of *.


Because 3 is right side of * in expression.


+
/ \
/ \
- (/)
/ / \
/ / \
* (/) +
/ \
/ \
$ 3
/ \
/ \
4 2


_____________________________________________________


Now take 8, Here 8 will be left child of /(devide sign no 2).


Because 8 is left side of / in expression. Here i am not saying that 8 will be right child of +. Because / is the right child of +. And operand won't be root in Tree. So i am taking Here 8 will be left child of /(devide sign no 2).


+
/ \
/ \
- (/)
/ \ / \
/ \ / \
* 3 (/) +
/ \ /
/ \ /
$ 3 8
/ \
/ \
4 2


________________________________________________________


Now take 4, Here 4 will be right child of /(devide sign no 2).


Because 4 is right side of / in the given expression.


+(first +)
/ \
/ \
- (/)
/ \ / \
/ \ / \
* 3 (/) + (second +)
/ \ / \
/ \ / \
$ 3 8 4
/ \
/ \
4 2


________________________________________________________


Now take 1, Here 1 will be left child of + (second +).


Because 1 is right side of /(devide sign no 2) in the given expression.


+ (first +)
/ \
/ \
- (/)
/ \ / \
/ \ / \
* 3 (/) + (second +)
/ \ / \ /
/ \ / \ /
$ 3 8 4 1
/ \
/ \
4 2




________________________________________________________


Now take 1, Here 1 will be right child of + (second +).


Because 1 is right side of + (second +) in the given expression.


+
/ \
/ \
- (/)
/ \ / \
/ \ / \
* 3 (/) +
/ \ / \ / \
/ \ / \ / \
$ 3 8 4 1 1
/ \
/ \
4 2


-------------------------------------------
The final Tree for an expression |
|
Infix :-> 4 $ 2 * 3 - 3 + 8 / 4 / (1 + 1) |
|
-------------------------------------------
|
+ |
/ \ |
/ \ |
- (/) |
/ \ / \ |
/ \ / \ |
* 3 (/) + |
/ \ / \ / \ |
/ \ / \ / \ |
$ 3 8 4 1 1 |
/ \ |
/ \ |
4 2 |
|
-------------------------------------------

easy and basic calculator in c programming language

easy and basic calculator.






just add your own detail.
Program
#include
#include
#define P printf
#define S scanf


int main(){

float x,y,sum,diff,prod,qou;


P("Enter first integer: ");
S("%f",&x);

P("Enter second integer: ");
S("%f",&y);

sum=x+y;
P("\nThe Sum is:%.2f",sum);

diff=x-y;
P("\nThe Difference is:%.2f",diff);

prod=x*y;
P("\nThe Product is:%.2f",prod);

qou=x/y;
P("\nThe Qoutient is:%.2f",qou);


getch();
return 0;

}

CLOCK program in c programming language

Description:
to find out the clock
Program
#include

#include
#include
#include

#include "i2c.h"
#include "lcd.h"

idata char g_strSetting[20] ;

unsigned char day, mon, year ;
unsigned char hh,mm,ss ;
unsigned char g_byLastSec ;

unsigned int iday, imon, iyear ;
unsigned int ihh,imm,iss ;

void main ()
{
/*
// initialize serial port
// 19200 baud assuming P89C51RD2 @ 12 MHz
RCAP2H = 0xff ;
RCAP2L = 0xd9 ;
T2CON = 0x34 ;
SCON = 0x70 ;
*/
TH1 = 0xfd ;
TL1 = 0xfd ;
PCON |= 0x80 ;
SCON = 0x70 ;
TMOD = 0x21 ;
TCON = 0x50 ;


lcd_init() ;
printf("Clock program for Mini51\n") ;
display_row(0,"CLOCK on Mini51") ;
if (RI)
{
// send CR during reset, to goto clock set mode
getchare() ;
while(1)
{
printf("\nEnter data and time in dd mm yy hh mm ss format\n") ;
scanf("%x%x%x%x%x%x",&iday,&imon,&iyear,&ihh,&imm,&iss) ;
day = iday ;
mon = imon ;
year = iyear ;
hh = ihh ;
mm = imm ;
ss = iss ;
printf("You entered:\n%02bx-%02bx-%02bx %02bx:%02bx:%02bx\n",day,mon,year,hh,mm,ss) ;
printf("Press Y to store these values to RTC, any other to discard\n") ;
if (getchare() == 'Y')
{
write_rtc_byte(DAY_ADDR,day) ;
write_rtc_byte(MONTH_ADDR,mon) ;
write_rtc_byte(YEAR_ADDR,year) ;
write_rtc_byte(HOURS_ADDR,hh) ;
write_rtc_byte(MINUTES_ADDR,mm) ;
write_rtc_byte(SECONDS_ADDR,ss) ;
break ;
}
}
}
g_byLastSec = 0xff ;
while(1)
{
day = read_rtc_byte(DAY_ADDR) ;
mon = read_rtc_byte(MONTH_ADDR) ;
year = read_rtc_byte(YEAR_ADDR) ;
hh = read_rtc_byte(HOURS_ADDR) ;
mm = read_rtc_byte(MINUTES_ADDR) ;
ss = read_rtc_byte(SECONDS_ADDR) ;
if (ss != g_byLastSec)
{
sprintf(g_strSetting,"%02bx-%02bx %02bx:%02bx:%02bx",day,mon,hh,mm,ss) ;
display_row(1,g_strSetting) ;
//printf("%s\n",g_strSetting) ;
printf("%02bx-%02bx-%02bx %02bx:%02bx:%02bx\n",day,mon,year,hh,mm,ss) ;
g_byLastSec = ss ;
}
}
}