My programming works, My code book.
Coming soon….
no database software required…. no built in packages required… no prior knowledge required…. its just as simple as writing couple of most simple programs….. all that you need is PHP enabled web server…..
Check this link after few days…. 🙂
Q.) Ignore ‘Spaces’ and split sentence into words and store into array
Function:
<?php
function splitIt($s){
$arr = array('.',',');
$s = str_replace($arr," ",$s);
$s = preg_replace("/( )+/", " ", $s);
$s = preg_replace("/^( )+/", "", $s);
$s = preg_replace("/( )+$/", "", $s);
$tokens = explode(" ",$s);
return $tokens;
}
?>
Input:
<?php
$answer = splitIt(" hello world. prt this sentence,please ");
print_r($answer);
?>
Output:
Array (
[0] => hello
[0] => world
[0] => prt
[0] => this
[0] => sentence
[0] => please
)
After many days of gap… back to blogging…….
Q.) An integer is said to be a palindrome if it is equal to its reverse. For example, 79197 and 324423 are palindromes. In this task you will be given an integer N, 1 ≤ N ≤ 1000000. You must find the smallest integer M ≥ N such that M is a prime number and M is a palindrome. For example, if N is 31 then the answer is 101.
Input format
A single integer N, (1 ≤ N ≤ 1000000), on a single line.
Output format
Your output must consist of a single integer, the smallest prime palindrome greater than or equal to N
Sol.)
#include<stdio.h>
void main()
{
long int a;
printf("Enter number: ");
scanf("%li",&a);
if(a <= 1000000)
while(1)
{
if(palindrome(a)){
if(prime(a)) {printf("Number is: %d",a);getch();return;}
}
a=a+1;
}
else printf("Plz enter number less than 1000000 next time");
getch();
}
int prime(long int a){
long int i;
for(i=2;i<=a/2;i++) if(a%i==0)return 0;
return 1;
}
int palindrome(long int a){
long int temp=a,reverse=0;
while( temp != 0 ){
reverse = reverse * 10;
reverse = reverse + temp%10;
temp = temp/10;
}
if(reverse == a) return 1;
else return 0;
}
Shivaji varma… 😀
#include<stdio.h>
#include<stdlib.h>
#include<errno.h>
int main()
{
int desc[2];
if(pipe(desc)==-1)
{
perror("pipe failed"); exit(0);
}
if(!fork())
{
write(desc[1],"hello",6);
}
else{
char temp[20];
read(desc[0],temp,6);
printf("\n%s\n",temp);
}
}
<html>
<head>
<style>
.navBtn{
background-color:#EAECED;
color:#565656;
width:80px;
font-family:"Times New Roman", Times, serif;
font-size:larger;
font-weight:bold;
text-align:center;
padding:6px;
cursor:pointer;
border-radius: 4px 4px 4px 4px;
}
.navBtn a,a:visited{
text-decoration:inherit;
color:inherit;
}
.navBtn:hover{
background-color:#1EB4DA;
color:#EAECED;
}
.navActive{
background-color:#FF3C3C;
color:#EAECED;
}
</style>
</head>
<body>
<div class="navBtn">1</div>
<br>
<div class="navBtn">2</div>
<br>
<div class="navBtn">3</div>
</body>
</html>
The following c program parser the given string for the given grammar::
E->E+T
E->T
T->T*F
T->F
F->(E)
F->i
Before writing writing recursive descent parser first eliminate left recursion and left factoring.
E->TEplus
Eplus->+TEplus / ∈ (epsilon)
T ->FTplus
Tplus->*FTplus / ∈ (epsilon)
F-> (E)
F-> i
#include<stdio.h>
#include<string.h>
char string[' '];
int len;
int pos=0;
int E();
int Eplus();
int T();
int Tplus();
int F();
int main()
{
printf("Enter the string:: ");
scanf("%s",&string);
len=strlen(string);
if(E()==0) printf("\nNot Accepted");
else if(pos<len) {printf("\nNot Accepted");}
else printf("\nAccepted");
printf("\n");
return 0;
}
int E()
{
if(T()==0) return 0;
if(Eplus()==0) return 0;
return 1;
}
int Eplus()
{
if(string[pos]=='+')
{ pos++;
if(T()==0) return 0;
if(Eplus()==0) return 0;
}
return 1;
}
int T()
{
if(F()==0) return 0;
if(Tplus()==0) return 0;
return 1;
}
int Tplus()
{
if(string[pos]=='*')
{ pos++;
if(F()==0) return 0;
if(Tplus()==0) return 0;
}
return 1;
}
int F()
{
if(string[pos]=='('){
pos++;
if(E()==0) return 0;
if(string[pos]==')') pos++;
else return 0;
}
else if(string[pos]=='i'){
pos++;}
else {return 0;}
return 1;
}
