Friday, July 28, 2006

Write a program to print all prime numbers from 1 to 300.
(Hint: Use nested loops, break and continue statements.)

Improvements:
The work done by the computer is considerably used..if we chuck out all the even numbers..cuz even numbers cannot be prime numbers (except the number 2)

Also, 1 is not a prime number. Instead..it is a Unique Number.



#include<stdio.h>

main()
{

int number, div, ifprime;



for (number=2;number<=300;number++)

{

for (div=2; div<number; div++)

{
if (number%div==0)
{

ifprime=0;
break;
}

ifprime=1;

}

if (ifprime)
{
printf("\n%d", number);

}
}
}




The File can be found at:
Download File

20 comments:

Anonymous said...

very cofusing code

Anonymous said...

what is ifprime?

khader basha sd said...

in if(ifprime), what is ifprime value?

khader basha sd said...

what is ifprime value in if(ifprime)?

Surbhi Bakshi said...

That's just a variable!
Used to give 0 or 1 value based on whether that no is prime or not!
If it is 1, it is a prime!

Tanya said...

its very confusing..... pls elaborate wid a simple method....

Abdul wali said...
This comment has been removed by the author.
Abdul wali said...

Take a simple mehthod
#include
void main()
{
int a,b;count=2;
for(a=3;a<=50;a++)
{
for(b=2;b<=a-1;b++)
if(a%b==0)
break;
else if(b==a-1)
{
count=count+1;
printf("%d is a prime number\n",a);
}
}
getche();
}

PRATIK ADARSH said...

@abdul

why did ya introduce count in here ????
whats the use ??

Anonymous said...

#include
int main()
{
int num,div;

for(num=2;num<=300;num++)
{
for(div=2;div<num;div++)
{
if((num%div==0))
{
break;
}
else if(div==num-1)
{
printf ("\n%d",num);
}
}
}

return(0);
}

Anonymous said...

tuhadi ma da phuda panchodo ya bnaya h q

sadik khan said...

even more simple is here...
#include
void main()
{
int i,num;

for(num=2;num<=1000 ;num++)
{
for(i=2;i<=num;i++)
{ if(num%i==0)
break;
}
if(i==num)
printf("\n%d is Prime",num);
}
}

sadik khan said...
This comment has been removed by the author.
Anonymous said...

realy good solution...!!!

Anonymous said...

realy good solution

Anonymous said...

Fucking solution.

Anonymous said...

int i,j;

for (i = 3; i <= 300; i++)
{
for (j = 2; j < i; j++)
{
if(i % j == 0)
//printf("%d\t",i);
break;
}
if (i == j)
printf("%d\t", i);
}

Anonymous said...

Sorry.. Here is the final one. There was a mistake in my earlier one :)

int i,j;

printf("2\t");

for (i = 3; i <= 300; i++)
{
for (j = 2; j < i; j++)
{
if(i % j == 0)
break;
}
if (i == j)
printf("%d\t", i);
}

Anantha Boudmanabhan said...

This may help!

#include
#include
main()
{
clrscr();
int i,n;

printf("The prime numbers b/w 1 and 300 are\n");

for(n=1;n<=300;n++)
{
i=2;
for(i=2;i<=n-1;i++)
{
if(n%i==0)
{break;
}
}
if(i==n)
printf("%d\n",n);
}
getch();
}


Anantha Boudmanabhan said...

This may help!

#include
#include
main()
{
clrscr();
int i,n;

printf("The prime numbers b/w 1 and 300 are\n");

for(n=1;n<=300;n++)
{
i=2;
for(i=2;i<=n-1;i++)
{
if(n%i==0)
break;
}
if(i==n)
printf("%d\n",n);
}
getch();
}