Friends i have little bit problem in making a c language program.

I have to make a program which prints prime numbers from 1 to 500.

I make this program but with while loop then i was told to make it using for loop, which i tried but failed.

So kindly help me. Thanks

from,

john(pakistan)

john_hasan
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## Recommended Answers

Jump to PostI dont know that high math, but I think I can do better.

`bool isPrime(int n) { bool prime ; int lim; if ( n == 0 || n == 1 || n % 2 == 0) return(false); prime = true; lim = (int)sqrt(n); for(int i=3; i<= …`

Jump to Post> for(int i=2; i<=((int)sqrt((double)n)); i++)

And it would be so much quicker if you didn't call sqrt() on every iteration of the loop!

n is constant (in this function), so it's root is constant also. Calculate it once and store in another variable to compare against.Also, since 2 …

Jump to PostThis should be considerably although I'd bet not noticeably faster.

`#include <iostream> using namespace std; int main (){ int prim=3, div,i=2; cout<<1<<" "<<2<<" "; for (i; i<=500; i++){ div=3; while(div<=prim/3){ if (prim%div!=0){div+=2;} else {prim+=2;div=3;}} cout<<prim<<" "; prim+=2;} system("pause");}`

Why do you care about the speed anyways?

Jump to PostWhy do you care about the speed anyways?

If the developers didnt care for speed then u wouldn't be conviniently using your favourite OS u are usign right now without atlest swearing atleast 100 times a day regarding the slow loading times.

Dont forget that this is a C/ …

Jump to PostTo "optimize" it a little further without getting too technical you could keep track of each prime along the way. Then to check if current number is prime just check to see if previous primes up to the square root of the current number are a factor rather than checking …

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hollystyles
commented:
good solution
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hollystyles
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good feedback
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