want to know the complexity of my program ,new way to find prime numbers
Code:
//simply we are going to use the approach that the numbers that are divisble by prime numbers are not prime and the others are prime.
#include<stdio.h>
#include<math.h>
int main()
{
int n,i,arr[100000],size=0,j,temp; // array is for storing prime numbers
scanf("%d",&n); // we will be finding prime numbers between 2 to n (both inclusive)
arr[0]=2; // we know that 2 is a prime number
size++; // we got 1 prime number so size of array gets incremented
for(i=3;i<=n;i++)
{
temp=pow(i,0.5); //square root of the number (for which we are going to check is it prime or not)
for(j=0;arr[j]<=temp;j++){
if(i%arr[j]==0)
break;
}
if(arr[j]>temp)
{
arr[size]=i;
size++;
}
}
for(i=0;i<size;i++)
printf("%d \n",arr[i]);
return 0;
}
Re: want to know the complexity of my program ,new way to find prime numbers
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Re: want to know the complexity of my program ,new way to find prime numbers
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Re: want to know the complexity of my program ,new way to find prime numbers
Quote:
Originally Posted by
VictorN
sorry but i didn't get what you mean to say ,this is a new problem and i really need the answer .
Re: want to know the complexity of my program ,new way to find prime numbers
Quote:
Originally Posted by
ujjawaltaleja
this is a new problem and i really need the answer .
Your algorithm is known as "trial division" (with primes as candidate factors). According to this paper (see page 4 about 10 lines from the bottom),
https://www.cs.hmc.edu/~oneill/papers/Sieve-JFP.pdf
the complexity is Θ(n√n / (log n)^2).
