Prime Number Generator in C - c

Question

This is the Link to the problem: http://www.spoj.com/problems/PRIME1/
Basically we get two limits and we have to print out all the primes between them...
Here is my Code (Language == C) :
#include <stdio.h>
void IsPrime(int test){
for(int i= 2; i<test; i++){
if(test%i==0){
return;
}
}
printf("%d\n",test);
}
int main(){
int T,lower,upper;
scanf("%d",&T);
while(T--){
scanf("%d",&lower);
scanf("%d",&upper);
for(int i = lower;i<=upper;i++){
if(i>1){
IsPrime(i);
}
}
}
return 0;
}
On my local machine I ran this and it works for the simple test cases... My message from the website is timeout error so I was wondering if there is a more efficient way to solve this problem because apparently I am not solving it fast enough?

Answer 1

To begin with, you don't have to go checking every number up to n to determine if n is prime, only to its square root (there is a mathematical proof, not going to give it now). So:
void IsPrime(int test){
// i <= sqrt(test)
// but to avoid sqrt you can do i * i <= test
for(int i= 2; i * i <= test; i++){
if(test%i==0){
return;
}
}
printf("%d\n",test);
}
Next, we know that after 2, all other prime numbers are odd, so we can loop by 2 if we treat 2 as special case:
// Do greater than one check only once
if (lower > 1) {
// Special case - lower is 2
if (lower == 2) {
printf("%d\n", 2);
++lower;
}
for(int i = lower; i <= upper; i += 2){
IsPrime(i);
}
}
However since you have to do it T times, you will end up doing the checks a lot more than needed. Also, the problem has limits on n and m so it's basically perfect for a sieve, as #HennoBrandsma said.
Using these optimizations, you should go find all prime numbers to the limit, and store them in a container. Then, when prompted with a range, simply traverse the sieve and print out the numbers.
(That will require you to change up the IsPrime function a bit more - instead of printing the number right away, let it return true or false, and then based on that, add the number to the container)

Answer 2

You can try the following which has a slight optimization on the number of tests as well as skipping any even values greater than 2:
int isprime (int v)
{
int i;
if (v < 0) v = -v; /* insure v non-negative */
if (v < 2 || !((unsigned)v & 1)) /* 0, 1 + even > 2 are not prime */
return 0;
if (v == 2) return 1;
for (i = 3; i * i <= v; i+=2)
if (v % i == 0)
return 0;
return 1;
}
If you can use the math library and math.h, the following may be faster:
int isprime (int v)
{
int i;
if (v < 0) v = -v; /* insure v non-negative */
if (v < 2 || !((unsigned)v & 1)) /* 0, 1 + even > 2 are not prime */
return 0;
if (v == 2) return 1;
for (i = 3; i <= sqrt (v); i+=2)
if (v % i == 0)
return 0;
return 1;
}
I timed both versions over the int range for values between 1-2 million and they are close.
Note: In actual testing with repetitive calls, the version with i * i <= v (isprime2 below) is consistently faster than the call with i <= sqrt (v) (isprime3 below). e.g.:
$ ./bin/isprimetst2
isprime (1.650138 sec) - 78497 primes
isprime2 (0.805816 sec) - 78497 primes
isprime3 (0.983928 sec) - 78497 primes
The short driver iterated over all primes from 0-2000000, e.g.:
r = 0;
t1 = clock ();
for (v = 0; v < 2000000 - 1; v++) r += isprime2 (v);
t2 = clock ();
printf (" isprime2 (%lf sec) - %u primes\n", (t2-t1)/CLOCKS_PER_SEC, r);
r = 0;
t1 = clock ();
for (v = 0; v < 2000000 - 1; v++) r += isprime3 (v);
t2 = clock ();
printf (" isprime3 (%lf sec) - %u primes\n", (t2-t1)/CLOCKS_PER_SEC, r);

Answer 3

You can use a library maths.h in C and use sqrt function to calculate the square root of given number. So the program might be like this:
#include <stdio.h>
#include <maths.h>
int isPrime(int number){
int i;
if(number % 2 == 0){
return;
}
for(i=3; i<=sqrt(number); i++){
if(number % i == 0){
return;
}
printf("%d\n",number);
}
int main(){
int lower,upper,i;
if(lower >1){
if(lower == 2){
printf("2\n");
}
for(i=lower; i<=upper; i++){
isPrime(i);
}
return 0;
}
In short you can use some extra checks (like if(number % 2 == 0)) using if-else condition to decrease the time complexity of the program.For example a new if condition may be if(number % 5 ==0) etc. ,so with the help of these conditions check won't go in for loop in many of the cases, and that would decrease the time of the program.

Answer 4

#include<stdio.h>
int main()
{
int low,high,j;
int prime(int);
int t;
scanf("%d",&t);
while (t>0)
{
scanf("%d %d",&low,&high);
while (low<=1)
{
low++;
continue;
}
for (j=low;j<=high;j++)
{
if (prime(j)){
printf("%d\n",j);
}
}
printf("\n");
t--;
}
return 0;
}
int prime(int n)
{
int i;
for (i=2;i*i<=n;i++)
{
if (n%i==0){
return 0;
}
}
return 1;
}