Help making this code run faster for spoj.
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by Josh Meredith
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Published on 2010-05-29T05:35:39Z
Indexed on
2010/05/29
5:42 UTC
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I've been doing a few of the challenges on the Sphere Online Judge, but I can't seem to get the second problem (the prime generator) to run within the time limit. Does anyone have any tips for increasing the speed of the following code?
#include <stdio.h>
#include <math.h>
int is_prime(int n);
void make_sieve();
void fast_prime(int n);
int primes[16000];
int main()
{
int nlines;
int m, n;
make_sieve();
scanf("%d", &nlines);
for (; nlines >= 1; nlines--) {
scanf("%d %d", &m, &n);
if (!(m % 2)) {
m++;
}
for ( ; m < n; m+=2) {
fast_prime(m);
}
printf("\n");
}
return 0;
}
/* Prints a number if it's prime. */
inline void fast_prime(int n)
{
int j;
for (int i = 0; ((j = primes[i]) > -1); i++) {
if (!(n % j)) {
return;
}
}
printf("%d\n", n);
}
/* Create an array listing prime numbers. */
void make_sieve()
{
int j = 0;
for (int i = 0; i < 16000; i++) {
primes[i] = -1;
}
for (int i = 2; i < 32000; i++) {
if (i % 2) {
if (is_prime(i)) {
primes[j] = i;
j++;
}
}
}
return;
}
/* Test if a number is prime. Return 1 if prime. Return 0 if not. */
int is_prime(int n)
{
int rootofn;
rootofn = sqrt(n);
if ((n <= 2) || (n == 3) || (n == 5) || (n == 7)) {
return 1;
}
if (((n % 2) == 0) || ((n % 3) == 0) || ((n % 5) == 0) || ((n % 7) == 0)) {
return 0;
}
for (int i = 11; i < rootofn; i += 2) {
if ((n % i) == 0) {
return 0;
}
}
return 1;
}
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