Hi

I made a program to answer question 1 from this paper

http://www.olympiad.org.uk/papers/20...round_one.html

I used a "sieve of eratosthenes". Just wondering if there was anything that I could do better.

Thanks!

Code:

#include <stdio.h>
int MAXINPUT=10000;
int primes[10000];
void set_up_table_of_primes();
int main()
{
extern int MAXINPUT;
extern int primes[10000];
int numeven;
printf("Type in your even number between 4 and 10000 inclusive:\n");
scanf("%d", &numeven);
set_up_table_of_primes();
int howmany = 0;
int i = 0;
if (numeven==4) {howmany++;} //If 4 was entered, howmany++
for (i=3; i<=numeven/2; i+=2) //Use odd primes to find answer
{
if ((primes[i]==1)&&(primes[numeven-i]==1))
{
//If two numbers add to make numeven *i+(n-i)=n* and are prime
howmany++;
printf("howmany = %d\n", howmany);
}
}
printf("%d can be expressed as the sum of primes in %d ways\n",numeven, howmany);
printf("finish\n");
return 0;
}
void set_up_table_of_primes()
{
//Sieve of Eratosthenes
extern int MAXINPUT;
extern int primes[10000];
int i, p, j;
for (i=0; i<MAXINPUT; i++)
{
//everything provisionally prime
primes[i] = 1;
}
printf("primes[15] = %d", primes[15]);
primes[0] = primes[1] = 0; //0 and 1 are not prime
p = 2;
int flag=0;
while (flag==0)
{
//Here we use a sieve of Eratosthenes to make all non-primes zeroed and leave primes as 1
while (primes[p] == 0) {p++;} //find next prime
if (p*p > MAXINPUT)
{
//Return to main function when done
return;
}
for (j=2*p; j<10000; j=j+p)
{
//multiples of p are not prime either
primes[j] = 0;
}
if(primes[p]==1) {p++;} //Move onto next prime number
}
}