What is the algorithm for finding the prime numbers between 1 and 100?

Answer 1

Code example:

/* ******************************************************************************** * FUNCTION: bIsAPrime ******************************************************************************** * AUTHORS: * Flaun * * DESCRIPTION: * Finds if a given number is a Prime number. * * PARAMETERS: * uiTestNumber: An unsigned integer to test. * * RETURNS: * bTRUE or bFALSE ******************************************************************************** */ #define bTRUE 1 #define bFALSE 0 #define iNO_REMAINDER 0 #define bIS_ODD(uiNumber) ((uiNumber) & 1) typedef int BOOL BOOL bIsAPrime( unsigned int uiTestNumber) { /* 4 is the first non prime number. */ if(uiTestNumber > 3) { /* The only even prime number is 2. */ if(!bIS_ODD(uiTestNumber)) { return bFALSE; } else { /* The only numbers left to divide against are 3, 5 and 7. */ /* All numbers that are divisible by a number > 7 are divisible */ /* by a number < 8. */ unsigned int uiDivisor = 0; for(uiDivisor = 3; uiDivisor <= 7; uiDivisor += 2) { /* 3, 5, 7 are prime numbers. */ if(uiTestNumber 0){ n++;

d=2;

}

}

return n;

}

int main()

{

int nPrime = 0;

char x;

//List the first 50 prime numbers starting from 0

for(int i =0; i<=50; i++){

cout<< (nPrime = NextPrime(nPrime))<<", ";

}

cin>>x;

exit(0);

}

--------------------------------------------------------------------------------------

Here is a relatively fast java algorithm to print out the primes between 1 and 100:

public static void main(String [] args)

{

boolean found=false;

double current=2;

while(current<100)

{

for (double count=2;count<=Math.sqrt(current);count++)

{

if (current/count2%1==0)

{

found=true;

break;

}

}

if (!found)

{

System.out.println(""+((int)current));

}

current++;

}

}