/*
C int nextprime(int n) ! Return next prime number >= n
C
C Return next prime number .GE. N
C
C For N .GE. 1,018,801 returns next number relatively prime to all
C primes below 1009.
C Note: 1009 is next prime after 997, and 1,018,801 = 1009^2.
C The database may be extended if larger primes are desired.
C
C Outline:
C 1. If number .LE. 3 then return number.
C Number is prime, zero, or negative.
C 2. If number between 4 and 997 then search PRIME array
C for next prime .GE. number. An approximating polynomial
C is used to give us a good initial guess for the index I
C so that PRIME(I) is very close to the prime we seek.
C The approximating polynomial is:
C
C I = -0.3034277E-04*X^2 + 0.1918667*X + 8.0918350
C
C The optimum coefficients for this polynomial were obtained
C using the OPTIMIZE program in the [.OPTIMIZE] directory.
C Look there for further details. Thus, this function is right
C on the money 452 times, is low by one 224 times, is low by two
C 40 times, and is low by three 1 time. It's high by one 185
C times, high by two 68 times, high by three 12 times, high by
C four 7 times, high by five 4 times, and high by 6 one time.
C
C 3. Else number is greater than 997. Do normal search for next prime.
C 3a. Increment NUMBER to an odd number. Since even numbers are
C not prime we skip over even numbers.
C 3b. Test if NUMBER is evenly divisible by any prime in the
C PRIME array less than SQRT(NUMBER).
C If NUMBER MOD PRIME(I) = 0, then NUMBER is not prime.
*/
#include <stdio.h>
#define NUMPRIMES 168
#define MIN(a,b) ( ((a) > (b)) ? (b) : (a) )
int nextprime( n ) /* Return next prime number >= n */
int n;
{
int i;
int number;
static int prime[NUMPRIMES+1] = { 1,
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41,
43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101,
103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167,
173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239,
241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313,
317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397,
401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467,
479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569,
571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643,
647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733,
739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823,
827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911,
919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997 };
number = n;
if (number <= 3) return(number); /* if number .le. 3 then it's prime */
if (number <= 997) /* just search through PRIME array */
{
i = (-0.3034277e-04*number + 0.1918667)*number + 8.0918350; /*(see comments above)*/
i = MIN(i,NUMPRIMES); /* don't let I go above 997*/
while (prime[i] < number) ++i; /* Search upward for first prime greater than NUMBER */
while (prime[i] >= number) --i; /* Search downward for first prime less than NUMBER, */
return( prime[i+1] ); /* then take next prime. */
}
else /* Else normal search for next prime */
{
if ((number % 2) == 0) ++number; /* rule out even numbers. They are not prime. */
i = 2; /* start with second prime since we only test odd numbers. */
while (prime[i]*prime[i] <= number) /* test for all primes .LE. SQRT(NUMBER) */
{
if ( (number % prime[i]) == 0 ) /* it's not prime */
{
number += 2; /* even numbers aren't prime */
i = 2; /* start over again with new number */
}
else
{
++i; /* test with next prime */
if ( i > NUMPRIMES) return(number); /* NUMBER is relatively prime to first 168 primes. */
}
}
return(number);
}
}
main()
{
int n1, ians;
for(;;)
{
printf("\nEnter a number: ");
scanf("%d", &n1);
ians = nextprime( n1 );
printf("%d\n",ians);
}
}
