#include <stdio.h> /* C Factor N C Test for factors of 2,3,5,... using NEXTPRIME function to get next prime. */ int factor( n ) int n; { int number, prime; number = n; prime = 2; /* start with 2 */ printf("\n"); for(;;) { if (( number % prime ) == 0 ) /* prime is a factor */ { printf("%d\n",prime); number /= prime; if (number == 1) return; } else { prime = nextprime(prime+1); } } } /* 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. */ #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); factor( n1 ); } }

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