Task: Brazilian numbers





Java: import java.math.BigInteger;
import java.util.List;

public class Brazilian {
    private static final List<Integer> primeList = List.of(
        2, 3, 5, 7, 9, 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, 169, 173, 179, 181,
        191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 247, 251, 257, 263, 269, 271, 277, 281,
        283, 293, 299, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 377, 379, 383, 389,
        397, 401, 403, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 481, 487, 491,
        499, 503, 509, 521, 523, 533, 541, 547, 557, 559, 563, 569, 571, 577, 587, 593, 599, 601, 607,
        611, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 689, 691, 701, 709, 719,
        727, 733, 739, 743, 751, 757, 761, 767, 769, 773, 787, 793, 797, 809, 811, 821, 823, 827, 829,
        839, 853, 857, 859, 863, 871, 877, 881, 883, 887, 907, 911, 919, 923, 929, 937, 941, 947, 949,
        953, 967, 971, 977, 983, 991, 997
    );

    public static boolean isPrime(int n) {
        if (n < 2) {
            return false;
        }

        for (Integer prime : primeList) {
            if (n == prime) {
                return true;
            }
            if (n % prime == 0) {
                return false;
            }
            if (prime * prime > n) {
                return true;
            }
        }

        BigInteger bi = BigInteger.valueOf(n);
        return bi.isProbablePrime(10);
    }

    private static boolean sameDigits(int n, int b) {
        int f = n % b;
        while ((n /= b) > 0) {
            if (n % b != f) {
                return false;
            }
        }
        return true;
    }

    private static boolean isBrazilian(int n) {
        if (n < 7) return false;
        if (n % 2 == 0) return true;
        for (int b = 2; b < n - 1; ++b) {
            if (sameDigits(n, b)) {
                return true;
            }
        }
        return false;
    }

    public static void main(String[] args) {
        for (String kind : List.of("", "odd ", "prime ")) {
            boolean quiet = false;
            int bigLim = 99_999;
            int limit = 20;
            System.out.printf("First %d %sBrazilian numbers:\n", limit, kind);
            int c = 0;
            int n = 7;
            while (c < bigLim) {
                if (isBrazilian(n)) {
                    if (!quiet) System.out.printf("%d ", n);
                    if (++c == limit) {
                        System.out.println("\n");
                        quiet = true;
                    }
                }
                if (quiet && !"".equals(kind)) continue;
                switch (kind) {
                    case "":
                        n++;
                        break;
                    case "odd ":
                        n += 2;
                        break;
                    case "prime ":
                        do {
                            n += 2;
                        } while (!isPrime(n));
                        break;
                    default:
                        throw new AssertionError("Oops");
                }
            }
            if ("".equals(kind)) {
                System.out.printf("The %dth Brazilian number is: %d\n\n", bigLim + 1, n);
            }
        }
    }
}






Fortran: 

      PROGRAM BRAZILIAN
      IMPLICIT NONE



      INTEGER , PARAMETER  ::  MAX_NUMBER = 2000000 , NUMVARS = 20



      LOGICAL , DIMENSION(1:MAX_NUMBER)  ::  b
      INTEGER  ::  bcount
      INTEGER  ::  bpos
      CHARACTER(15)  ::  holder
      CHARACTER(100)  ::  outline
      LOGICAL , DIMENSION(1:MAX_NUMBER)  ::  p






      CALL BRAZILIANSIEVE(b , MAX_NUMBER)
      WRITE(6 , 34)"The first 20 Brazilian numbers:"
      bcount = 0
      outline = ''
      holder = ''
      bpos = 1
      DO WHILE ( bcount<NUMVARS )
         IF( b(bpos) )THEN
            bcount = bcount + 1
            WRITE(holder , *)bpos
            outline = TRIM(outline) // " " // ADJUSTL(holder)
         END IF
         bpos = bpos + 1
      END DO
 
      WRITE(6 , 34)outline
      WRITE(6 , 34)"The first 20 odd Brazilian numbers:"
      outline = ''
      holder = ''
      bcount = 0
      bpos = 1
      DO WHILE ( bcount<NUMVARS )
         IF( b(bpos) )THEN
            bcount = bcount + 1
            WRITE(holder , *)bpos
            outline = TRIM(outline) // " " // ADJUSTL(holder)
         END IF
         bpos = bpos + 2
      END DO
      WRITE(6 , 34)outline
      WRITE(6 , 34)"The first 20 prime Brazilian numbers:"
      CALL ERATOSTHENES(p , MAX_NUMBER)
      bcount = 0
      outline = ''
      holder = ''
      bpos = 1
      DO WHILE ( bcount<NUMVARS )
         IF( b(bpos) .AND. p(bpos) )THEN
            bcount = bcount + 1
            WRITE(holder , *)bpos
            outline = TRIM(outline) // " " // ADJUSTL(holder)
         END IF
         bpos = bpos + 1
      END DO
      WRITE(6 , 34)outline
      WRITE(6 , 34)"Various Brazilian numbers:"
      bcount = 0
      bpos = 1
      DO WHILE ( bcount<1000000 )
         IF( b(bpos) )THEN
            bcount = bcount + 1
            IF( (bcount==100) .OR. (bcount==1000) .OR. (bcount==10000) .OR.               &
              & (bcount==100000) .OR. (bcount==1000000) )WRITE(* , *)bcount ,             &
               &"th Brazilian number: " , bpos
         END IF
         bpos = bpos + 1
      END DO
      STOP
 34   FORMAT(/ , a)
      END PROGRAM BRAZILIAN

      SUBROUTINE BRAZILIANSIEVE(B , N)
      IMPLICIT NONE



      INTEGER  ::  N
      LOGICAL , DIMENSION(*)  ::  B
      INTENT (IN) N
      INTENT (OUT) B



      INTEGER  ::  b11
      INTEGER  ::  base
      INTEGER  ::  bn
      INTEGER  ::  bnn
      INTEGER  ::  bpower
      INTEGER  ::  digit
      INTEGER  ::  i
      LOGICAL  ::  iseven
      INTEGER  ::  powermax

      iseven = .FALSE.
      B(1:6) = .FALSE.                 
      DO i = 7 , N
         B(i) = iseven
         iseven = .NOT.iseven
      END DO
      DO base = 2 , (N/2)
         b11 = base + 1
         bnn = b11
         DO digit = 3 , base - 1 , 2
            bnn = bnn + b11 + b11
            IF( bnn>N )EXIT
            B(bnn) = .TRUE.
         END DO
      END DO
      DO base = 2 , INT(SQRT(FLOAT(N)))
         powermax = HUGE(powermax)/base             
         IF( powermax>N )powermax = N               
         DO digit = 1 , base - 1 , 2
            bpower = base*base
            bn = digit*(bpower + base + 1)
            DO WHILE ( (bn<=N) .AND. (bpower<=powermax) )
               IF( bn<=N )B(bn) = .TRUE.
               bpower = bpower*base
               bn = bn + (digit*bpower)
            END DO
         END DO
      END DO
      RETURN
      END SUBROUTINE BRAZILIANSIEVE

      SUBROUTINE ERATOSTHENES(P , N)
      IMPLICIT NONE



      INTEGER  ::  N
      LOGICAL , DIMENSION(*)  ::  P
      INTENT (IN) N
      INTENT (INOUT) P



      INTEGER  ::  i
      INTEGER  ::  ii
      LOGICAL  ::  oddeven
      INTEGER  ::  pr

      P(1) = .FALSE.
      P(2) = .TRUE.
      oddeven = .TRUE.
      DO i = 3 , N
         P(i) = oddeven
         oddeven = .NOT.oddeven
      END DO
      DO i = 2 , INT(SQRT(FLOAT(N)))
         ii = i + i
         IF( P(i) )THEN
            DO pr = i*i , N , ii
               P(pr) = .FALSE.
            END DO
         END IF
      END DO
      RETURN
      END SUBROUTINE ERATOSTHENES