对于一个大学项目，我想实现 Bellard 算法，用 Fortran 计算 pi 的第 n 位。我在 math.stackexchange 上偶然发现了这个问题:https://math.stackexchange.com/questions/1776840/confusion-with-bellards-algorithm-for-pi
有了这个问题的答案，我设法实现了代码，但我没有得到结果，而且我不知道我做错了什么:
PROGRAM pi
IMPLICIT NONE
INTEGER :: find_number_of_primes, number_primes, last_digit, digit, N, &
eps, i, j, multiplicity, test
REAL :: log2, base, v_max, m, s, v, b, A, pi_sum
INTEGER, ALLOCATABLE :: primes(:)
digit = 3
eps = 2
base = 10
N = INT((digit+eps)*log2(base))
pi_sum = 0
number_primes = find_number_of_primes(2*N)
ALLOCATE(primes(number_primes))
CALL get_primes(number_primes, primes)
DO i=1,SIZE(primes)
v_max = INT(log(REAL(2*N))/log(REAL(primes(i))))
m = primes(i)**v_max
s = 0
v = 0
b = 1
A = 1
DO j = 1,(N)
b = MOD((j/(primes(i)**multiplicity(digit, j)) * b), m)
A = MOD((2*j-1)/(primes(i)**multiplicity(primes(i), (2*j-1)))*A, m)
v = v - multiplicity(primes(i),j)+multiplicity(primes(i),(2*j-1))
IF (v > 0) THEN
s = MOD(s*j*b*(A**(-1))*a**(v_max-v), m)
END IF
END DO
s = MOD(s * base**(digit-1), m)
pi_sum = pi_sum + MOD((s/m), REAL(1))
END DO
PRINT*, pi_sum
END PROGRAM
自定义函数 find_number_of_primes、get_primes、log2 和 multiplicity 正在按预期工作。第一个查找给定区间内素数的个数，第二个返回数组中的素数，第三个计算
log_2(x) = log(x)/log(2)
最后一个通过检查第二个数字除以第一个数字的频率来计算重数(我猜这就是它的名字)，直到除法的其余部分不再为零。以下是源代码:
对数:
real function log2(x)
implicit none
real, intent(in) :: x
log2 = log(x) / log(2.)
end function
求区间内素数的个数:
integer function find_number_of_primes(limit) result(number_primes)
implicit none
INTEGER, INTENT(IN) :: limit
INTEGER :: i, j
LOGICAL :: is_prime
number_primes = 0
DO i = 3,(limit-1)
is_prime = .TRUE.
DO j = 2, (i-1)
IF (MODULO(i, j) == 0) THEN
is_prime = .FALSE.
EXIT
END IF
END DO
IF (is_prime .EQV. .TRUE.) THEN
number_primes = number_primes + 1
END IF
END DO
end function find_number_of_primes
获取实际素数:
SUBROUTINE get_primes(number_primes, primes)
IMPLICIT NONE
INTEGER, INTENT(IN) :: number_primes
INTEGER :: i, found_primes, j
LOGICAL:: is_prime
INTEGER, INTENT(OUT) :: primes(number_primes)
i = 3
found_primes = 0
DO
is_prime = .TRUE.
DO j = 2, (i-1)
IF (MODULO(i,j) == 0) THEN
is_prime = .FALSE.
END IF
END DO
IF (is_prime .EQV. .TRUE.) THEN
found_primes = found_primes + 1
primes(found_primes) = i
IF (found_primes == number_primes) EXIT
END IF
i = i + 1
END DO
end SUBROUTINE get_primes
integer function multiplicity(a, b)
implicit none
INTEGER, INTENT(IN) :: a, b
multiplicity = 0
DO
multiplicity = multiplicity + 1
IF (MOD(b, (a**(multiplicity+1))) /= 0) THEN
EXIT
END IF
END DO
end function multiplicity
整个文件的 Pastebin:https://pastebin.com/0F4zQYaH
现在，在我链接的问题中，在内循环中计算b时，有a^{v(n, k)}分母，但在问题的答案中，分母中的项是a^{v(a, k)}。此外，问题中的内部循环从 1 到 N，而答案中的循环从 1 到 2N.
最终结果是 NaN，以下是 digit = 2 和 eps = 1 的一些控制台输出:
************ i = 1 ***************************
v_max = 2.00000000
m = 9.00000000
------------- j = 1 -------------------------------
b = 0.00000000
A = 0.00000000
v = 0.00000000
------------- j = 2 -------------------------------
b = 0.00000000
A = 0.00000000
v = 0.00000000
------------- j = 3 -------------------------------
b = 0.00000000
A = 0.00000000
v = 0.00000000
------------- j = 4 -------------------------------
b = 0.00000000
A = 0.00000000
v = 0.00000000
------------- j = 5 -------------------------------
b = 0.00000000
A = 0.00000000
v = 1.00000000
v > 0 --> s = NaN
------------- j = 6 -------------------------------
b = 0.00000000
A = 0.00000000
v = 1.00000000
v > 0 --> s = NaN
如此一来，完整输出:https://pastebin.com/ucJNg6Vi
有人可以帮我找出我做错了什么吗？
最佳答案
您的代码中有一个错误:
v = v - multiplicity(primes(i),j)+multiplicity(primes(i),(2*j-1))
应该是
v = v - multiplicity(primes(i),j)- multiplicity(primes(i),(2*j-1))
另外，
s = MOD(s*j*b*(A**(-1))*a**(v_max-v), m)
应该是
s = MOD(s + j*b*(A**(-1))*a**(v_max-v), m)
关于algorithm - 计算 Pi 的贝拉德算法，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/52835997/