我正在研究一个以纯lua运行的计算器,但我需要帮助将小数点输出到分数
答案 0 :(得分:2)
此解决方案使用连续分数精确恢复分母,分母高达10 7
local function to_frac(num)
local W = math.floor(num)
local F = num - W
local pn, n, N = 0, 1
local pd, d, D = 1, 0
local x, err, q, Q
repeat
x = x and 1 / (x - q) or F
q, Q = math.floor(x), math.floor(x + 0.5)
pn, n, N = n, q*n + pn, Q*n + pn
pd, d, D = d, q*d + pd, Q*d + pd
err = F - N/D
until math.abs(err) < 1e-15
return N + D*W, D, err
end
local function print_frac(numer,denom)
print(string.format("%.14g/%d = %d/%d + %g", numer, denom, to_frac(numer/denom)))
end
print_frac(1, 4) --> 1/4 = 1/4 + 0
print_frac(12, 8) --> 12/8 = 3/2 + 0
print_frac(4, 2) --> 4/2 = 2/1 + 0
print_frac(16, 11) --> 16/11 = 16/11 + 5.55112e-17
print_frac(1, 13) --> 1/13 = 1/13 + 0
print_frac(math.sqrt(3), 1) --> 1.7320508075689/1 = 50843527/29354524 + -4.44089e-16
print_frac(math.pi, 1) --> 3.1415926535898/1 = 80143857/25510582 + 4.44089e-16
print_frac(0, 3) --> 0/3 = 0/1 + 0
print_frac(-10, 3) --> -10/3 = -10/3 + -1.11022e-16
答案 1 :(得分:1)
这是不可能的。你需要一个存储分数的类。
您可以获得approximate solution。它可以很好地用于可以表示为分数的东西,并且可以用于其他所有东西
local function gcd(a, b)
while a ~= 0 do
a, b = b%a, a;
end
return b;
end
local function round(a)
return math.floor(a+.5)
end
function to_frac(num)
local integer = math.floor(num)
local decimal = num - integer
if decimal == 0 then
return num, 1.0, 0.0
end
local prec = 1000000000
local gcd_ = gcd(round(decimal*prec), prec)
local numer = math.floor((integer*prec + round(decimal*prec))/gcd_)
local denom = math.floor(prec/gcd_)
local err = numer/denom - num
return numer, denom, err
end
function print_frac(numer,denom)
print(string.format("%d/%d = %d/%d + %g", numer, denom, to_frac(numer/denom)))
end
print_frac(1,4)
print_frac(12,8)
print_frac(4,2)
print_frac(16,11)
print_frac(1,13)
输出:
1/4 = 1/4 + 0
12/8 = 3/2 + 0
4/2 = 2/1 + 0
16/11 = 290909091/200000000 + 4.54546e-10
1/13 = 76923077/1000000000 + 7.69231e-11