' Puzzlet #036
' Find 3 consecutive primes, all less than 1000,
' whose sum is a perfect cube.
' By Dave Ellis.

declare IsPrime(p: int)
declare IsCube (c: int)
declare NextPrime()
declare PrintResult()
def i, count, prime, sum: int
def primeArray[4]: int
openconsole
print "Searching ...": print
primeArray = 0, 2, 3, 5: ' pre-load first 3 primes
count = 1

do
    count = count + 1: 'counts primes
    sum = 0: 'initialise
    ' find sum of current 3 primes
    for i = 1 to 3
        sum = sum + primeArray[i]
    next i
    ' is the sum a perfect cube?
    if IsCube(sum)
        PrintResult()
    endif
    ' discard lowest prime, get a new highest one
    primeArray[1] = primeArray[2]
    primeArray[2] = primeArray[3]
    primeArray[3] = NextPrime()
until primeArray[3] > 1000

print: print "No more!"
print: print "Press a key to exit ..."   
do: until inkey$ <> ""
closeconsole
end

sub IsPrime(p)
' is p a prime?
def factor, flag, root: int
root = sqrt(p): ' limits test range
factor = 3: 'first factor to test
flag = 1
do
    if p % factor = 0
        flag = 0
    else
        factor = factor + 2
    endif
until flag = 0 | factor > root
return flag

sub IsCube(c)
' if c is a perfect cube, returns
' 1, else returns 0
def possCube: float
possCube = exp(log(c)/3)
if possCube = int(possCube)
    return 1
else
    return 0
endif

sub NextPrime
' get the next higher prime
' and returns it
def flag, try: int
try = primeArray[3] + 2
flag = 0
do
    if IsPrime(try)
        flag = 1
    else
        try = try + 2
    endif
until flag
return try


sub PrintResult
' pretty printer
def n: int
for n = 1 to 3
    print "prime #",
    print ltrim$(str$(count + n - 1)), ": ",
    print primeArray[n]
next n 
print "Sum: ", sum
print "Cube Root: ",
print exp(log(sum)/3)
return