欧拉计划第 87 题，主要三元组

Question

The problem goes: How many numbers below fifty million can be expressed as the sum of a prime square, prime cube, and prime fourth power? And here is the link to the problem: https://projecteuler.net/problem=87

I'm having a problem when trying to get the right answer for higher inputs. My code in python is below. The function primestill(n) returns a generator of all primes below n and works just fine. I would appreciate any help.

def many2(x):
    counter = 0
    s1 = int(x**(1/4))+1
    for i in primestill(s1):
        s2 = int((x-i**4)**(1/3)) + 1
        for j in primestill(s2):
            s3 = int((x - i**4 - j**3)**(1/2))+1
            for k in primestill(s3):
                if i**4 + j**3 + k**2 < x:
                   counter += 1
    return counter

Answer 1

You should generate a list of primes up to the largest one that will fit under the 50 million mark. This would be the largest prime with a square below 50M.

There are only 908 primes up to that largest prime so your loops can just combine these primes.

Note, however, that there are several combinations of primes that can produce the same sum so you will need to ensure that you count each sum only once (using a set for example).

To optimize the process you can break the nested loops once you reach a prime that would go beyond the maximum (50M):

# find eligible primes (Eratosthenes sieve)
N        = 50_000_000
maxPrime = int(N**0.5)
sieve    = [0,0]+[1]*maxPrime
p,np     = 1,2
primes   = []
while np<=maxPrime:
    p,np = np,p+2
    if not sieve[p]: continue
    primes.append(p)
    sieve[p*p::p] = [0]*len(sieve[p*p::p])

# nested loops to count distinct sums made from a**2 + b**3 + c**4
sums = set()
for c in primes:
    for b in primes:
        b3c4 = b**3 + c**4
        if b3c4 >= N: break
        for a in primes:
            s = a*a+b3c4
            if s >= N: break
            sums.add(s)

print(len(sums))