Quadratic primes
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https://projecteuler.net/problem=27
Quadratic primes
Problem 27
Euler discovered the remarkable quadratic formula:
n² + n + 41
It turns out that the formula will produce 40 primes for the consecutive valuesn = 0 to 39. However, when n = 40, 402 + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and certainly whenn = 41, 41² + 41 + 41 is clearly divisible by 41.
The incredible formula n² − 79n + 1601 was discovered, which produces 80 primes for the consecutive valuesn = 0 to 79. The product of the coefficients, −79 and 1601, is −126479.
Considering quadratics of the form:
n² + an + b, where |a| < 1000 and |b| < 1000where |n| is the modulus/absolute value ofn
e.g. |11| = 11 and |−4| = 4
Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values ofn, starting with n = 0.
首先b一定大于零,其次本想第一个公式算出n,第二个公式只要去验证n+1即可,不需要从0开始计算,不过真正计算的时候,发现事实不是这样的。
看来还是每一个公式从0开始计算的实在,而且速度也不慢。
def QuadraticPrimes(): coefficients = 0 maxnum = 0 for i in range(-999,1000): for j in range(1,1000): n = 0 tmp = j #算出最大的连续n值 while tmp > 0 and isPrime(tmp): n += 1 tmp = n * n + i * n + j if maxnum < n: maxnum = n coefficients = i * j return coefficients#判断n是否是质数,记录一下字典,加快速度primedict = {}import mathdef isPrime(n): if n in primedict:return primedict[n] temp = int(math.sqrt(n)) + 1 for i in range(2,temp): if n % i == 0: primedict[n]=False return False primedict[n]=True return Trueprint(QuadraticPrimes())
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