projecteuler---->problem=27----Quadratic primes

最新推荐文章于 2018-10-17 13:31:49 发布

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最新推荐文章于 2018-10-17 13:31:49 发布

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分类专栏： projecteuler Python 文章标签： projectEuler python

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本文探讨了一种特殊形式的二次方程，该方程在连续的n值下能产生大量的质数。通过算法计算，最终确定了产生最多质数的系数，并计算了这些系数的乘积。

Euler discovered the remarkable quadratic formula:

n² + n + 41

It turns out that the formula will produce 40 primes for the consecutive values n = 0 to 39. However, when n = 40, 40² + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and certainly when n = 41, 41² + 41 + 41 is clearly divisible by 41.

The incredible formula n² − 79n + 1601 was discovered, which produces 80 primes for the consecutive values n = 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| < 1000

where |n| is the modulus/absolute value of n
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 of n, starting with n = 0.

import math,time
def isOk(temp):
	if temp < 0: return False
	for i in range(2,int(math.sqrt(temp))+1):
		if temp%i==0 :
 			return False
	return True

begin = time.time()
maxValue=0
lastValue=1
for a in range(-1000,1001):
	for b in range(-1000, 1001):
		n = 0	
		resu = 0
		while True:
			temp = n*n + a*n + b
			if(isOk(temp)):
				resu+=1
				n+=1
			else :
				break
		if resu>maxValue:
			maxValue = resu
			lastValue = a*b
print lastValue
end = time.time()
print end-begin