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

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,timedef isOk(temp):if temp < 0: return Falsefor i in range(2,int(math.sqrt(temp))+1):if temp%i==0 : return Falsereturn Truebegin = time.time()maxValue=0lastValue=1for a in range(-1000,1001):for b in range(-1000, 1001):n = 0resu = 0while True:temp = n*n + a*n + bif(isOk(temp)):resu+=1n+=1else :breakif resu>maxValue:maxValue = resulastValue = a*bprint lastValueend = time.time()print end-begin