Project Euler——Problem 27

原题与答案：

Problem 27

Euler published 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.

Using computers, 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.

Answer:-59231
 

分析与解答：

题目是求两个系数a和b，使得他们构成式子，然后将n从0开始取值，直到不是素数为止，现在我们要在a和b的绝对值都在1000内的范围中找出能够令n取到最大的a和b，输出他们的乘积。本身这个题目不难，重点在于枚举和素数的判定，我们列一个素数表即可，考虑到题目中的提示，79的n已经是比较大的了，我们根据这个范围大概列了90000以内的素数表，然后穷举a和b，整个过程十分简单，我的源程序代码如下：