Project Euler：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.

a,b分别从[-999,999]之间取值，找到满足条件的a,b使得n按顺序取0,1,2,3,4,...的情况下连续生成的质数序列最长。

#include <iostream>
using namespace std;

bool prim(int a)
{
	if (a < 2)
		return false;
	for (int i = 2; i*i <= a; i++)
	{
		if (a%i == 0)
			return false;
	}
	return true;

}

int main()
{
	int maxcount = 0;
	int res = 0;
	for (int a = -999; a <= 999; a++)
	{
		for (int b = -999; b <= 999; b++)
		{
			int n = 0;
			while (true)
			{
				int tmp = n*n + a*n + b;
				if (prim(tmp))
					n++;
				else
					break;
			}
			if (n > maxcount)
			{
				maxcount = n;
				res = a*b;
			}
		}
	}
	cout << res << endl;
	system("pause");
	return 0;
}

Project Euler：Problem 27 Quadratic primes

原文：http://blog.csdn.net/youb11/article/details/46324639