Problem 27

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本文探讨了一种通过二次多项式生成连续素数的方法，并提供了一段Java代码实现。通过对不同系数的二次方程进行遍历，寻找能够生成最多连续素数的公式及其系数乘积。

问题描述：

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:

解决问题：

利用题目本身的限制条件，因为a和b是1000以下，那么先使用boolean[] prime = new boolean[1000001];

public static boolean IsPrime(int number){
		boolean result = true;
		
		if(number%2==0){
			result = false;
		}else{
			int middle = (int)Math.sqrt(number);
			for(int i=3; i<=middle; i+=2){
				if(number%i==0){
					result = false;
					break;
				}
			}
		}
		
		return result;
	}
	
	public static void find(){
		int a =0, b=0;
		int max_len = 0;
		//先找出1000以内的所有素数
		boolean[] prime = new boolean[1000001];
		Arrays.fill(prime, false);
		for(int i=2;i<=1000001;i++){
			if(IsPrime(i)){
				prime[i] = true;
			}
		}
		
		for(int i=-1000; i<=1000; i++){ //循环a
			for(int j=-1000; j<=1000; j++){ //循环b
				int n;
				for( n=0; n<Math.abs(j); n++){
					int value = n*n+ i*n + j;
					if(value>0&&prime[value]){
						;
					}else{
						break;
					}
				}
				if(max_len<n){
					a = i;
					b = j;
					max_len = n;
				}
			}
		}
		System.out.println("a:"+a+",b:"+b+",len"+max_len);
		System.out.println(a*b);
	}