Problem 27 Quadratic primes （暴力枚举）

Quadratic primes

Problem 27

Euler discovered the remarkable quadratic formula:

It turns out that the formula will produce 40 primes for the consecutive integer values . However, when  is divisible by 41, and certainly when  is clearly divisible by 41.

The incredible formula  was discovered, which produces 80 primes for the consecutive values . The product of the coefficients, −79 and 1601, is −126479.

Considering quadratics of the form:

, where  and 

where  is the modulus/absolute value of 
e.g.  and

Find the product of the coefficients,  and , for the quadratic expression that produces the maximum number of primes for consecutive values of , starting with .

题意：二次质数。欧拉发现了著名的二次方程：n² + n + 41。当n=0~39时，这个方程能够连续产生40个质数。但是，当n=40时，402 + 40 + 41 = 40(40 + 1) + 41可以被41整除，当然，还有当n = 41, 41² + 41 + 41 可以被41整除。

发现了更奇妙的方程： n² – 79n + 1601，当n=1~79时，它能够 连续产生80个质数。方程系数的积为 –79 * 1601 = -126479.

考虑二次方程的形式：

n² + an + b, |a| < 1000 且 |b| < 1000

|n| 是n 的绝对值。

找到系数a和b，当n从0开始连续递增时，可以产生最多的质数。给出a*b的值。

暴力枚举吧....

代码：

#include<bits/stdc++.h> 
using namespace std;
int  isprime(int n)
{
    if (n<=1)return 0;
    for(int i=2;i<=sqrt(n);i++)
    if(n%i==0)return 0;
    return 1;
    
}
int gao(int a, int b) //连续素数的个数 
{
    int cnt = 0;
    for (int i=0;;i++)
    {
        if (isprime(i*i+a*i+b)) //二元产生的是素数 
        {
            cnt++;
        }
        else break;
    }
    return cnt;
}
 
int main()
{
    int maxa,maxb,maxprimes=0;
    for (int a=-1000;a<1000;a++)
    {
        for(int b=-1000;b<1000;b++)
        {
            if(gao(a,b)>maxprimes)
            {
                maxa = a;
                maxb = b;
                maxprimes = gao(a,b);
            }
        }
    }
    cout<<maxa*maxb<<endl;
    return 0;
}