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;
}