Project Euler：Problem 58 Spiral primes

Starting with 1 and spiralling anticlockwise in the following way, a square spiral with side length 7 is formed.

37 36 35 34 33 32 31

38 17 16 15 14 13 30

39 18 5 4 3 12 29

40 19 6 1 2 11 28

41 20 7 8 9 10 27

42 21 22 23 24 25 26

43 44 45 46 47 48 49

It is interesting to note that the odd squares lie along the bottom right diagonal, but what is more interesting is that 8 out of the 13 numbers lying along both diagonals are prime;
that is, a ratio of 8/13 ≈ 62%.

If one complete new layer is wrapped around the spiral above, a square spiral with side length 9 will be formed. If this process is continued, what is the side length of the square
spiral for which the ratio of primes along both diagonals first falls below 10%?

这题是28题的一个扩展，相同找规律，然后推断质数即可了

#include <iostream>
#include <string>
using namespace std;

int cp[100000000];

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

void count_prime(unsigned long long n)
{
cp
= cp[n - 1];
int a[3];
a[0] = (2 * n + 1)*(2 * n + 1) - 4 * n;
a[1] = (2 * n + 1)*(2 * n + 1) - (2 * n + 1) + 1;
a[2] = (2 * n + 1)*(2 * n + 1) - 6 * n;
for (int i = 0; i < 3; i++)
{
if (isPrime(a[i]))
cp
++;
}
}

int main()
{
memset(cp, 0, sizeof(cp));
cp[0] = 0;
unsigned long long ans;
double a, b, res;
for (unsigned long long i = 1; i < 100000000; i++)
{
count_prime(i);
a = cp[i] * 1.0;
b = (4 * i + 1)*1.0;
res = a / b*1.0;
cout << res << endl;
if (res < 0.10)
{
ans = 2 * i + 1;
break;
}
}
cout << ans << endl;
system("pause");
return 0;
}