Project Euler task 58
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 
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%?
------------------------------------------------------------------------------------#include < iostream >
using namespace std;
bool isPrime(long int n) {
if (n <= 1) return false;
if (n == 2) return true;
if (n % 2 == 0) return false;
if (n < 9) return true;
if (n % 3 == 0) return false;
long counter = 5;
while ((counter * counter) <= n) {
if (n % counter == 0) return false;
if (n % (counter + 2) == 0) return false;
counter += 6; // we have checked the %2 and %3
}
return true;
}
int main() {
int iPrimes = 0;
int iNumber = 1;
int iStep = 2;
do {
for (int i = 0; i < 3; i++) {
iNumber += iStep;
iPrimes += (int) isPrime(iNumber);
}
iNumber += iStep;
if ((double) iPrimes / (2 * iStep + 1) < 0.1) {
cout << "Verified" << endl;
break;
}
iStep += 2;
} while (true);
cout << iStep + 1 << endl;
return 0;
}
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