[Project Euler] Problem 58

Problem Description

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%?

C++

This problem is similar with problem 28, not too hard, just brute force.

const double THRESHOLD = 0.1;

void Problem_58()
{

int firstNum = 0;
int lastNum = 1;
int layer = 0;

double ratio = 1;
int whole = 1;
int prime = 0;
while(ratio >= THRESHOLD)
{
layer++;
int increment = layer * 2;
int firstNum = lastNum + increment;
lastNum = firstNum;
if(IsPrime(lastNum))
{
prime++;
}
for(int i=0; i<3; i++)
{
lastNum += increment;
if(IsPrime(lastNum))
{
prime++;
}
}
whole += 4;
ratio = ((double)prime) / whole;
printf("%d/%d, %f,%d\n", prime, whole, ratio, increment + 1);
}
printf("%f,%d\n", ratio, 2 * layer + 1);

}