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); }