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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%?

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这题是28题的一个扩展，同样找规律，然后判断质数就行了

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#include 
  
   
#include 
   
     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[n] = 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[n]++; } } 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; }