PE37

/*The number 3797 has an interesting property. Being prime itself, it is possible to continuously remove digits from left to right, and remain prime at each stage: 3797, 797, 97, and 7. Similarly we can work from right to left: 3797, 379, 37, and 3.Find the sum of the only eleven primes that are both truncatable from left to right and right to left.NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes.*/#include <Windows.h>#include <iostream>#include <vector>using namespace std;typedef unsigned long DWORD;const DWORD size = 1000000;bool prime[size] = {false};vector<DWORD> primes;bool IsPrime(long long int n) { if (n==1) return false;if (n==2) return true;if(n%2==0) return false; long long int hd=8; long long int lg=6; long long int lh=8; long long int va=n-1; while(va>=hd) { if(!((va-hd)%lg)) return false; lh+=8; lg+=4; hd+=lh; } return true; } bool Problem37(DWORD N){DWORD tmp = N;int i = 10;int d = 10;while (tmp!=0){d*=10;tmp /= i;if (!prime[tmp] && tmp!=0)return false;}for (int i=10; i<=d; i*=10)if (!prime[N%i])return false;return true;}int main(){DWORD start = GetTickCount();for (int n=1; n<size; n++){if (IsPrime(n)){primes.push_back(n);prime[n] = true;}}int count = 0;int sum11 = 0; for (size_t i = 0; i<primes.size(); i++) {if (primes[i]<10)continue;if (Problem37(primes[i])){cout << primes[i] << endl;sum11 += primes[i];count++;if (count == 11){cout << "result = " << sum11<< endl;DWORD end = GetTickCount();cout << "the running time = " << end - start << " ms." << endl;return 0;}} }return 0;}