标签:c++ project euler
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 <iostream> #include <string> using namespace std; bool prim(int a) { if (a == 1) return false; for (int i = 2; i*i <= a; i++) { if (a%i == 0) return false; } return true; } bool tr_prim(int a) { int num = a; int count = 0; int tmp[10] = { 0 }; while (a) { if (!prim(a)) return false; count++; tmp[count] = a % 10; a /= 10; } for (int i = count; i > 1; i--) { num = num - tmp[i] * pow(10, i - 1); if (!prim(num)) return false; } return true; } int main() { int sum = 0; for (int i = 10; i <= 1000000; i++) { if (tr_prim(i)) { //cout << i << endl; sum += i; } } cout << sum << endl; system("pause"); return 0;
Project Euler:Problem 37 Truncatable primes
标签:c++ project euler
原文地址:http://blog.csdn.net/youb11/article/details/46356929