Project Euler：Problem 37 Truncatable primes

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

原文：http://blog.csdn.net/youb11/article/details/46356929