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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;