Project Euler：Problem 37 Truncatable primes

最新推荐文章于 2020-12-22 12:34:54 发布

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最新推荐文章于 2020-12-22 12:34:54 发布

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分类专栏： C++ 欧拉计划文章标签： C++ Project Euler

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本文探讨了一个有趣的数学问题：找到所有同时可以从左到右和从右到左截断数字序列仍然保持质数属性的唯一十一组质数，并计算它们的总和。

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;