Project Euler：Problem 60 Prime pair sets

The primes 3, 7, 109, and 673, are quite remarkable. By taking any two primes and concatenating them in any order the result will always be prime. For example, taking 7 and 109,
both 7109 and 1097 are prime. The sum of these four primes, 792, represents the lowest sum for a set of four primes with this property.
Find the lowest sum for a set of five primes for which any two primes concatenate to produce another prime.

太暴力了不忍心贴代码了(╯‵□′)╯︵┻━┻

#include <iostream>
#include <string>
#include <vector>
#include <map>
using namespace std;

bool isp[100000000];
int prime[1500];

bool isPrime(int n)
{
	for (int i = 2; i*i < n; i++)
	{
		if (n%i == 0)
			return false;
	}
	return true;
}

int cnt;
void prim()
{
	int count = 0;
	for (int i = 2; i < 100000000; i++)
	{
		if (isPrime(i))
		{
			isp[i] = 1;
			if (i<10000)
				prime[count++] = i;
		}
	}
	cnt = count;
}

bool check(int a, int b, int c, int d, int e)
{
	int num[5] = { a, b, c, d, e };
	for (int i = 0; i < 5; i++)
	{
		for (int j = 0; j < 5; j++)
		{
			if (i == j)
				continue;

			int tp1 = num[i];
			int tp2 = num[j];
			while (tp2)
			{
				tp1 *= 10;
				tp2 /= 10;

			}
			tp1 += num[j];
			if (isp[tp1] == 0)
				return false;

		}
	}
	return true;
}

bool twocheck(int a, int b)
{
	int tp1 = a;
	int tp2 = b;
	while (tp2)
	{
		tp1 *= 10;
		tp2 /= 10;
	}
	tp1 += b;
	if (isp[tp1])
		return true;
	return false;
}

int main()
{
	memset(isp, 0, sizeof(isp));
	prim();
	int res = 0;
	int flag = 0;
	for (int a = 0; a < cnt; a++)
	{
		if (flag == 1)
			break;
		for (int b = a + 1; b < cnt; b++)
		{
			if (!twocheck(prime[a], prime[b]) || !twocheck(prime[b], prime[a]))
				continue;
			if (flag == 1)
				break;
			for (int c = b + 1; c < cnt; c++)
			{
				if (!twocheck(prime[c], prime[b]) || !twocheck(prime[b], prime[c]) || !twocheck(prime[a], prime[c]) || !twocheck(prime[c], prime[a]))
					continue;
				if (flag == 1)
					break;
				for (int d = c + 1; d < cnt; d++)
				{
					if (!twocheck(prime[d], prime[b]) || !twocheck(prime[b], prime[d]) || !twocheck(prime[a], prime[d]) || !twocheck(prime[d], prime[a])
						|| !twocheck(prime[c], prime[d]) || !twocheck(prime[d], prime[c]))
						continue;
					if (flag == 1)
						break;
					for (int e = d + 1; e < cnt; e++)
					{
						if (check(prime[a], prime[b], prime[c], prime[d], prime[e]))
						{
							res = prime[a] + prime[b] + prime[c] + prime[d] + prime[e];
							cout << res << endl;
							flag = 1;
							break;
						}
					}
				}
			}
		}
	}

	system("pause");
	return 0;
}