Sicily 1002. Anti-prime Sequences dfs

         一开始卡时卡得厉害，后来优化了，还是TLM, 看了别人的解题报告, 发现自己写的check函数复杂了用了3个循环，其实两个循环就可以了，还学会了筛选法求素数

#include<iostream>

#include <memory.h>

#include <math.h>

#include <stdio.h>

using namespace std;



int result[1001];

bool used[1001];

bool isprime[10010];

int d, m, n, num;

int dfs(int);



bool isfound;



bool check(int);

void prime_list();



int main()

{

	

	prime_list();



	while (cin >> n >> m >> d && !(n== 0 && m == 0 && d==0))

	{

		isfound = false;

		num = m-n+1;

		memset(used, false, 1001*sizeof(bool));

		

		dfs(0);

		if (isfound)

		{

			for (int i = 0; i < num-1; i++)

			    printf("%d,", result[i]);

			printf("%d\n", result[num-1]);

		}

		else

			printf("No anti-prime sequence exists.\n");

	}

	return 0;

}



int dfs(int current)

{

	if (isfound) return 0;

	if (!check(current)) return 0;



	if (current == num) {isfound = true; return 0;}



	for (int i = 0; i < num && !isfound; i++)

	{

		if (!used[i])

		{

			used[i] = true;

			result[current] = n+i;

			dfs(current+1);

			used[i] = false;

		}

	}



	return 0;

}



bool check(int current)

{

	int sum = 0;

	if (current <= 1) return true;

	for (int i = 0; i < current; i++)

	{

		sum = 0;

		if (i+d < current)

		for (int j = 0; j < d; j++)

		{

			sum += result[i+j];		

			if (j > 1 && isprime[sum]) return false;

		} 

		else

		{

			for (int j = i; j < current; j++)

			{

				sum += result[j];

				if (j - i > 0 && isprime[sum]) return false;

			}



		}

	}

	return true;

}



//筛选法求1~10000的素数

void prime_list()

{

	int i , j;

	memset(isprime, true, sizeof(isprime));

	isprime[0] = false;

	isprime[1] = false;



	//注意：合数N的不等于1的最小正因数不大于N的平方根

	for ( i = 2; i*i <= 10000; i++)

	{

		j = 2;

		if (isprime[i])

		{

			while (i*j <= 10000)

			{

				isprime[i*j] = false;

				j++;

			}

		}

	}

}