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++;
			}
		}
	}
}