DFS（8）——poj2034Anti-prime Sequences

一、题目回顾

题目链接：Anti-prime Sequences

Sample Input

1 10 2

1 10 3

1 10 5

40 60 7

0 0 0

 

Sample Output

1,3,5,4,2,6,9,7,8,10

1,3,5,4,6,2,10,8,7,9

No anti-prime sequence exists.

40,41,43,42,44,46,45,47,48,50,55,53,52,60,56,49,51,59,58,57,54

 

题意：求n到m的一个排列，满足任意连续的k(2<=k<=d)个数的和都不为素数。

 

二、解题思路

• dfs+素数打表

 

三、代码

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn = 10001;
int ans[1001];
int n,m,d,flag;
int prime[maxn];						
bool vis[1001];

void isPrime()
{
/*	memset(prime,1,sizeof(prime));  
    prime[1] = 0;  
    for (int i=2;i<=10000;i++)  
        if (prime[i]) 
			for (int j=i+i;j<=10000;j+=i) 
				prime[j] = 0;*/
	for(int i=2;i<maxn;i++)
		for(int j=2;i*j<maxn;j++)
	    	prime[i*j]=1;
}

void dfs(int now)
{
	if(flag)	return;
	if(now>m-n+1){
		flag = 1;
		return;
	}
	for(int i=n;i<=m;i++){
		int biaoji = 0;
		if(!vis[i]){
			for(int j=2;j<=d&&now-j>=0;j++){
				if(!prime[ans[now-1]+i-ans[now-j]])  
            		biaoji = 1;  
			}
			if(biaoji)	continue;
			ans[now] = ans[now-1] + i;
			vis[i] = 1;
			dfs(now+1);
			if(flag)	return;
			vis[i] = 0;
		}
	}
}

int main()
{
	isPrime();
	while(cin>>n>>m>>d && !(n==0&&m==0&&d==0)){
		memset(vis,0,sizeof(vis));
		flag = 0;
		dfs(1);
		if(flag==0)	printf("No anti-prime sequence exists.
");
		else{
			printf("%d",ans[1]);  
        	for(int i=2;i<=m-n+1;i++)  
        		printf(",%d",ans[i]-ans[i-1]);  
        	printf("
");  
		}
	}
	return 0;
}