hdoj 1016 Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32717    Accepted Submission(s): 14482

Problem Description

A ring is compose of n circles as shown in diagram. Put natural number 1, 2, ..., n into each circle separately, and the sum of numbers in two adjacent circles should be a prime.

Note: the number of first circle should always be 1.

 

 

Input

n (0 < n < 20).

 

 

Output

The output format is shown as sample below. Each row represents a series of circle numbers in the ring beginning from 1 clockwisely and anticlockwisely. The order of numbers must satisfy the above requirements. Print solutions in lexicographical order.

You are to write a program that completes above process.

Print a blank line after each case.

 

 

Sample Input

6

8

 

 

Sample Output

Case 1:

1 4 3 2 5 6

1 6 5 2 3 4

Case 2:

1 2 3 8 5 6 7 4

1 2 5 8 3 4 7 6

1 4 7 6 5 8 3 2

1 6 7 4 3 8 5 2

 

和nyoj上的素数环做法一样

#include<stdio.h>

#include<string.h>

#define MAX 21

int a[MAX];

int vis[MAX];

int prime[50];

int k;

void biao()

{

	int i,j;

	for(i=0;i<=50;i++)

	prime[i]=1;

	for(i=2;i<=50;i++)

	{

		if(prime[i])

		{

			for(j=2*i;j<=50;j+=i)

			{

				prime[j]=0;

			}

		}

	}

	prime[1]=0;

}

void dfs(int cur,int n)

{

	int i,j;

	if(cur==n+1&&prime[1+a[n]])

	{

		for(i=1;i<=n;i++)

		{

			if(i==1)

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

		    else

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

		}

		printf("\n");

	}

	else

	{

		for(i=2;i<=n;i++)

		{

			if(!vis[i]&&prime[i+a[cur-1]])

			{

				a[cur]=i;

				vis[i]=1;

				dfs(cur+1,n);

				vis[i]=0;

			}

		}

	}

}

int main()

{

	int n,m,j,i,t;

	biao();

	k=1;

	while(scanf("%d",&n)!=EOF)

	{

		memset(a,0,sizeof(a));

		memset(vis,0,sizeof(vis));

		a[1]=1;vis[1]=1;

		printf("Case %d:\n",k++);

		if(n==1)

		printf("1\n");	

		else			

		dfs(2,n);

		printf("\n");

	}

	return 0;

}