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("
");
	}
	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:
",k++);
		if(n==1)
		printf("1
");	
		else			
		dfs(2,n);
		printf("
");
	}
	return 0;
}