杭电acm 1016 素数环

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
格式变态，一直以为最后不输出空行，PE6次，简单的dfs.由于数据较小，直接素数手动打表了，懒得写函数。
显然奇数是不成立的，所以只能输入偶数。


#include<stdio.h>
#include<string.h>
int value[40]={0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0}; //素数打表，1素数，0非素数。从0-40. 
int v[25],a[25];                        //v作为使用的标记，a储存n个数 
void dfs(int x,int y)
{
	int i;
	if(x==y+1&&value[a[1]+a[y]]==1)    //x递归到y+1尸，且最后一个数跟1的和是素数   结束 
	{
		printf("1");                   //1每次都作为开头输出
        for(i=2;i<=y;i++)
        printf(" %d",a[i]);            //注意空格 
        printf("\n");                  //空行 
        return ;
	}
	for(i=2;i<=y;i++)                  //从2开始计数 
	{
		if(v[i]==0&&value[i+a[x-1]])   //i没被利用且这个数加数组a中的前一个数是素数 
		{
			v[i]=1;                    //标记 
			a[x]=i;                    //储存 
			dfs(x+1,y);                //回溯 
			v[i]=0;                    //恢复原状态  
		}
	}
}
int main()
{
	int T=1,i,n;
	while(scanf("%d",&n)!=EOF)
	{
		printf("Case %d:\n",T++);
		if(n==1)                      //1单独处理 
		{
			printf("1\n");
			continue;
		}
//		if(n%2==1)
//        {
//            printf("No Answer\n");   //奇数不成立 
//            flag=1;
//            continue;
//        }
        memset(v,0,sizeof(v));         //清空数组v为0 
        v[1]=a[1]=1;                   //当n=1时，成立 
        dfs(2,n);
        printf("\n");                  //狗血的换行 
	}
	return 0;
}