*寒假水106——素数环 搜索

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. 


 
Inputn (0 < n < 20). 
OutputThe 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

#include<iostream>
#include<string.h>
#include<stdio.h>

using namespace std;

int sushu[40]={0,1,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0};//素数打表，因为n最大是20，所以只要打到40
int xin[21],pei[21];
int n;

void dfs(int a)
{
int i;
if(a==n&&sushu[pei[a-1]+pei[0]])
{
for(i=0;i<a-1;i++)
cout<<pei[i]<<" ";
cout<<pei[a-1]<<endl;
}
else
{
for(i=2;i<=n;i++)
{
if(!xin[i])
{
if(sushu[i+pei[a-1]])
{
xin[i]=-1;
pei[a++]=i;
dfs(a);
xin[i]=0;
a--;
}
}
}
}
}

int main()
{
int num=0;
while(cin>>n)
{
num++;
printf("Case %d:\n",num);
memset(xin,0,sizeof(xin));
pei[0]=1;
dfs(1);
cout<<endl;
}
return 0;
}

题解：看到题目时我就按捺不住内心的激动，没错，王岐叔叔让我们写过这道题。想当初，那是一个温暖的阳光洒进教室来的午后，我爬在桌子上想这道题，嗯，我果然睡着了，一觉起来夕阳都没了。好在很快写下了这道题的框架。