九度·oj 1459

题目描述：

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.



输入：

n (1 < n < 17).
输出：

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.

样例输入：

6
8

样例输出：

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

提示：

用printf打印输出。

#include<stdio.h>
#include<string.h>
#define true 1
#define false 0
using namespace std;
int ans[22];
int n;
bool hash[22];
int prime[]={2,3,5,7,11,13,17,19,23,29,31,37,41};
bool judge(int x)
{
for(int i=0;i<13;i++)
{
if(prime[i]==x)
{
return true;
}
}
return false;

}
void check()
{
if(judge(ans
+ans[1])==false)
return;
for(int i=1;i<=n;i++)
{
if(i!=1) printf(" ");

printf("%d",ans[i]);
}
printf("\n");
}
void DFS(int num)
{
if(num>1)
{
if(judge(ans[num]+ans[num-1])==false)
{
return;
}
}
if(num==n)
{
check();
return;
}
for(int i=2;i<=n;i++)
{
if(hash[i]==false)
{
hash[i]=true;
ans[num+1]=i;
DFS(num+1);
hash[i]=false;
}
}
}

int main()
{
int cas=0;
while(scanf("%d",&n)!=EOF)
{ cas++;
for(int i=0;i<22;i++)
{
hash[i]=false;
}
ans[1]=1;
printf("Case %d:\n",cas);
hash[1]=true;
DFS(1);
printf("\n");
}
}