joj 1195

1195: Prime Ring Problem

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	10s	8192K	1973	499	Standard

A ring is composed of n circles as shown in diagram. Put natural numbers 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 <= 16)

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.

You are to write a program that completes above process.

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 <cmath>

using namespace std;
const int maxn = 33;

int isp[maxn], n, a[10000],vis[10000];
bool judge(int x)
{
int i;
for (i=2; i<=sqrt((double)x); i++)
{
if (x % i == 0) return false;
}
return true;
}

void makeprime()
{
int i;
for (i=2; i<=maxn; i++)
if (judge(i)) isp[i] = 1;

}

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

}

int main()
{
makeprime();
int count=0;
while (cin>>n)
{
count++;
cout<<"Case "<<count<<":"<<endl;
if(n%2==0)
{
memset (a,0,sizeof(a));
memset (vis,0,sizeof(vis));
vis[1] = 1;
a[0] = 1;
dfs(1);
}
cout<<endl;

}
return 0;
}

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