UVA

  Prime Ring Problem 

A ring is composed of n (even number) 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 <bits/stdc++.h>
#define MAXN 102400
using namespace std;

int n;
bool isp[MAXN];
int res[32];
int vis[32];

void init_isp()
{
    isp[0] = isp[1] = false;
    for(int i = 2; i < MAXN; i++)
        isp[i] = true;
    for(int i = 2; i < sqrt(MAXN)+1; i++)
        if(isp[i])
            for(int j = 2; i*j < MAXN; j++)
                isp[i*j] = false;
}

void dfs(int cur)
{
    if(cur == n && isp[res[0]+res[n-1]]){
        printf("%d", res[0]);
        for(int i = 1; i < n; i++)
            printf(" %d", res[i]);
        printf("
");
    }
    else{
        for(int i = 2; i <= n; i++){
            if(!vis[i] && isp[res[cur-1]+i]){
                res[cur] = i;
                vis[i] = 1;
                dfs(cur+1);
                vis[i] = 0;
            }
        }
    }
}

int main()
{
    int kase = 0;
    init_isp();
    while(cin >> n)
    {
        if(kase++)
            printf("
");
        printf("Case %d:
",kase);
        res[0] = 1;
        memset(vis, 0, sizeof(vis));
        vis[1] = 1;
        dfs(1);
    }
    return 0;
}