Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35961    Accepted Submission(s): 15867

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

Source

Asia 1996, Shanghai (Mainland China)

/*
ID: LinKArftc
PROG: 1016.cpp
LANG: C++
*/

#include <map>
#include <set>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <cstdio>
#include <string>
#include <utility>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define eps 1e-8
#define randin srand((unsigned int)time(NULL))
#define input freopen("input.txt","r",stdin)
#define debug(s) cout << "s = " << s << endl;
#define outstars cout << "*************" << endl;
const double PI = acos(-1.0);
const double e = exp(1.0);
const int inf = 0x3f3f3f3f;
const int INF = 0x7fffffff;
typedef long long ll;

int n;
int vis[30];
int ans[30];

bool isPrime(int x) {
    for (int i = 2; i * i <= x; i ++) {
        if (x % i == 0) return false;
    }
    return true;
}

void dfs(int cur, int cnt) {
    if (cnt > n) {
        for (int i = 1; i <= n; i ++) printf("%d%c", ans[i], i == n ? ‘\n‘ : ‘ ‘);
        return ;
    }
    for (int i = 2; i <= n; i ++) {
        if (!vis[i] && isPrime(i + cur)) {
            if (cnt == n) {
                if (!isPrime(i + 1)) continue;
            }
            ans[cnt] = i;
            vis[i] = true;
            dfs(i, cnt + 1);
            vis[i] = false;
        }
    }
}

int main() {

    int _t = 1;
    while (~scanf("%d", &n)) {
        printf("Case %d:\n", _t ++);
        memset(vis, 0, sizeof(vis));
        vis[1] = true;
        ans[1] = 1;
        dfs(1, 2);
        printf("\n");
    }

    return 0;
}