UVa524

524 Prime Ring Problem
A ring is composed of n (even number) circles as shown in diagram. Put
natural numbers 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  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
68
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

题意：
       给出一个偶数N，需要将1到N这N个数等距排列在一个圆周上，使得相邻两个数的和总是一个素数，将所有的解（镜面对称的两个解视作不同解）都按顺时针顺序打印出来。

输入：

       正整数N。

输出：

       所有解。

分析：

使用深搜的方式进行枚举即可。从1开始打印解。

#include<iostream>
#include<cstdio>
#define MAX_N 50
using namespace std;
int n,A[MAX_N] = {1},ispe[MAX_N],vis[MAX_N];
void dfs(int cur){
    if(cur == n && ispe[A[0] + A[n - 1]]){
        for(int i = 0; i < n; i++)
            i ? printf(" %d", A[i]) : printf("%d", A[i]);
        printf("
");
    }else for(int i = 2; i <= n; i++)
        if(!vis[i]&& ispe[i + A[cur - 1]]){
            A[cur] = i;
            vis[i] = 1;
            dfs(cur + 1);
            vis[i] = 0;
        }
}
int main(){
    for(int i = 2; i <= 50; i++) ispe[i] = 1;
    for(int i = 2; i <= 50; i++)
        for(int j = i + i; j + i <= 50; j += i)
            ispe[j] = 0;
    int kase = 0;
    while(cin >> n){
        if(kase++)
            printf("
");
        printf("Case %d:
",kase);
        dfs(1);
    }
    return 0;
}