HDUOJ-1016-Prime Ring Problem

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)

思路：循环加和，递归求解

代码：

#include<iostream>
#include<math.h>
#include<string.h>
using namespace std;
int vis[50];//标记访问数组
int ans[50];//定义循环数组
int N;
bool is_prime(int n)//判断是否为素数，可以使用打表的方式
{
    int i, j;
    if (n < 2)
    {
        return false;
    }
    for (i = 2; i <= sqrt(n); i++)
    {
        if (n%i== 0)
        {
            return false;
        }
    }
    return true;
}
void out_put()//输出满足循环和为素数
{
    int i;
    for (i = 1; i < N; i++)
    {
        cout << ans[i] << " ";
    }
    cout << ans[i] << endl;
}
void DFS(int step)//深度优先  递归调用
{
    if (step == N&&is_prime(ans[N] + 1))//到达递归出口
    {
        out_put();
        return;
    }
    else
    {
        for (int i = 2; i <= N; i++)
        {
            if (vis[i] == 0 && is_prime(ans[step] + i))//未访问，并且相邻加和为1
            {
                ans[step + 1] = i;//赋值给输出数组
                vis[i] = 1;//标记为访问
                DFS(step + 1);//深度递归下去
                vis[i] = 0;//标记为未访问
            }
        }
    }
}
int main()
{
    int t = 1;
    while (cin >> N)
    {
        //cout << is_prime(N) << endl;
        memset(vis,0, sizeof(vis));//访问数组初始化
        ans[1] = 1;
        cout << "Case " << t <<":"<< endl;//注意:输出一定要按照题目所给出的格式进行，否则报错
        DFS(1);
        t++;
        cout << endl;
    }
    return 0;
}