Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 34799    Accepted Submission(s): 15411

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

 

 

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这是我第一道接触的搜索题，当时参考学长的代码写，却始终不能理解代码的意思，于是就一遍一遍第敲，直到整个代码都背下来了，后来接触的搜索逐渐增加，终于理解了这道题的代码，每次看到这道题，就能想起当年苦逼的敲代码。。。

 

题意：输入一个数n，代表从1到n个数，排序排成一个环，使得相邻两个数之和为素数（第一个和最后一个数也要满足），如果有多组情况，必须按从小到大的顺序输出，

 

附上代码:

 

1 #include 
      
        2 using namespace std; 3 int a[30]= {
    0,1},b[30],sushu[50],n;  //a表示素数环中的数字，b用来标记，sushu标记是否是素数 4 void DFS(int s,int k)   //s代表当前循环到第几个数，k标记此时是否是第一次进人深搜函数 5 { 6     int i,j; 7     if(s>n) 8     { 9         if(sushu[a[1]+a[n]])   //最后一个数字和第一个数字的和10         {11             for(i=1; i<=n; i++)  //全部满足，输出这组数据的全部数12             {13                 if(i>1) cout<<" ";14                 cout<
       
        >n)45     {46         cout<<"Case "<
        
         <<":"<