Prime Ring Problem

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

Total Submission(s): 53931    Accepted Submission(s): 23862

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

 

 

直接dfs就可以了

 

 

 

#include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            using namespace std;int vis[25],n;bool prime(int k){    for(int i=2;i<=sqrt(k);i++)    {        if(k%i==0)        return false;    }    return true;}void dfs(int k){    if(vis[k]==n&&prime(k+1))    {        printf("1");        for(int i=2;i<=n;i++)        {            for(int j=1;j<=n;j++)            {                if(vis[j]==i)                printf(" %d",j);            }        }        printf("\n");        return ;    }    for(int i=1;i<=n;i++)    {        if(prime(i+k)&&!vis[i])        {            vis[i]=vis[k]+1;            dfs(i);            vis[i]=0;        }    }}int main() {    int Case=0;    while(~scanf("%d",&n))    {        printf("Case %d:\n",++Case);        memset(vis,0,sizeof(vis));        vis[1]=1;        dfs(1);        printf("\n");    }    return 0;}