hdu 1016 Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28064    Accepted Submission(s): 12487

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

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

 

Source

Asia 1996, Shanghai (Mainland China)

 

 1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstdlib> 5 #include<cstring> 6 using namespace std; 7 int prime[40],vis[40]; 8 int a[40],n; 9 10 int dfs(int x)11 {12     if(x==n && !prime[a[n]+a[1]])13     {14         for(int i=1;i<=n;i++)15         {16             if(i==1)17             printf("%d",a[i]);18             else19             printf(" %d",a[i]);20         }21         printf("\n");22     }23     else24     {25         for(int i=2;i<=n;i++)26         {27             if(!vis[i] && !prime[i+a[x]])28             {29                 vis[i]=1;30                 a[x+1]=i;31                 dfs(x+1);32                 vis[i]=0;33             }34         }35     }36 }37 38 int main()39 {40     int k=0,j;41     while(cin >> n)42     {43         memset(prime,0,sizeof(prime));44         memset(vis,0,sizeof(vis));45         k++;46 47         prime[1]=1;48         for(int i=2;i<=n*2;i++)49         {50             if(!prime[i])51             for(j=i+i;j<=n*2;j+=i)52             {53                 prime[j]=1;54             }55         }56         printf("Case %d:\n",k);57         a[1]=1;58         dfs(1);59         printf("\n");60     }61     return 0;62 }

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