Prime Ring Problem（杭电1016）（DFS）

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28204    Accepted Submission(s): 12561

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 61 6 5 2 3 4Case 2:1 2 3 8 5 6 7 41 2 5 8 3 4 7 61 4 7 6 5 8 3 21 6 7 4 3 8 5 2

 

Source

Asia 1996, Shanghai (Mainland China)

#include<stdio.h>#include<string.h>#include<math.h>int visit[40],a[40],n;int fun(int x){for(int i=2;i<=sqrt(x);i++)    if(x%i==0)        return 0;return 1;}void dfs(int k){if(fun(a[0]+a[n-1])&&k==n){printf("%d",a[0]);for(int i=1;i<n;i++){printf(" %d",a[i]);}printf("\n");return;}for(int i=2;i<=n;i++){if((!visit[i])&&fun(i+a[k-1])){visit[i]=1;a[k]=i;dfs(k+1);visit[i]=0;}}}int main(){int kase=1;while(scanf("%d",&n)!=EOF){printf("Case %d:\n",kase++);memset(visit,0,sizeof(visit));a[0]=1;    dfs(1);    printf("\n");    }    return 0;}