HDU Prime Ring Problem （DFS+素数打表）

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

 

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int n,vis[40],ans[40],prime[100];void init_prime(){int i,j;for(i=2;i<100;i++) {if(!prime[i]){for(j=2*i;j<=100;j+=i) prime[j]=1;}}}void dfs(int num,int cnt){int i;if(cnt==n) {for(i=0;i<n;i++) {if(i==0) printf("%d",ans[i]);else printf(" %d",ans[i]);}printf("\n");}for(i=2;i<=n;i++) {if(vis[i]==1) continue;if(prime[i+num]==0) {if(cnt==n-1 && prime[i+1]) continue;vis[i]=1;ans[cnt]=i;dfs(i,cnt+1);vis[i]=0; }}}int main(){int number,i,j;number=0;init_prime();while(scanf("%d",&n)!=EOF){printf("Case %d:\n",++number);memset(vis,0,sizeof(vis));ans[0]=1;vis[1]=1;dfs(1,1);printf("\n");}return 0;}