HDU1016:Prime Ring Problem(DFS)

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
素数筛选法+深搜上代码：
#include<iostream>#include<cstring>using namespace std;int cnt;int a[22];int book[22];int prime[50];int n;int Case;void isPrime(){for(int i=1;i<50;i++)prime[i]=1;prime[0]=prime[1]=0;for(int i=2;i<50;i++){if(prime[i]){for(int j=2*i;j<50;j+=i)prime[j]=0;}} }void dfs(int step){if(step==n+1&&prime[a[n]+1]){cnt++;//cout<<"Case "<<Case<<":"<<endl;for(int i=1;i<n;i++){cout<<a[i]<<" ";}cout<<a[n]<<endl;}for(int i=2;i<=n;i++){if(!book[i]&&prime[i+a[step-1]]){a[step]=i;book[i]=1;dfs(step+1);book[i]=0;}}return ;}int main(){isPrime();while(cin>>n){memset(book,0,sizeof(book));Case++;cout<<"Case "<<Case<<":"<<endl;a[1]=1;dfs(2);cout<<endl;}return 0;}