Prime Ring Problem

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 

暴力做肯定超时，所以用深搜做。每次判断是否满足素数，如果满足继续搜下去，直到有n个数，同时把前40个数是否是素数保存起来可省不少时间

#include <iostream>#include <algorithm>#include <cstring>using namespace std;int a[21],n,b[21];//a数组用来存数，b数组用来判断数有没有用过bool isprime(int t){for(int i=2; i*i <= t; ++i){if(t%i==0)return false;}return true;}bool ans[41];void dfs(int k){if(k > n) return ;if(k==n){for(int i=0; i < n-1; ++i) cout << a[i] << " ";cout << a[n-1] << endl;return ;}for(int i=2; i <= n; ++i){if(b[i]==1) continue;if(k+1==n){if(ans[i+a[k-1]]&&ans[i+1]){a[k]=i;b[i]=1;dfs(k+1);b[i]=0;}}else{if(ans[i+a[k-1]]){a[k]=i;b[i]=1;dfs(k+1);b[i]=0;}}}}int main(){int k=1;for(int i=0; i < 41; ++i) ans[i]=isprime(i);while(cin >> n){cout << "Case " << k << ":" << endl;k++;memset(a,0,sizeof(a));memset(b,0,sizeof(b));a[0]=b[0]=1;dfs(1);cout << endl;}}