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

素数环，要注意第一个和最后一个也要为素数，因为n为20以内的数，所以可以直接打表写出40以内的所有素数，开头是1，然后就分别判断相邻两个数和是否为素数，并且是否都没用过即可。

#include<cstdio>#include<iostream>#include<algorithm>#include<cmath>#include<string.h>using namespace std;int sushu[13]={2,3,5,7,11,13,17,19,23,29,31,37};int n;int a[22];int b[22];int su(int x,int y){int sum=x+y;for(int i=0;i<12;i++){if(sum==sushu[i])return 1;}return 0;}void dg(int m,int s,int a[]){a[s]=m;if(s==n&&su(a[1],a[n])){for(int i=1;i<n;i++)printf("%d ",a[i]);printf("%d\n",a[n]);return;}for(int i=1;i<=n;i++){int flag=0;for(int j=1;j<=s;j++){if(i==a[j])flag=1;}if(!flag&&su(m,i)){dg(i,s+1,a);}}return;}int main(){int k=1;while(~scanf("%d",&n)){printf("Case %d:\n",k++);int s=1;memset(a,0,sizeof(a));dg(1,s,a);printf("\n");}return 0;}//2 3 5 7 11 13 17 19 23 29 31 37 41