joj 1195

1195: Prime Ring Problem

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ResultTIME LimitMEMORY LimitRun TimesAC TimesJUDGE10s8192K1973499Standard

A ring is composed of n circles as shown in diagram. Put natural numbers 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 <= 16)

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.

You are to write a program that completes above process.

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 <cmath>using namespace std;const int maxn = 33;int isp[maxn], n, a[10000],vis[10000];bool judge(int x){int i;for (i=2; i<=sqrt((double)x); i++){if (x % i == 0) return false;}return true;}void makeprime(){int i;for (i=2; i<=maxn; i++)if (judge(i)) isp[i] = 1;}void dfs(int cur){if (cur == n && isp[a[0]+a[n-1]]==1){int i;cout<<a[0];for (i=1;i<n; i++){cout<<" "<<a[i];}cout<<endl;}else{int i;for (i=2; i<=n; i++){if (!vis[i] && isp[i+a[cur-1]] == 1){a[cur] = i;vis[i] = 1;dfs(cur+1);vis[i] = 0;}}}}int main(){makeprime();int count=0;while (cin>>n){count++;cout<<"Case "<<count<<":"<<endl;if(n%2==0){memset (a,0,sizeof(a));memset (vis,0,sizeof(vis)); vis[1] = 1;a[0] = 1;dfs(1);}cout<<endl;}return 0;}