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.
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.
You are to write a program that completes above process.
Print a blank line after each case.
Sample Input
6 8
Sample Output
Case 1: 1 4 3 2 5 6 1 6 5 2 3 4 Case 2: 1 2 3 8 5 6 7 4 1 2 5 8 3 4 7 6 1 4 7 6 5 8 3 2 1 6 7 4 3 8 5 2
import java.util.Scanner; public class Main { static int n; static int arr[] = new int[16]; static boolean vis[] = new boolean[16]; //记录是否使用该数字 static int N=0; static boolean falg = true; public static void main(String[] args) { Scanner cin = new Scanner(System.in); while(cin.hasNextInt()){ n = cin.nextInt(); N++; arr = new int[16]; for (int i = 2; i <= n; i++) { vis[i] = true; //设置为每一个数都未使用过 } arr[0] = 1; falg = true; DFS(1); System.out.println(); } } private static void DFS(int cur) { if(cur==n && judge(arr[0]+arr[cur-1])){ if(falg){ System.out.println("Case "+N+":"); falg = false; } for (int i = 0; i < n-1; i++) { System.out.print(arr[i]+" "); } System.out.print(arr[n-1]); System.out.println(); }else{ for (int i = 2; i <= n; i++) { if(vis[i] && judge(arr[cur-1]+i)){ //判断此时的数是否被使用了 并且判断该数字相加是否为素数 //System.out.println(" cur"+cur+ " "+i); arr[cur] = i; vis[i] = false; //标记此时这个数已经被使用 DFS(cur+1); vis[i] = true; //取消标记 供下一种情况使用 } } } } //判断此时相加的数是否为素数 private static boolean judge(int num) { int i; for (i = 2; i <=num; i++) { if(num%i==0) break; } if(i>=num){ return true; }else{ return false; } } }