HDU 1016 Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 18300 Accepted Submission(s): 8189

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

Source

Asia 1996, Shanghai (Mainland China) 

/*    一道很经典的搜索基础题，运用了DFS的思想*/#include<cstdio>#include<cmath>#include<cstring>using namespace std;int n;int vis[25] = {0}, num[25] = {0};int prime(int k){for(int i = 2; i * i <= k; i++)if(k % i == 0) return 0;return 1;}void dfs(int k){int i;      //如果k大于了n  说明已经完成了dfs  是否输出 最后还要判断一下第一个数和最后一个数之和是否为素数if(k > n && prime(num[1]+num[n])) {for(i = 1; i < n; i++)printf("%d ", num[i]);printf("%d\n", num[i]);}//否则进行dfselse{    for(i = 2; i <= n; i++){      if(!vis[i] && prime(i+num[k-1]))   //判断该数是否被访问  并且该数与上一个num[]数组里面的数之和为素数  {  num[k] = i;     vis[i] = 1;   //设置已经访问  dfs(k+1);  vis[i] = 0;  //回头时  将该处重新设置为没有访问  }}}}int main (void){int t = 1;while(scanf("%d", &n)!=EOF){memset(vis,0,sizeof(vis));memset(num,0,sizeof(num));printf("Case %d:\n",t++);num[1] = 1;dfs(2);printf("\n");}return 0;}