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): 53427    Accepted Submission(s): 23627

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

#include <stdio.h>int a[100],b[100];          //全局变量的数组默认初始化为0 int n;int prime(int k){ int f=0;   if(k==1)  return 1;for(int i=2; i<=k-1; i++){if(k%i==0)  f=1;}return f;}void dfs(int x){if(x==n)                       //当搜索到最后一个数时 {if(prime(a[x]+1)==0)        //判断这个数与1的和是否为素数 {int i; printf("1");for( i=2; i<=n; i++){printf(" %d",a[i]);   //如果是的话则从1开始输出排好序的数组 }      printf("\n");    }return;}for(int j=2; j<=n; j++)         {if(b[j]==0 && prime(a[x]+j)==0)  //如果找到一个未被标记的数并满足其与前一个数之和为素数 {b[j]=1;                  //标记这个数并从这个数开始进行下一轮搜索 a[x+1]=j;                  dfs(x+1);b[j]=0;                  //回溯时重新标记为0 }}return;}int main(){int y=1;while(scanf("%d",&n)!=EOF){printf("Case %d:\n",y++);a[1]=1;b[1]=1;                       //b数组用来对1之后的每个数进行标记，如果搜索一次后标记为1，回溯时再标记为0 dfs(1);printf("\n");}return 0;}