HDU1016 Prime Ring Problem

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Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28668 Accepted Submission(s): 12753

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

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)

思路：首先对【1，n】之间的数进行深搜，这就相当于放扑克牌的意思，然后用一个数【j从0开始计数，当然也可以从其他的数，不过末值要变。】标记a[i]+a[I+1]是素数的次数，当次数满足n-1时输出一种情况，（因为i最大为n-1）；

代码：

#include<stdio.h>int a[20];             //n>0&&n<20;int book[20];           //标记；int n;int fun(int x)           用来判断一个数是否是素数；{int i;for(i=2;i<x;i++){if(x%i==0){return 0;          //如果不是，返回0；}}return 1;                         //如果是素数，返回1；}void dfs(int step)                            //开搜：{int i;if(step>2)                               //需要判断一下step是否大于2；因为i从2开始。如果不判断；step-2可能为0，出现问题；{if(fun(a[step-1]+a[step-2])==0)     //不满足，返回；return ;}if(step==n+1)                                     //如果值到达n，进行判断；{int j=0;for(i=1;i<n;i++)                          {if(fun(a[i]+a[i+1])==1)               //如果相邻的两个数相加是素数；j++;

j++;}if(j==(n-1)&&fun(a[1]+a[n])==1)           //如果j达到n-1（因为j从0 开始的），并且a【1】和啊【n】的和也是素数，（因为是环状嘛）{                                                     //开始输出。       printf("%d",a[1]);           for(i=2;i<=n;i++)   {         printf(" %d",a[i]);   }              printf("\n");}         return ;}for(i=2;i<=n;i++)                              //这里相当于放扑克牌的搜法；{if(book[i]==0){a[step]=i;book[i]=1;dfs(step+1);book[i]=0;}}return ;}int main()                                        //主函数{int k=1;                                        //用来输出第几组数据；while(scanf("%d",&n)!=EOF){a[1]=1;int i;for(i=1;i<=n;i++){book[i]=0;}printf("Case %d:\n",k);dfs(2);k++;printf("\n");                     //数据之间空行隔了两行；}return 0;}

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