Prime Ring Problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 39923 Accepted Submission(s): 17610
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.

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 4Case 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
题目大概要义:
一个圈由n个数组成,自然数1,2……n,相邻的两个数总和应该是一个素数,最后一格与第一格相加为素数。
第一格为1。
输入:n
输出:格式如下,每一行代表在环圈数从1开始顺时针,以字典顺序打印解决方案。
实现C语言代码
#AC
#include<stdio.h>
int n;
int a[25];
int visit[25];
int sum=1;
int su(int x)
{
int i;
for(i=2;i*i<x+1;i++)
if(!(x%i)) return 0;
return 1;
}
int DFS(int x)
{
if(x==n&&su(a[n-1]+1))
{
for(int i=0;i<n-1;i++)
printf("%d ",a[i]);
printf("%d",a[n-1]);
printf("\n");
return 0;
}
for(int i=2;i<=n;i++) //此处的i代表圈内的自然数;
{
if(visit[i]==0)
{
if(su(a[x-1]+i))
{
visit[i]=1; //数i的使用;
a[x]=i;
DFS(x+1);
visit[i]=0; //释放数i;
}
}
}
}
int main()
{
while(scanf("%d",&n)!=EOF)
{
a[0]=1;
printf("Case %d:\n",sum++);
DFS(1);
printf("\n");
}
}