题目：
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.
Inputn (0 < n < 20).
OutputThe 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
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
题意描述：
输入n (0 < n < 20)
计算并输出从1到n可以组成多少个素数环
解题思路：
属于搜索题，进行DFS搜索的时候进行筛选即可。
代码实现：
1 #include<stdio.h> 2 #include<string.h> 3 #include<math.h> 4 int r[30],book[30]; 5 void dfs(int step); 6 int n; 7 int isp(int n); 8 int main() 9 { 10 int c=1; 11 while(scanf("%d",&n) != EOF) 12 { 13 printf("Case %d: ",c++); 14 memset(book,0,sizeof(book)); 15 r[1]=1; 16 dfs(2); 17 printf(" "); 18 } 19 return 0; 20 } 21 void dfs(int step) 22 { 23 int i; 24 if(step==n+1) 25 { 26 printf("%d",r[1]); 27 for(i=2;i<=n;i++) 28 printf(" %d",r[i]); 29 printf(" "); 30 } 31 for(i=2;i<=n;i++) 32 { 33 if(book[i]==0) 34 { 35 if(step != n) 36 { 37 if(isp(r[step-1]+i)) 38 { 39 r[step]=i; 40 book[i]=1; 41 dfs(step+1); 42 43 book[i]=0; 44 } 45 else 46 continue; 47 } 48 else 49 { 50 if(isp(i+r[n-1])&&isp(1+i)) 51 { 52 r[step]=i; 53 book[i]=1; 54 dfs(step+1); 55 56 book[i]=0; 57 } 58 else 59 continue; 60 } 61 } 62 } 63 return ; 64 } 65 int isp(int n) 66 { 67 int i,k,flag=0;; 68 k=(int)sqrt(n); 69 for(i=2;i<=k;i++) 70 { 71 if(n % i==0) 72 { 73 flag=1; 74 break; 75 } 76 } 77 if(flag||n==1) 78 return 0; 79 else 80 return 1; 81 }
易错分析：
1、注意素数环，末尾的数不光跟前一个有关还和第一个数有关