# Prime Ring Problem **Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 37427    Accepted Submission(s): 16521** 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 ~~~ 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 ~~~ 1.自然数（natural number），是非负整数（1, 2, 3, 4……）。认为自然数不包含[零](http://baike.baidu.com/subview/45676/8420704.htm)的其中一个理由是因为人们在开始学习数字的时候是由“一、二、三...”开始，而不是由“零、一、二、三...”开始, 因为这样是非常不自然的。在全球范围内，目前针对0是否属于自然数的争论依旧存在。在中国大陆，2000年左右之前的中小学教材一般不将0列入自然数之内，或称其属于“扩大的自然数列”。在2000年左右之后的新版中小学教材中，普遍将0列入自然数。 2.代码： ~~~ #include<cstdio> #include<cstring> using namespace std; int n; int vis[30]; int a[30]; int p[20]={2,3,5,7,11,13,17,19,23,29,31,37,41}; int c=1; int cnt; bool isPrime(int x) { for(int i=0;i<13;i++) { if(x==p[i]) { return 1; } } return 0; } void dfs(int level,int num) { a[level]=num; vis[num]=1; if(level==n) { if(isPrime(a[n]+a[1])) { if(cnt==0) printf("Case %d:\n",c++); cnt++; for(int i=1;i<=n;i++) { if(i==1) printf("%d",a[i]); else printf(" %d",a[i]); } printf("\n"); } return; } for(int i=1;i<=n;i++) { if(vis[i]==1||isPrime(i+a[level])==0) continue; dfs(++level,i); level--; vis[i]=0; } } int main() { while(scanf("%d",&n)==1) { memset(vis,0,sizeof(vis)); cnt=0; dfs(1,1); printf("\n"); } return 0; } ~~~