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
题意：给你一个n，得到一个数组，1~n；让你写出一个素数环，要求相邻两个数（顺逆时针）和是素数，输出的时候第一个永远是一；
解题思路：按照递归求全排列的思想，从第2个开始递归一直到最后一位，并且首尾也是素数才算是搜索完成；
感悟：一开始我还以为得剪纸，但是只有20个数，一次就过了；
代码：
#include #include #include #include #define maxn 25 using namespace std; int ans[maxn],visit[maxn],n; int Prime(int a) { for(int i=2;i<=sqrt(a);i++) if(a%i==0) return 0; return 1; } void dfs(int cur) { if(cur==n&&Prime(ans[0]+ans[n-1]))//递归到最后一位，并且首尾也能是素数 { for(int i=0;i printf("%d ",ans[i]); printf("%d\n",ans[n-1]); } else { for(int i=2;i<=n;i++) { if(!visit[i]&&Prime(i+ans[cur-1]))//这个数没用过并且相邻的是素数 { ans[cur]=i; visit[i]=1; dfs(cur+1); visit[i]=0; } } } } int main() { //freopen("in.txt", "r", stdin); int s=1; memset(visit,0,sizeof(visit)); while(~scanf("%d",&n)&&n) { for(int i=0;i ans[i]=i+1; printf("Case %d:\n",s++); dfs(1); printf("\n"); } }