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524 Prime Ring Problem
A ring is composed of n (even number) circles as shown in diagram. Put
natural numbers 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 16)
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.
You are to write a program that completes above process.
Sample Input
68
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：
给出一个偶数N，需要将1到N这N个数等距排列在一个圆周上，使得相邻两个数的和总是一个素数，将所有的解（镜面对称的两个解视作不同解）都按顺时针顺序打印出来。
输入：
正整数N。
输出：
所有解。
分析：
使用深搜的方式进行枚举即可。从1开始打印解。
1 #include<iostream> 2 #include<cstdio> 3 #define MAX_N 50 4 using namespace std; 5 int n,A[MAX_N] = {1},ispe[MAX_N],vis[MAX_N]; 6 void dfs(int cur){ 7 if(cur == n && ispe[A[0] + A[n - 1]]){ 8 for(int i = 0; i < n; i++) 9 i ? printf(" %d", A[i]) : printf("%d", A[i]); 10 printf("\n"); 11 }else for(int i = 2; i <= n; i++) 12 if(!vis[i]&& ispe[i + A[cur - 1]]){ 13 A[cur] = i; 14 vis[i] = 1; 15 dfs(cur + 1); 16 vis[i] = 0; 17 } 18 } 19 int main(){ 20 for(int i = 2; i <= 50; i++) ispe[i] = 1; 21 for(int i = 2; i <= 50; i++) 22 for(int j = i + i; j + i <= 50; j += i) 23 ispe[j] = 0; 24 int kase = 0; 25 while(cin >> n){ 26 if(kase++) 27 printf("\n"); 28 printf("Case %d:\n",kase); 29 dfs(1); 30 } 31 return 0; 32 }
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原文地址：http://www.cnblogs.com/cyb123456/p/5792815.html
