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Description
A ring is composed of n (even number) circles as shown in diagram. Put natural numbers 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.
You are to write a program that completes above process.
6 8
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
因为起点已经确定为1；
只要用dfs回溯就行；
同是要注意使用vis[]数组判重；
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 6 int is_prime(int x) { 7 for(int i = 2; i*i <= x; i++) 8 if(x % i == 0) return 0; 9 return 1; 10 } 11 12 int n, A[50]; 13 bool isp[50],vis[50]; 14 void dfs(int cur) { 15 if(cur == n && isp[A[0]+A[n-1]]){ 16 for(int i = 0; i < n; i++){ 17 if(i != 0) printf(" "); 18 printf("%d", A[i]); 19 } 20 printf("\n"); 21 } 22 else for(int i = 2; i <= n; i++) 23 if(!vis[i] && isp[i+A[cur-1]]){ 24 A[cur] = i; 25 vis[i] = 1; 26 dfs(cur+1); 27 vis[i] = 0; 28 } 29 } 30 31 int main() { 32 int kase = 0; 33 while(scanf("%d", &n) == 1 && n > 0){ 34 if(kase > 0) printf("\n"); 35 printf("Case %d:\n", ++kase); 36 for(int i = 2; i <= n*2; i++) isp[i] = is_prime(i); 37 memset(vis, 0, sizeof(vis)); 38 A[0] = 1; 39 dfs(1); 40 } 41 return 0; 42 }
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原文地址：http://www.cnblogs.com/demodemo/p/4690596.html
