Prime Ring Problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 29577 Accepted Submission(s): 13188
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
我的第一百题和进杭电前一万名，，全贡献给了水题o(?□?)o 竟然格式错误两次！！ 入门DFS。 下面代码：
#include
#include
#define MAX 25 bool visited[MAX] ; int n , ans[MAX] ; bool prime[2*MAX]; void DFS(int num) { if(num == n) { if(!prime[ans[num-1]+1]) return ; for(int i = 0 ; i < n ; ++i) { printf("%d",ans[i]); if(i != n-1) { printf(" "); } } printf("\n") ; return ; } else { for(int i = 1 ; i <= n ; ++i) { if(!visited[i] && prime[i+ans[num-1]]) { visited[i] = true ; ans[num] = i; DFS(num+1); visited[i] = false ; } } } } int main() { for(int i = 2 ; i < 51 ; ++i) prime[i] = true ; prime[0]=prime[1]=false; prime[2] = true ; for(int i = 2 ; i < 51; ++i) { for(int j = 2 ; j*i < 51 ; ++j) { prime[i*j] = false ; } } ans[0] = 1 ; int c=1; while(~scanf("%d",&n)) { memset(visited,0,sizeof(visited)); visited[1] = true ; printf("Case %d:\n",c++); DFS(1); printf("\n"); } return 0 ; }