Prime Ring Problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34108 Accepted Submission(s): 15098
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.
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.
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
Source
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RE:素数环， 字典序输出；
1 #include <cmath> 2 #include <cstdio> 3 #include <cstring> 4 #include <iostream> 5 using namespace std; 6 int dis[30], vis[30], my[30]; 7 int n; 8 int is_prime(int a) //判断素数； 9 { 10 11 if(a == 1 || a == 0) 12 return 0; 13 if(a == 2) 14 return 1; 15 else 16 { 17 for(int i=2; i <= sqrt(a); i++) //不要忘记等号； 18 { 19 if(a % i == 0) 20 return 0; 21 } 22 return 1; 23 } 24 } 25 void Dfs(int a) 26 { 27 int i; 28 if(a == n && is_prime(dis[n-1] + dis[0])) 29 { 30 for(i = 0; i < n; i++) 31 { 32 if(i > 0) 33 printf(" "); 34 printf("%d", dis[i]); 35 } 36 printf(" "); 37 return; //子循环结束； 38 } 39 for(i=2; i<=n; i++) 40 { 41 if(!vis[i] && is_prime(dis[a - 1] + i)) 42 { 43 vis[i] = 1; 44 dis[a] = i; 45 Dfs(a + 1); 46 vis[i] = 0; 47 } 48 } 49 } 50 int main() 51 { 52 int i, j = 1; 53 while(~scanf("%d", &n)) 54 { 55 memset(vis, 0 , sizeof(vis)); 56 printf("Case %d: ", j++); 57 vis[0] = 1; 58 dis[0] = 1; 59 Dfs(1); 60 printf(" "); 61 } 62 return 0; 63 }