Prime Ring Problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23458 Accepted Submission(s):
10465
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
这一题的意思是输入一个数n，然后让你用1~n的数围成一个圈，但是每相邻的两个数的和必须为素数。
下面是代码：
1 #include <iostream>
2 #include <cmath>
3 using namespace std;
4 #define MAX 22
5 int a[MAX]; //标记数组
6 int b[MAX];
7 int n;
8 bool prime(double n)
9 {
10 int i, m;
11 m = (int)sqrt(n);
12 for (i=2; i<=m; i++)
13 if ((int)n%i == 0)
14 return false;
15 return true;
16 }
17 void dfs(int i)
18 {
19 int j;
20 if (i>=n)
21 {
22 if (prime(double(b[n-1]+1))) //判断最后一个和第一个数的和是不是素数
23 {
24 cout<<b[0];
25 for (j=1; j<n; j++)
26 cout<<" "<<b[j];
27 cout<<endl;
28 }
29 }
30 else
31 {
32 for (j=2; j<=n; j++)
33 {
34 if (a[j] || !prime(double(b[i-1]+j))) //a[j]已经被加入到圈中或者相邻两个数和不是素数，则continue
35 continue;
36 a[j] = 1; //如果j已经加入圈中,则标记为1
37 b[i] = j;
38 dfs(i+1);
39 a[j] = 0;
40 }
41 }
42 }
43 int main()
44 {
45 int i=1, j;
46 while (cin>>n)
47 {
48 memset(a,0,sizeof(a)); //全置为0
49 cout<<"Case "<<i++<<":"<<endl;
50 b[0] = 1;
51 dfs(1);
52 cout<<endl;
53 }
54 return 0;
55 }