Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32872 Accepted Submission(s): 14544


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

Source Asia 1996, Shanghai (Mainland China)
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題意：要求1-n之間的所有數組成一個圓環，而且要求相鄰的兩個數相加的和是素數。 思路：因為n的數據范圍只是1-20，所以相加的和最大不會超過40，所以把40以內的素數標記出來。然後DFS深搜，符合題意的直接輸出就可以了。 、

代碼：

#include
#include
#include
#include
#include
#include

using namespace std;

int pv[50];
int n;
int p[50];
int v[50];

void getprime()
{
    for(int i=1;i<50;i++)
    {
        pv[i] = 1;
    }
    for(int i=1;i<=41;i++)
    {
       int pi = sqrt(i);
       for(int j=2;j<=pi;j++)
       {
           if(i%j == 0)
           {
               pv[i] = 0;
               break;
           }
       }
    }
}

void DFS(int cnt,int x)
{
    if(cnt == n && pv[p[n-1]+p[0]] == 1)
    {
        for(int i=0;i