Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30395    Accepted Submission(s): 13525

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)

 

//第一次写dfs，既然发现和小白上的代码一样，开心
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int prime[]= {0,0,1,1,0,1,0,1,0,0,0,
              1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,
              0,0,1,0,1,0,0,0,0,0,1,0,0,0
             };//方便判断一个数是否素数
int flag[25];//标志数组
int c[25];//结果数组
int n;
void dfs(int cur)
{
    int i,j;
    if(cur==n+1&&prime[1+c[cur-1]])//别忘记第一个数和最后一个数的和也应该是素数
    {
        for(j=1; j<n; j++) printf("%d ",c[j]);
        printf("%d
",c[n]);
    }
    else for(i=2; i<=n; i++)//每个数都应是从2——n选择
            if(!flag[i]&&prime[i+c[cur-1]])//是否满足条件
            {
                c[cur]=i;//尝试把i作为第cur个满足条件的数
                flag[i]=1;//设置使用标志
                dfs(cur+1);
                flag[i]=0;//谨记清除标记
            }
}
int main()
{
    int k=1;
    while(~scanf("%d",&n))
    {
        printf("Case %d:
",k++);
        memset(flag,0,sizeof(flag));
        c[1]=1;
        dfs(2);//第一个数就是1，从第二个数开始
        printf("
");
    }
}
不知什么鬼，G++提交超时，c++ 256ms