Prime Ring Problem

Prime Ring Problem

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 27   Accepted Submission(s) : 10

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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）
输出素数环，每两个相邻的数的和必须是素数，首尾两个数相邻，输出的第一个数是1；
需要注意的数，每一个样例直接需要输出一个空行；
一年后再去做这一题，觉得真的是水题了的，觉得自己以前怎么写得那么复杂的。。。
  
 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 #include <string.h>
 4 int num[25];
 5 int depend(int num[],int sum)
 6 {
 7     int jj,sign;
 8     for(jj=2,sign=0;jj*jj<=sum;jj++)
 9         if(sum%jj==0)sign++;
10     return sign;
11 }
12 
13 int depend2(int num[],int N)
14 {
15     int jj=0,sum[25];
16     memset(sum,0,sizeof(sum));
17     for(jj=0;jj<N;jj++)
18     {
19         sum[num[jj]]+=1;
20             if(sum[num[jj]]>1)
21                 return 0;
22     }
23     return 1;
24 }
25 
26 void DFS(int num[],int i,int N)
27 {
28     int flat,sum,ii,j;
29     if(i==N-1)
30     {
31         sum=num[i]+num[i-1];
32         flat=depend(num,sum);
33         if(flat!=0)return ;
34         sum=num[0]+num[N-1];
35         flat=depend(num,sum);
36         if(flat!=0)return ;
37         for(ii=0;ii<N;ii++)
38         {
39             printf("%d",num[ii]);
40             if(ii!=N-1)
41                 putchar(' ');
42         }
43         putchar('\n');
44         return;
45     }
46     if(i>0&&i<N)
47     {
48         sum=num[i]+num[i-1];
49         flat=depend(num,sum);
50         if(flat!=0)return ;
51     }
52     for(j=1,flat=1;j<=N;j++)
53     {
54         num[i+1]=j;
55         flat=depend2(num,i+2);
56             if(flat!=1)continue;
57         DFS(num,i+1,N);
58     }
59 }
60 
61 int main()
62 {
63     int N,times,i,j;
64     times=1;
65     while(scanf("%d",&N)!=EOF)
66     {
67         memset(num,0,sizeof(num));
68         printf("Case %d:\n",times++);
69         num[0]=1;
70         DFS(num,0,N);
71         putchar('\n');
72     }
73     return 0;
74 
75 }
View Code

 修改后：2015.5.8

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <stdlib.h>
 4 using namespace std;
 5 int N;
 6 int ID[100];
 7 int Num[100];
 8 int Pr[42]={0,1,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0};
 9 int Jude(int sum)
10 {
11     int i;
12     for(i=2;i*i<=sum;i++)
13     {
14         if(sum%i==0)return 0;
15     }
16     return 1;
17 }
18 void Put()
19 {
20     for(int i=0;i<N;i++)
21     {
22         printf("%d",Num[i]);
23         if(i!=N-1)putchar(' ');
24     }putchar(10);
25 }
26 void DFS(int d)
27 {
28     int i,sum;
29     if(d==N)
30     {
31         sum=Num[d-1]+1;
32         if(Pr[sum])
33         {
34             Put();
35         }
36         return ;
37     }
38     for(i=2;i<=N;i++)
39     {
40         if(ID[i])
41         {
42             sum=Num[d-1]+i;
43             if(Pr[sum])
44             {
45                 Num[d]=i;ID[i]=0;
46                 DFS(d+1);ID[i]=1;
47             }
48         }
49     }
50     return ;
51 }
52 int main()
53 {
54     int i,j=1;
55     while(scanf("%d",&N)!=EOF)
56     {
57         for(i=1;i<=N;i++)ID[i]=1;
58         printf("Case %d:\n",j++);
59         if(N%2==0)
60         {Num[0]=1;ID[1]=0;DFS(1);}
61         putchar(10);
62     }
63     return 0;
64 }
View Code

 

posted @ 2014-05-24 20:22  Wurq  阅读(153)  评论(0)  编辑  收藏  举报