HDU-1016-Prime Ring Problem

Posted 2020-12-24 1625--h

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

 

 

分析：

1. 从1-n个数组以1为开头，构成一个数列，保证相邻两数的和为质数，并且首尾之和也同样满足，我们可以遍历整个数组来检测这个情况是否满足。

2.n的最大值是20，可想而知上述方法在n等于20的时候复杂度之高，是不可行的，我们要在过程中直接筛掉一部分，第一个可以想得到的就是，要是质数，必须是奇数，两数之和为奇数，所以n必须是偶数才可以。

3.在构成数列的时候，如果满足相邻两数之和不为奇数则直接判输。

4.依然要遍历每一种可能，只是我们不把数组一次性建好，而是填充一个元素，检测一下是否满足条件，满足则继续，不满足则判输。

5.当数组填充满了的时候，要检测首尾数字是否满足情况。

6.如果有一种情况满足则直接输出。

 

优化：n数值小，判断是否是质数可以直接用数组来保存。

 1 #include <iostream>
 2 using namespace std;
 3 int a[20];
 4 bool isprime[38] = {
 5     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
 6 };
 7 
 8 void print(int n)
 9 {
10     for (int i = 0; i < n; i++)
11     {
12         if (i==0)
13         printf("%d", a[i]);
14         else
15         {
16             printf(" %d", a[i]);
17         }
18     }
19     printf("
");
20 }
21 bool ok(int cur)
22 {
23     if (!isprime[a[cur] + a[cur - 1]])
24         return false;
25     for (int i = 1; i < cur; i++)
26     {
27         if (a[i] == a[cur])
28             return false;
29     }
30     return true;
31 }
32 void dfs(int n, int cur)
33 {
34     if (n & 1)
35         return;
36     if (cur == n)
37     {
38         if (isprime[1 + a[cur - 1]])
39             print(n);
40     }
41     else
42     {
43         for (int i = 2; i <= n; i++)
44         {
45             a[cur] = i;
46             if (ok(cur))
47                 dfs(n, cur + 1);
48         }
49     }
50 }
51 int main()
52 {
53     int n,k=0;
54     while (cin >> n)
55     {
56         k++;
57         printf("Case %d:
", k);
58         a[0] = 1;
59         dfs(n, 1);
60         printf("
");
61     }
62     return 0;
63 }