HDOJ 1016 Prime Ring Problem (素数环问题)

最新推荐文章于 2020-07-29 21:27:22 发布

转载最新推荐文章于 2020-07-29 21:27:22 发布 · 84 阅读

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原文链接：http://www.cnblogs.com/biggan/p/7457789.html

本文介绍了一个关于构造由自然数组成的环形结构的问题，其中每个相邻节点的和必须为素数。通过使用递归回溯算法，文章提供了一种解决方案来生成所有可能满足条件的环形序列，并按字典序输出。

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

#include <iostream>
#include <cmath>
#include <cstring>
using namespace std;
int sushu[50];
int a[20],vis[20];

int check(int n)
{
int t=sqrt(n);
for(int i=2;i<=t;i++)  //小于等于开房数 
   { if(n%i==0) 
        return 0;
   }
return 1;   
}

void dp(int m,int k)  //k相当于步骤 
{
    if(k==m)          //k=m-1时也要算，加一 
      {
        if(sushu[a[m-1]+a[0]])
          {
              cout<<1;
            for(int i=1;i<m;i++)
                  cout<<' '<<a[i];
              cout<<endl;
          }    
        return;
      }
    for(int i=2;i<=m;i++)
       {
             if(!vis[i]&&sushu[a[k-1]+i])
                {
                a[k]=i;
                vis[i]=1;//三者顺序不能改
                dp(m,k+1);
                vis[i]=0; //不可放在括号外面 
             }
          
       }
}


int main()
{
a[0]=1;
memset(sushu,0,sizeof(sushu));
memset(vis,0,sizeof(vis));
for(int i=1;i<=32;i++)
   {
       sushu[i]=check(i);
   }
int n;
int num=0;
while(cin>>n)
    {
        num++;
        printf("Case %d:\n",num);
        dp(n,1);
        cout<<endl;
    }
return 0; 
}
 