HDU 1016 Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20806    Accepted Submission(s): 9300

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)

 

Recommend

JGShining

素数环要求，每两个数的和为素数，第一个和最后一个数也是素数。。。。。。这里我在写深搜集的时候 。没输出 。。纠结半天。。出来了。。。

#include<iostream>
#include<cstring>
using namespace std;
int prime[1000];
int a[21];
int vis[21];
void dfs(int n,int t)
{
	int i;
	if(t>n&&!prime[a[n]+1])
	{
		for(i=1;i<=n;i++)
			if(i<n)
				cout<<a[i]<<' ';
			else 
				cout<<a[i]<<endl;
			return;

	}
	for(i=2;i<=n;i++)
	{
		if(!vis[i]&&!prime[a[t-1]+i])
        {
             a[t]=i;
			 vis[i]=1;
			 dfs(n,t+1);
			 vis[i]=0;
		}
	}
}
int main()
{
	int n;	
	memset(prime,0,sizeof(prime));
    prime[0]=prime[1]=1;
    for(int i=2;i<1000;i++)
		if(!prime[i])
		{
			for(int j=i*2;j<1000;j+=i)
				 prime[j]=1;
		}
		i=1;
	while(cin>>n)
	{ 
		 
		cout<<"Case "<<i++<<':'<<endl;
		memset(vis,0,sizeof(vis));
		a[0]=a[1]=1;
		vis[0]=vis[1]=1;
		dfs(n,2);//从2开始
		cout<<endl;
	}
	return 0;
}

#include <iostream>
#include <string.h>
#include <math.h>
#include <stdio.h>
using namespace std;
bool prim[40];
int num[25];
bool used[25];
//筛法
void is_prim()
{
    memset(prim,0,sizeof(prim));
    prim[0]=prim[1]=1;
    int sq=sqrt((double)40);
    for(int i=2;i<sq;i++)
        for(int j=i*i;j<40;j+=i)
            prim[j]=1;
}
void dfs(int root,int n,int t)
{
    //数组从1开始，工n个数，当t>n时，说明n个数已排好序
    if(t>n&&!prim[num[n]+num[1]])
    {
        for(int j=1;j<n;j++)
            printf("%d ",num[j]);
        printf("%d\n",num[n]);
        return;
    }
    for(int i=1;i<=n;i++)
    {
        if(!prim[root+i]&&!used[i])
        {
            num[t]=i;
            used[i]=1;
            dfs(i,n,t+1);
            used[i]=0;
        }

    }
}
int main()
{
    is_prim();
    int n;
    int cas=1;
    while(scanf("%d",&n)!=EOF)
    {
        memset(used,0,sizeof(used));
        printf("Case %d:\n",cas++);
        num[1]=1;//第一个数一定是1
        used[1]=1;
        dfs(1,n,2);
        used[1]=0;
        printf("\n");

    }
    return 0;
}