hdu 1016Prime Ring Problem

http://acm.hdu.edu.cn/showproblem.php?pid=1016

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13648    Accepted Submission(s): 6220

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<stdio.h>
#include<string.h>
int a[22];
int vis[22];
int n;
int prime(int x)
{
    int i;
    int flag=1;
    if(x>=2)
    {
        if(x==2) return 1;
        else
        {
            for(i=2;i*i<=x;i++)
            {
                if(x%i==0)
                {
                    flag=0;
                    break;
                }
            }
            if(flag) return 1;
            else return 0;
        }
    }
    else return 0;
}

void dfs(int x)
{
    int i;
    if(x==n&&prime(a[0]+a[n-1]))
    {
        printf("1");
        for(i=1;i<n;i++)
        printf(" %d",a[i]);
        printf("\n");
    }
    else
    {
        for(i=2;i<=n;i++)
        {
            if(!vis[i]&&prime(a[x-1]+i))
            {
                vis[i]=1;
                a[x]=i;
                dfs(x+1);
                vis[i]=0;
            }
        }
    }
    
}
int main()
{
    int count=0;
    while(~scanf("%d",&n))
    {
        count++;
        memset(vis,0,sizeof(vis));
        a[0]=1;
        printf("Case %d:\n",count);
        dfs(1);
        printf("\n");
    }
}

数据小，列举出来：

#include<stdio.h>
#include<string.h>
int a[22];
int vis[22];
int n;
int prime[38] = {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};
void dfs(int x)
{
    int i;
    if(x==n&&prime[a[0]+a[n-1]])
    {
        printf("1");
        for(i=1;i<n;i++)
        printf(" %d",a[i]);
        printf("\n");
    }
    else
    {
        for(i=2;i<=n;i++)
        {
            if(!vis[i]&&prime[a[x-1]+i])
            {
                vis[i]=1;
                a[x]=i;
                dfs(x+1);
                vis[i]=0;
            }
        }
    }
    
}
int main()
{
    int count=0;
    while(~scanf("%d",&n))
    {
        count++;
        memset(vis,0,sizeof(vis));
        a[0]=1;
        printf("Case %d:\n",count);
        dfs(1);
        printf("\n");
    }
}