hdu 1016 Prime Ring Problem

B - Prime Ring Problem
Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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<stdio.h>
using namespace std;
bool flag[100]={false};
int ans[25];
int n;
int vis[25]={0};
int t;
void dfs(int num)
{
    if(num==n&&(flag[ans[0]+ans[num-1]]==0))
    {
        for(int i=0;i<n-1;i++)
            printf("%d ",ans[i]);
        printf("%d\n",ans[n-1]);
    }
    else
    {
        if(num==n) return;
        for(int i=2;i<=n;i++)
        {
           // cout<<"i: "<<endl;
            if(!vis[i]&&(flag[ans[num-1]+i]==0))
            {
                //cout<<"i: "<<i<<endl;
                vis[i]=1;
                ans[num]=i;
                dfs(num+1);
                vis[i]=0;
            }
        }
    }
}
int main()
{
    ans[0]=1;
    for(int i=2;i<100;i++)
    {
        if(!flag[i])
        {
             for(int j=i<<1;j<100;j+=i)
                flag[j]=true;
        }

    }
     t=1;
   // for(int i=0;i<23;i++) cout<<flag[i]<<endl;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<=n;i++) vis[i]=0;

        printf("Case %d:\n",t++);
        dfs(1);
        printf("\n");
    }

}
View Code

 

 
posted @ 2016-07-28 09:49  超级学渣渣  阅读(174)  评论(0编辑  收藏  举报