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
",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:
",t++);
        dfs(1);
        printf("
");
    }

}

View Code