Prime Ring Problem

Prime Ring Problem

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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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<algorithm>
#include<stack>
#include<cstring>
using namespace std;
int mark[21];
int prinum[]={2,3,5,7,11,13,17,19,23,29,31,37};
int n,rlt[21];
int isprinum(int nn)
{
    for(int i=0;i<12;i++)
    {
        if(nn==prinum[i])
            return 1;
    }
    return 0;
}
int dfs(int a, int b, int flag)
{
    if(!isprinum(a+b))
        return 0;
    rlt[flag]=b;
    if(flag==n && isprinum(b+1))
    {
        for(int i=1;i<n;i++)
        {
            cout<<rlt[i]<<" ";
        }
        cout<<rlt[n]<<endl;
        return 1;
    }
    mark[b]=1;
    int i;
    for(i=2;i<=n;i++)
    {
        if(mark[i]==0 && dfs(b,i,flag+1))
            break;
    }
    mark[b]=0;//？为什么在这里要还原
    return 0;
    
}

int main()
{
    int tt=1;
    rlt[1]=1;
    memset(mark,0,sizeof(mark));
    while(cin>>n && n>0 && n<20)
    {
        printf("Case %d:\n",tt++);
        if(n==1)
        {
            printf("1\n");
        }
        else
        {
            for(int i=2;i<=n;i++)
            {
                dfs(1,i,2);
            }
        }
        cout<<endl;
    }
    return 0;
    
}