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Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 20363    Accepted Submission(s): 9115

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)

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <cassert>
#include <set>
#include <sstream>
#include <map>
using namespace std ;
#ifdef DeBUG
#define bug assert
#else
#define bug //
#endif
#define zero {0}
#define INF 2000000000
#define eps 1e-6
int cnt=1;
int a[30];//记录序列 
int used[30];//记录i是否使用 
int prime[]={
0,1,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,
1,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,
0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0
};
int n;
void print()//打印数组 
{
    for(int i=1;i<n;i++)
    printf("%d ",a[i]);
    printf("%d
",a[n]);
}
void dfs(int depth)
{
    if(depth==n+1)
    {
        if(prime[a[n]+1])//判断最后一个元素与第一个元素是否为素数 
        {
            print();
        }
        return;
    }
    else
    {
        for(int i=1;i<=n;i++)
        {
            if(!used[i]&&prime[a[depth-1]+i])
            {
                a[depth]=i;
                used[i]=true;
            //    print();
                dfs(depth+1);
                used[i]=false;
                a[depth]=0;
            }
        }
    }
}
int main()
{


    #ifdef DeBUGs 
        freopen("C:\Users\Sky\Desktop\1.in","r",stdin);
//    freopen("C:\Users\Sky\Desktop\打表.txt","w",stdout);//PLEASE DELETE IT!!!!!!!!!!!!!!!!!!!!!!!!
    #endif
    while(scanf("%d",&n)+1)
    {
        memset(a,0,sizeof(a));
        memset(used,0,sizeof(used));
        printf("Case %d:
",cnt++);
        a[1]=1;
        used[1]=1;
        dfs(2);
        printf("
");
    }
    return 0;
}