4.3.2 Prime Ring Problem

 

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)

 

Recommend

JGShining

 思路：dfs，我是预处理，将每一个数可以与之匹配的质数先找出来，再dfs

#include <cstdio>
#include <cstring>
using namespace std;
const int b[]={2,3,5,7,11,13,17,19,23,29,31,37};
int a[22],x[22][10],c[22];
int n,t;
bool prime[40],f[40];

bool judge(int k)
{
    if (prime[k]) return true;
    return false;
}

void w(int k)
{
    prime[k]=true;
}

void dfs(int k)
{
int t;
if (n==k)
{
//    printf("a[k]:%d a[k-1]:%d k:%d\n",a[k],a[k-1],k);
    if (judge(a[k-1]+1))
    {
        for (int i=0;i<n-1;i++)
            printf("%d ",a[i]);
        printf("%d\n",a[n-1]);
    //    printf("\n");
   }
    return;
}
for (int i=0;i<c[a[k-1]];i++)
{
    t=x[a[k-1]][i];
    if (t>n) break;
    if (not f[t])
    {
        f[t]=true;
        a[k]=t;
        dfs(k+1);
       a[k]=0;
        f[t]=false;
    }
}
}

void init()
{
memset(prime,false,sizeof(prime));
memset(c,false,sizeof(c));
for (int i=0;i<11;i++)
    w(b[i]);
//for (int i=0;i<41;i++)
//    if (prime[i]) printf("%d \n",i);

for (int i=0;i<20;i++)
    for (int j=1;j<20;j++)
        if (i!=j && judge(i+j))
        {
            x[i][c[i]]=j;
            c[i]++;
        }
int cnt=0;
   while (scanf("%d",&n)!=EOF)
    {
        cnt++;
        printf("Case %d:\n",cnt);
        memset(f,false,sizeof(f));
        f[1]=true;
        a[0]=1;
       dfs(1);
        printf("\n");
    }
}

int main ()
{
init();
return 0;
}