Prime Ring Problem(回溯)

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

68

 

思路：经典题目。

Sample Output

Case 1:1 4 3 2 5 61 6 5 2 3 4Case 2:1 2 3 8 5 6 7 41 2 5 8 3 4 7 61 4 7 6 5 8 3 21 6 7 4 3 8 5 2

#include <stdio.h>#include <math.h>bool isprime[38],vis[20];int n,A[20];void trackback(int cur){    int i;    if(cur==n+1 && isprime[A[1]+A[n]])    {        for(i=1;i<=n;i++)        {            printf("%d",A[i]);            if(i<n) printf(" ");            else printf("\n");        }    }    else if(cur<=n)    {        for(i=2;i<=n;i++)        {            if(isprime[A[cur-1]+i] && !vis[i])            {                A[cur]=i;                vis[i]=1;                trackback(cur+1);                vis[i]=0;            }        }    }}int main(){    int i,j,count=1;    for(i=2;i<=37;i++)//初始化素数表    {        isprime[i]=1;        for(j=2;j<=sqrt(i);j++)        {            if(i%j==0)            {                isprime[i]=0;                break;            }        }    }    for(i=1;i<=19;i++) vis[i]=0;//标志数组清零    A[1]=1;    while(~scanf("%d",&n))    {        printf("Case %d:\n",count++);        trackback(2);        printf("\n");    }}