hdu 1016 Prime Ring Problem

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 51048    Accepted Submission(s): 22596

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

 

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 2
1 6 7 4 3 8 5 2
题目大意：给你一个数字n，输出素数环（即环中任何两个数字之和是素数）。
题目解析：
这是一个简单的模拟题；
素数打表+DFS
代码：
#include<cstdio>#include<cstring>using namespace std;//判断是否为素数，是返回1，不是返回0int a[20],vis[20],isprime[45]={0},n;void get_prime(){int i,j;for(i=2;i<8;i++)    if(!isprime[i])    for(j=i*i;j<45;j+=i)    isprime[j]=1;}int dfs(int step){    int i;    if(step==n+1&&!isprime[a[n]+a[1]])//搜索结束条件    {        for(i=1;i<n;i++)        printf("%d ",a[i]);        printf("%d\n",a[n]);        return 0;    }    for(i=2;i<=n;i++)    {        if(!vis[i]&&!isprime[i+a[step-1]])        {            a[step]=i;            vis[i]=1;            dfs(step+1);            vis[i]=0;        }    }}int main(){    int cas=1;    a[1]=1;    get_prime();    while(~scanf("%d",&n))    {        memset(vis,0,sizeof(vis));        printf("Case %d:\n",cas++);        dfs(2);        printf("\n");    }    return 0;}