HDU：Prime Ring Problem

Prime Ring Problem

Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 35   Accepted Submission(s) : 23

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 21 6 7 4 3 8 5 2

 

翻译：

有一个整数n，把从1到n的数字无重复的排列成环，且使每相邻两个数（包括首尾）的和都为素数，称为素数环。

为了简便起见，我们规定每个素数环都从1开始。

直接dfs就出来了。

源代码：

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int t,c[22];bool lu[22];//标记已用过的数bool su(int n)//判断素数{    for(int i=2;i<n;i++)        if(n%i==0)return false;    return true;}void DFS(int n){    if(t==n)    {        if(su(c[t-1]+1))        {            printf("%d",c[0]);            for(int i=1;i<t;i++)            {                printf(" %d",c[i]);            }            printf("\n");        }    }    for(int i=2;i<=t;i++)    {        if(lu[i]) continue;        if(su(i+c[n-1]))        {            lu[i]=1;            c[n]=i;            DFS(n+1);            lu[i]=0;        }    }}int main(){    int f=1;    while(cin>>t)    {        printf("Case %d:\n",f++);        memset(lu,0,sizeof(lu));        c[0]=1;        DFS(1);        printf("\n");    }    return 0;}