杭电 HDU 1016 Prime Ring Problem

http://acm.hdu.edu.cn/showproblem.php?pid=1016

Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19395    Accepted Submission(s): 8654

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

 

用的深搜

AC代码：

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int nub[50];int out[50];int n;void Dfs(int b,int s){    int now,next;    int i,sum,j,p;    now = b;       //当前第时在几个数    out[now] = s;  //把满足条件的数存到输出数组中 //   printf("b:%d s:%d\n",b,s);    if(b < n)      //判断数是否排满    {        for(i = 2; i <= n; i++)   //把还没有到的数加进去尝试一下        {            next = i;            if(nub[next] == 0)   //判断此数没有被用            {                p = 0;                sum = s + next;                for(j = 2; j < sum; j++) //判断和是否为素数                {                    if(sum%j==0)                    {                        break;                    }                    else                    {                        p++;                    }                }                if(p == sum-2)                {                    nub[next] = 1;                    Dfs(b+1,next);                    nub[next] = 0;                }            }        }    }    else    {        sum = out[1] + out[n];        p = 0;        for(j = 2; j < sum; j++)        {            if(sum%j==0)            {                break;            }            else            {                p++;            }        }        if(p == sum-2)        {            printf("%d",out[1]);            for(i = 2; i <= n; i++)            {                printf(" %d",out[i]);            }            printf("\n");        }        return ;    }}int main(){    int i,x,y;    i = 0;    while(scanf("%d",&n)!=EOF)    {        x = y = 1;        i++;        memset(nub,0,sizeof(nub));        nub[1] = 1;        printf("Case %d:\n",i);        Dfs(x,y);        printf("\n");    }    return 0;}