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
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
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;
}