Prime Ring Problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 25156    Accepted Submission(s): 11234

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
</pre><pre name="code" class="cpp">#include<cstdio>
#include<stdlib.h>
#include<cstring>
#include<algorithm>
#include<iostream> using namespace std; int date[25]= {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20};
int ans[25];
int n, cnt;
int flag; void dfs(int step)
{
int temp;
if(step==n) //深搜究竟时注意推断首尾是否符合要求
{
flag=1;
temp = ans[0]+ans[n-1];
if(flag)
{
for(int i=2; i*i<=temp; i++)
if(temp%i==0)
{
flag=0;
break;
}
}
if(flag)
{
printf("%d", ans[0]);
for(int i=1; i<n; i++)
printf(" %d", ans[i]);
printf("\n");
}
}
else
{
for(int i=1; i<n; i++)
{
if(date[i])
{
temp=ans[cnt-1]+date[i];
flag=1;
for(int j=2; j*j<=temp; j++)
{
if(temp%j==0)
{
flag=0;
break;
}
}
if(flag)
{
ans[cnt++]=date[i];
date[i]=0;
dfs(step+1);
date[i]=ans[--cnt]; //记得将数据的还原处理
ans[cnt]=0;
}
}
}
}
} int main()
{
int count=1;
while(scanf("%d", &n)!=EOF)
{
memset(ans, 0, sizeof(ans));
cnt=1;
ans[0]=1;
printf("Case %d:\n",count++);
dfs(1); //从1開始进入搜索
printf("\n"); //每次输出后记得加个换行
} return 0;
}

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