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.
Inputn (0 < n < 20).
OutputThe 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.
OutputThe 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 代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
#include <cmath>
int n,ans[]={};
using namespace std;
int vis[];
bool isnum(int x)
{
if(x<=)return false;
for(int i=;i<=sqrt(x);i++)
{
if(x%i==)return false;
}
return true;
}
void dfs(int k,int last)
{
if(k==n)
{
if(!isnum(ans[]+ans[n-]))return;
printf("%d",ans[]);
for(int i=;i<n;i++)
printf(" %d",ans[i]);
putchar('\n');
return;
}
for(int i=;i<=n;i++)
if(!vis[i]&&isnum(last+i))
{
vis[i]=;
ans[k]=i;
dfs(k+,i);
vis[i]=;
}
}
int main()
{
int k=;
while(cin>>n)
{
vis[]=;
printf("Case %d:\n",++k);
dfs(,);
putchar('\n');
}
}