9.I - DFS剪枝作业题

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

Inputcopy	Outputcopy
6 8	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

#include<bits/stdc++.h>
#define endl '\n'
#define ll long long
#define int ll
using namespace std;
const int N=1e6+7;
const int M=2e3+7;
int prime[N];
int n,sum=0;
bool vis[N];
int a[N];
bool isprime(int x)
{
	if(x==1) return 0;
	if(x==2) return 1;
	for(int i=2;i*i<=x;i++)
	{
		if(x%i==0) return 0;
	}
	return 1;
}
void dfs(int x)
{
	if(x==n+1&&isprime(a[n]+a[1]))
	{
		for(int i=1;i<=n;i++) cout<<a[i]<<' ';
		cout<<endl;
		return;
	}
	for(int i=2;i<=n;i++)
	{
		if(vis[i]==0&&isprime(i+a[x-1]))
		{
			vis[i]=1;
			a[x]=i;
			dfs(x+1);
			vis[i]=0;
		}
	}
}
signed main()
{
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	int cnt=0;
	while(cin>>n)
	{
		cout<<"Case "<<++cnt<<":"<<endl;
		a[1]=1;
		dfs(2);
		cout<<endl;
		memset(vis,0,sizeof vis);
	}
	return 0;
}