Description
输入正整数n,把整数1,2,…,n组成一个环，使得相邻两个整数之和均为素数。输出时，从整数1开始逆时针排列。同一个环恰好输出一次。n<=16.
A ring is composed of n (even number) circles as shown in diagram.
Put natural numbers1,2……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<=16)
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.
You are to write a program that completes above process.
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
题解
n<=16，数据很小，可以暴搜，途中用stack记录经历的点
这道题的难点是输出格式啊QAQ！！！
example:
Case 1:
1 4 3 2 5 6
1 6 5 2 3 4
Case 2://这里没有空格
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//这里也没有空格
//这里有回车//luogu的原题中输出格式是崩坏的啊！！！QAQ
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int n;
bool sf[21];
int stack[21];
int flag;
bool prime(int x)
{
for(int i=2;i<=sqrt(x);++i)
if(x%i==0)return 0;
return 1;
}
int tot=0;
void print()
{
if(!prime(1+stack[n]))return;
if(!flag){
if(tot)cout<<endl;
flag++;
printf("Case %d:",++tot);
}
cout<<endl;
cout<<1;
for(int i=2;i<=n;++i)
cout<<" "<<stack[i];
}
void search(int x)
{
for(int i=2;i<=n;++i)
if(!sf[i]&&prime(stack[x-1]+i))
{
stack[x]=i;
sf[i]=1;
if(x==n)print();
else search(x+1);
sf[i]=0;
}
return;
}
int main()
{
while(cin>>n)
{
memset(sf,0,sizeof(sf));
flag=0;
stack[1]=1;
sf[1]=1;
search(2);
cout<<endl;
}
}