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.<br><br>Note: the number of first circle should always be 1.<br><br><img src=../../data/images/1016-1.gif><br>
Input
n (0 < n < 20).<br>
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.<br><br>You are to write a program that completes above process.<br><br>Print a blank line after each case.<br>
Sample Input
6<br>8<br>
Sample Output
Case 1:<br>1 4 3 2 5 6<br>1 6 5 2 3 4<br><br>Case 2:<br>1 2 3 8 5 6 7 4<br>1 2 5 8 3 4 7 6<br>1 4 7 6 5 8 3 2<br>1 6 7 4 3 8 5 2<br>
Source
Asia 1996, Shanghai (Mainland China)
題意：素數環
思路：DFS，比較好想
感想：代碼短的時候就好開心
AC代碼：
#include<iostream>
#include<cstdio>
#include<stdio.h>
#include<cstring>
#include<cmath>
using namespace std;
int prime[40]={0,0,1,1,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0};
int n,cou;
int num[25];
int vis [25];
int cc[25];
void dfs(int now)
{
int i;
if(now==n&&prime[1+cc[n]])
{
for(i=1;i<=n;i++)
{
if(i==1)
cout<<cc[i];
else
cout<<" "<<cc[i];
}
cout<<endl;
}
else
{
for(i=2;i<=n;i++)
{
if(prime[i+cc[now]]&&!vis[i])
{
vis[i]=1;
now++;
cc[now]=i;
//cout<<cc[now]<<" "<<cc[now-1]<<endl;
dfs(now);
now--;
vis[i]=0;
}
}
}
}
int main()
{
//freopen("r.txt","r",stdin);
int i,j;
cou=1;
while(cin>>n)
{
cout<<"Case "<<cou<<":"<<endl;
num[1]=cc[1]=1;
for(i=2;i<=n;i++)
num[i]=i;
memset(vis,0,sizeof(vis));
vis[1]=1;
dfs(1);
cou++;
if(cou!=1)
cout<<endl;
}
}
