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
首先看到这个问题，我想了一下暴力枚举出所有符合条件的，但是过不了，超时了，然后我就看了一下别人的，用的是DFS。
因为不熟，我借鉴了一下，换成了我的风格。
#include <bits/stdc++.h>
using namespace std;
int n,a[20],b[40],c[20];
void fn()
{
memset(b,0,sizeof(b));
for(int i=2; i<40; i++)
if(!b[i])
{
for(int j=i*2; j<40; j+=i)
b[j]=1;
}
}
void dfs(int num)
{
if(num==n&&b[a[num-1]+a[0]]==0)
{
for(int i=0; i<n; i++)
printf("%d%c",a[i],i==n-1?'\n':' ');
}
else
{
for(int i=2; i<=n; i++)
{
if(c[i]==0)
{
if(!b[a[num-1]+i])
{
a[num++]=i;
c[i]=1;
dfs(num);
c[i]=0;
num--;
}
}
}
}
}
int main()
{
int t=1;
fn();
while(cin>>n)
{
memset(c,0,sizeof(c));
printf("Case %d:\n",t++);
a[0]=1;
dfs(1);
printf("\n");
}
return 0;
}