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.

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.
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
题意:给一个数n,用1到n中的所有数围成一个环,并且每相邻的两个数相加为素数,问这样的环有多少个,并输出每个环。每个从1开始输出。
这题用dfs搜索,每次从1开始搜。一要如何搜,那就看代码吧。
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,v[25],a[25];
int ssb[100],zj[100],pn;
void f()
{
int i,j;
zj[0]=1;
zj[1]=1;
pn=0;
for (i=2;i<=100;i++)
{
if (zj[i]==0) ssb[pn++]=i;
for (j=0;j<pn;j++)
{
if (i*ssb[j]>100) break;
zj[i*ssb[j]]=1;
if (i%ssb[j]==0) break;
}
}
}
void dfs(int x,int k)
{
int i,j;
for (i=2;i<=n;i++)
{
if (!v[i]&&zj[i+x]==0)
{
v[i]=1;
a[k+1]=i;
if (k+1==n)
{
if (zj[1+a[n]]==0){
for (j=1;j<n;j++) printf("%d ",a[j]);
printf("%d\n",a[n]);}
}
dfs(i,k+1);
v[i]=0;
}
}
}
int main()
{
int cut=1;
a[1]=1;
memset(zj,0,sizeof(zj));
f();
while (~scanf("%d",&n))
{
printf("Case %d:\n",cut);
cut++;
memset(v,0,sizeof(v));
dfs(1,1);
printf("\n");
}
return 0;
}
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