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 ",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: ",cut); cut++; memset(v,0,sizeof(v)); dfs(1,1); printf(" "); } return 0; }