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.
Inputn (0 < n < 20).
OutputThe 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
题解:
一般的DFS问题,需要使用回溯法,不然会超时。
代码如下:
#include<iostream> #include<cstring> using namespace std; bool isp[50],vis[20]; int n,a[20]; bool is_prime(int n) { for(int i=2;i*i<=n;i++) if(n%i==0) return false; return n!=1; } void dfs(int cur) { if(cur==n&&isp[a[0]+a[n-1]]) { for(int i=0;i<n-1;i++) printf("%d ",a[i]); printf("%d ",a[n-1]); } else for(int i=2;i<=n;i++)//放置第i个数 { if(!vis[i]&&isp[i+a[cur-1]]) { a[cur]=i; vis[i]=true; dfs(cur+1); vis[i]=false;//回溯后对标记进行复原 } } } int main() { memset(isp,false,sizeof(isp)); for(int i=2;i<2*20;i++) isp[i]=is_prime(i);//生成素数表 a[0]=1; int cas=1; while(cin>>n) { memset(vis,false,sizeof(vis)); printf("Case %d: ",cas++); dfs(1); printf(" "); } return 0; }