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.
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 OutpuCase 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 题目意思:给你一个20以内的正整数n,将1到n围成一个圈,每个数都和旁边数的和为素数输出所以可能的情况 解法:普通的应用DFS的题目,注意跳出条件就可以了;
1 #include <iostream> 2 #include <string.h> 3 using namespace std; 4 5 const int MAX = 100; 6 int n; 7 int visit[MAX]; 8 int Map[MAX]; 9 int a[MAX] = {3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61}; 10 11 void print() 12 { 13 for(int i = 1;i <=n;i++) 14 if(i==1) 15 cout<<Map[i]; 16 else 17 cout<<" "<<Map[i]; 18 cout<<endl; 19 } 20 21 int pand(int x) 22 { 23 for(int i = 0;i<16;i++) 24 if(x==a[i]) 25 return 1; 26 return 0; 27 } 28 29 void dfs(int x,int ti) 30 { 31 if(ti==n) 32 { 33 if(pand(x+1)) 34 { 35 print(); 36 } 37 } 38 39 for(int i = 2;i <= n;i++) 40 { 41 if(visit[i]==0 &&pand(i+x)) 42 { 43 visit[i] = 1; 44 Map[ti+1] = i; 45 dfs(i,ti+1); 46 visit[i] = 0; 47 } 48 } 49 50 } 51 52 int main() 53 { 54 int N=0; 55 while(cin>>n) 56 { 57 cout<<"Case "<<++N<<":"<<endl; 58 if(n==1) 59 { 60 cout<<1<<‘\n‘<<endl; 61 continue; 62 } 63 64 memset(visit,0,sizeof(visit)); 65 visit[1] = 1; 66 Map[1] = 1; 67 dfs(1,1); 68 cout<<endl; 69 70 } 71 72 return 0; 73 }
时间: 2024-10-24 18:04:46