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
给你一个数n 1-n组成一个环 相邻的两个数之和要是素数（所以同时要保证头和尾的和是素数）
DFS 基础题
核心代码就一丢丢 很容易看懂
1 #include <iostream> 2 using namespace std; 3 #include<string.h> 4 #include<set> 5 #include<stdio.h> 6 #include<math.h> 7 #include<queue> 8 #include<map> 9 #include<algorithm> 10 #include<cstdio> 11 #include<cmath> 12 #include<cstring> 13 #include <cstdio> 14 #include <cstdlib> 15 #include<stack> 16 int n; 17 int a[25]; 18 int b[25]; 19 int sushu(int x) 20 { 21 int s=(int)sqrt(x); 22 for(int i=2;i<=s;i++) 23 { 24 if(x%i==0) 25 return 0; 26 } 27 return 1; 28 } 29 void dfs(int x) 30 { 31 if(x==n&&sushu(a[1]+a[n])) 32 { 33 int flag=0; 34 for(int i=1;i<=n;i++) 35 { 36 if(!flag) 37 { 38 flag=1; 39 printf("%d",a[i]); 40 } 41 else 42 { 43 printf(" %d",a[i]); 44 } 45 } 46 printf(" "); 47 return; 48 } 49 for(int i=2;i<=n;i++) 50 { 51 if(b[i]==0&&sushu(a[x]+i)) 52 { 53 //cout<<i<<endl; 54 b[i]=1; 55 a[x+1]=i; 56 dfs(x+1); 57 b[i]=0; 58 } 59 } 60 } 61 int main() 62 { 63 int add=0; 64 while(~scanf("%d",&n)) 65 { 66 memset(a,0,sizeof(a)); 67 memset(b,0,sizeof(b)); 68 a[1]=1; 69 printf("Case %d: ",++add); 70 dfs(1); 71 printf(" "); 72 } 73 return 0; 74 }
讲道理 输入是奇数的时候直接跳过dfs就好 所以我加了个判断，结果时间还变长了一点 ，编程就像女人，永远搞不懂她。
贴上加判断的AC
